分类目录归档:知识问答

**(双星号/星号)和*(星号/星号)对参数有什么作用?

问题:**(双星号/星号)和*(星号/星号)对参数有什么作用?

在以下方法定义中,***param2什么?

def foo(param1, *param2):
def bar(param1, **param2):

In the following method definitions, what does the * and ** do for param2?

def foo(param1, *param2):
def bar(param1, **param2):

回答 0

*args**kwargs是一种常见的成语,以允许参数,以作为部分所述功能的任意数量的多个上定义函数 Python文档英寸

*args给你的所有函数参数为一个元组

def foo(*args):
    for a in args:
        print(a)        

foo(1)
# 1

foo(1,2,3)
# 1
# 2
# 3

**kwargs会给你所有的 关键字参数除了那些与作为字典的形式参数。

def bar(**kwargs):
    for a in kwargs:
        print(a, kwargs[a])  

bar(name='one', age=27)
# age 27
# name one

这两个习惯用法都可以与普通参数混合使用,以允许使用一组固定参数和一些可变参数:

def foo(kind, *args, **kwargs):
   pass

也可以以其他方式使用此方法:

def foo(a, b, c):
    print(a, b, c)

obj = {'b':10, 'c':'lee'}

foo(100,**obj)
# 100 10 lee

*l习惯用法的另一种用法是在调用函数时解压缩参数列表

def foo(bar, lee):
    print(bar, lee)

l = [1,2]

foo(*l)
# 1 2

在Python 3中,可以*l在分配的左侧使用(Extended Iterable Unpacking),尽管在这种情况下它提供的是列表而不是元组:

first, *rest = [1,2,3,4]
first, *l, last = [1,2,3,4]

Python 3还添加了新的语义(请参阅PEP 3102):

def func(arg1, arg2, arg3, *, kwarg1, kwarg2):
    pass

该函数仅接受3个位置参数,之后的所有内容*只能作为关键字参数传递。

The *args and **kwargs is a common idiom to allow arbitrary number of arguments to functions as described in the section more on defining functions in the Python documentation.

The *args will give you all function parameters as a tuple:

def foo(*args):
    for a in args:
        print(a)        

foo(1)
# 1

foo(1,2,3)
# 1
# 2
# 3

The **kwargs will give you all keyword arguments except for those corresponding to a formal parameter as a dictionary.

def bar(**kwargs):
    for a in kwargs:
        print(a, kwargs[a])  

bar(name='one', age=27)
# age 27
# name one

Both idioms can be mixed with normal arguments to allow a set of fixed and some variable arguments:

def foo(kind, *args, **kwargs):
   pass

It is also possible to use this the other way around:

def foo(a, b, c):
    print(a, b, c)

obj = {'b':10, 'c':'lee'}

foo(100,**obj)
# 100 10 lee

Another usage of the *l idiom is to unpack argument lists when calling a function.

def foo(bar, lee):
    print(bar, lee)

l = [1,2]

foo(*l)
# 1 2

In Python 3 it is possible to use *l on the left side of an assignment (Extended Iterable Unpacking), though it gives a list instead of a tuple in this context:

first, *rest = [1,2,3,4]
first, *l, last = [1,2,3,4]

Also Python 3 adds new semantic (refer PEP 3102):

def func(arg1, arg2, arg3, *, kwarg1, kwarg2):
    pass

Such function accepts only 3 positional arguments, and everything after * can only be passed as keyword arguments.


回答 1

另外值得一提的是,你可以使用***调用功能,以及时。这是一个快捷方式,允许您使用列表/元组或字典将多个参数直接传递给函数。例如,如果您具有以下功能:

def foo(x,y,z):
    print("x=" + str(x))
    print("y=" + str(y))
    print("z=" + str(z))

您可以执行以下操作:

>>> mylist = [1,2,3]
>>> foo(*mylist)
x=1
y=2
z=3

>>> mydict = {'x':1,'y':2,'z':3}
>>> foo(**mydict)
x=1
y=2
z=3

>>> mytuple = (1, 2, 3)
>>> foo(*mytuple)
x=1
y=2
z=3

注意:中的键mydict必须完全像function参数一样命名foo。否则会抛出TypeError

>>> mydict = {'x':1,'y':2,'z':3,'badnews':9}
>>> foo(**mydict)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() got an unexpected keyword argument 'badnews'

It’s also worth noting that you can use * and ** when calling functions as well. This is a shortcut that allows you to pass multiple arguments to a function directly using either a list/tuple or a dictionary. For example, if you have the following function:

def foo(x,y,z):
    print("x=" + str(x))
    print("y=" + str(y))
    print("z=" + str(z))

You can do things like:

>>> mylist = [1,2,3]
>>> foo(*mylist)
x=1
y=2
z=3

>>> mydict = {'x':1,'y':2,'z':3}
>>> foo(**mydict)
x=1
y=2
z=3

>>> mytuple = (1, 2, 3)
>>> foo(*mytuple)
x=1
y=2
z=3

Note: The keys in mydict have to be named exactly like the parameters of function foo. Otherwise it will throw a TypeError:

>>> mydict = {'x':1,'y':2,'z':3,'badnews':9}
>>> foo(**mydict)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() got an unexpected keyword argument 'badnews'

回答 2

单个*表示可以有任意数量的额外位置参数。foo()可以像这样调用foo(1,2,3,4,5)。在foo()主体中,param2是一个包含2-5的序列。

双**表示可以有任意数量的额外命名参数。bar()可以像这样调用bar(1, a=2, b=3)。在bar()的主体中,param2是一个包含{‘a’:2,’b’:3}的字典。

使用以下代码:

def foo(param1, *param2):
    print(param1)
    print(param2)

def bar(param1, **param2):
    print(param1)
    print(param2)

foo(1,2,3,4,5)
bar(1,a=2,b=3)

输出是

1
(2, 3, 4, 5)
1
{'a': 2, 'b': 3}

The single * means that there can be any number of extra positional arguments. foo() can be invoked like foo(1,2,3,4,5). In the body of foo() param2 is a sequence containing 2-5.

The double ** means there can be any number of extra named parameters. bar() can be invoked like bar(1, a=2, b=3). In the body of bar() param2 is a dictionary containing {‘a’:2, ‘b’:3 }

With the following code:

def foo(param1, *param2):
    print(param1)
    print(param2)

def bar(param1, **param2):
    print(param1)
    print(param2)

foo(1,2,3,4,5)
bar(1,a=2,b=3)

the output is

1
(2, 3, 4, 5)
1
{'a': 2, 'b': 3}

回答 3

这是什么**(双星)和*(明星)的参数做

它们允许定义函数以接受并允许用户传递任意数量的参数,位置(*)和关键字(**)。

定义功能

*args允许任意数量的可选位置参数(参数),这些参数将分配给名为的元组args

**kwargs允许任意数量的可选关键字参数(参数),这些参数将位于名为的字典中kwargs

您可以(并且应该)选择任何适当的名称,但是如果目的是使参数具有非特定的语义,args并且kwargs是标准名称。

扩展,传递任意数量的参数

您还可以分别使用*args**kwargs传入列表(或任何可迭代的)和字典(或任何映射)的参数。

接收参数的函数不必知道它们正在扩展。

例如,Python 2的xrange并不明确期望*args,但是因为它使用3个整数作为参数:

>>> x = xrange(3) # create our *args - an iterable of 3 integers
>>> xrange(*x)    # expand here
xrange(0, 2, 2)

再举一个例子,我们可以在下面使用dict扩展str.format

>>> foo = 'FOO'
>>> bar = 'BAR'
>>> 'this is foo, {foo} and bar, {bar}'.format(**locals())
'this is foo, FOO and bar, BAR'

Python 3的新功能:使用仅关键字参数定义函数

您可以在- 之后添加仅关键字参数*args -例如,在此处,kwarg2必须将其作为关键字参数-而不是位置:

def foo(arg, kwarg=None, *args, kwarg2=None, **kwargs): 
    return arg, kwarg, args, kwarg2, kwargs

用法:

>>> foo(1,2,3,4,5,kwarg2='kwarg2', bar='bar', baz='baz')
(1, 2, (3, 4, 5), 'kwarg2', {'bar': 'bar', 'baz': 'baz'})

同样,*可以单独用于表示仅关键字参数跟随,而不允许无限的位置参数。

def foo(arg, kwarg=None, *, kwarg2=None, **kwargs): 
    return arg, kwarg, kwarg2, kwargs

在这里,kwarg2再次必须是一个明确命名的关键字参数:

>>> foo(1,2,kwarg2='kwarg2', foo='foo', bar='bar')
(1, 2, 'kwarg2', {'foo': 'foo', 'bar': 'bar'})

而且我们不再可以接受无限的位置参数,因为我们没有*args*

>>> foo(1,2,3,4,5, kwarg2='kwarg2', foo='foo', bar='bar')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() takes from 1 to 2 positional arguments 
    but 5 positional arguments (and 1 keyword-only argument) were given

再次,更简单地说,在这里我们需要kwarg使用名称,而不是位置:

def bar(*, kwarg=None): 
    return kwarg

在此示例中,我们看到如果尝试通过kwarg位置传递,则会收到错误消息:

>>> bar('kwarg')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: bar() takes 0 positional arguments but 1 was given

我们必须显式传递kwarg参数作为关键字参数。

>>> bar(kwarg='kwarg')
'kwarg'

兼容Python 2的演示

*args(通常说“ star-args”)和**kwargs(可以通过说“ kwargs”来暗示星号,但是用“ double-star kwargs”来明确表示)是使用***表示法的Python的常见用法。这些特定的变量名称不是必需的(例如,您可以使用*foos**bars),但是背离约定可能会激怒您的Python编码人员。

当我们不知道函数将要接收什么或我们可能传递多少个参数时,我们通常会使用它们,有时甚至即使分别命名每个变量也会变得非常混乱和多余(但这是通常显式的情况比隐式更好)。

例子1

以下功能描述了如何使用它们,并演示了行为。请注意,命名b参数将由前面的第二个位置参数使用:

def foo(a, b=10, *args, **kwargs):
    '''
    this function takes required argument a, not required keyword argument b
    and any number of unknown positional arguments and keyword arguments after
    '''
    print('a is a required argument, and its value is {0}'.format(a))
    print('b not required, its default value is 10, actual value: {0}'.format(b))
    # we can inspect the unknown arguments we were passed:
    #  - args:
    print('args is of type {0} and length {1}'.format(type(args), len(args)))
    for arg in args:
        print('unknown arg: {0}'.format(arg))
    #  - kwargs:
    print('kwargs is of type {0} and length {1}'.format(type(kwargs),
                                                        len(kwargs)))
    for kw, arg in kwargs.items():
        print('unknown kwarg - kw: {0}, arg: {1}'.format(kw, arg))
    # But we don't have to know anything about them 
    # to pass them to other functions.
    print('Args or kwargs can be passed without knowing what they are.')
    # max can take two or more positional args: max(a, b, c...)
    print('e.g. max(a, b, *args) \n{0}'.format(
      max(a, b, *args))) 
    kweg = 'dict({0})'.format( # named args same as unknown kwargs
      ', '.join('{k}={v}'.format(k=k, v=v) 
                             for k, v in sorted(kwargs.items())))
    print('e.g. dict(**kwargs) (same as {kweg}) returns: \n{0}'.format(
      dict(**kwargs), kweg=kweg))

我们可以检查函数的签名的在线帮助,以help(foo),它告诉我们

foo(a, b=10, *args, **kwargs)

让我们用 foo(1, 2, 3, 4, e=5, f=6, g=7)

打印:

a is a required argument, and its value is 1
b not required, its default value is 10, actual value: 2
args is of type <type 'tuple'> and length 2
unknown arg: 3
unknown arg: 4
kwargs is of type <type 'dict'> and length 3
unknown kwarg - kw: e, arg: 5
unknown kwarg - kw: g, arg: 7
unknown kwarg - kw: f, arg: 6
Args or kwargs can be passed without knowing what they are.
e.g. max(a, b, *args) 
4
e.g. dict(**kwargs) (same as dict(e=5, f=6, g=7)) returns: 
{'e': 5, 'g': 7, 'f': 6}

例子2

我们还可以使用另一个函数来调用它a

def bar(a):
    b, c, d, e, f = 2, 3, 4, 5, 6
    # dumping every local variable into foo as a keyword argument 
    # by expanding the locals dict:
    foo(**locals()) 

bar(100) 印刷品:

a is a required argument, and its value is 100
b not required, its default value is 10, actual value: 2
args is of type <type 'tuple'> and length 0
kwargs is of type <type 'dict'> and length 4
unknown kwarg - kw: c, arg: 3
unknown kwarg - kw: e, arg: 5
unknown kwarg - kw: d, arg: 4
unknown kwarg - kw: f, arg: 6
Args or kwargs can be passed without knowing what they are.
e.g. max(a, b, *args) 
100
e.g. dict(**kwargs) (same as dict(c=3, d=4, e=5, f=6)) returns: 
{'c': 3, 'e': 5, 'd': 4, 'f': 6}

示例3:装饰器中的实际用法

好的,所以也许我们还没有看到该实用程序。因此,假设您在区分代码之前和/或之后有多个带有冗余代码的功能。为了说明的目的,以下命名函数只是伪代码。

def foo(a, b, c, d=0, e=100):
    # imagine this is much more code than a simple function call
    preprocess() 
    differentiating_process_foo(a,b,c,d,e)
    # imagine this is much more code than a simple function call
    postprocess()

def bar(a, b, c=None, d=0, e=100, f=None):
    preprocess()
    differentiating_process_bar(a,b,c,d,e,f)
    postprocess()

def baz(a, b, c, d, e, f):
    ... and so on

我们也许可以用不同的方式处理此问题,但是我们当然可以用装饰器提取冗余,因此下面的示例演示了如何*args并且**kwargs非常有用:

def decorator(function):
    '''function to wrap other functions with a pre- and postprocess'''
    @functools.wraps(function) # applies module, name, and docstring to wrapper
    def wrapper(*args, **kwargs):
        # again, imagine this is complicated, but we only write it once!
        preprocess()
        function(*args, **kwargs)
        postprocess()
    return wrapper

现在,由于我们考虑了冗余性,每个包装函数都可以更加简洁地编写:

@decorator
def foo(a, b, c, d=0, e=100):
    differentiating_process_foo(a,b,c,d,e)

@decorator
def bar(a, b, c=None, d=0, e=100, f=None):
    differentiating_process_bar(a,b,c,d,e,f)

@decorator
def baz(a, b, c=None, d=0, e=100, f=None, g=None):
    differentiating_process_baz(a,b,c,d,e,f, g)

@decorator
def quux(a, b, c=None, d=0, e=100, f=None, g=None, h=None):
    differentiating_process_quux(a,b,c,d,e,f,g,h)

通过分解*args**kwargs允许我们这样做的代码,我们减少了代码行,提高了可读性和可维护性,并且在程序中具有唯一的规范逻辑位置。如果需要更改此结构的任何部分,则可以在一个位置进行每次更改。

What does ** (double star) and * (star) do for parameters

They allow for functions to be defined to accept and for users to pass any number of arguments, positional (*) and keyword (**).

Defining Functions

*args allows for any number of optional positional arguments (parameters), which will be assigned to a tuple named args.

**kwargs allows for any number of optional keyword arguments (parameters), which will be in a dict named kwargs.

You can (and should) choose any appropriate name, but if the intention is for the arguments to be of non-specific semantics, args and kwargs are standard names.

Expansion, Passing any number of arguments

You can also use *args and **kwargs to pass in parameters from lists (or any iterable) and dicts (or any mapping), respectively.

The function recieving the parameters does not have to know that they are being expanded.

For example, Python 2’s xrange does not explicitly expect *args, but since it takes 3 integers as arguments:

>>> x = xrange(3) # create our *args - an iterable of 3 integers
>>> xrange(*x)    # expand here
xrange(0, 2, 2)

As another example, we can use dict expansion in str.format:

>>> foo = 'FOO'
>>> bar = 'BAR'
>>> 'this is foo, {foo} and bar, {bar}'.format(**locals())
'this is foo, FOO and bar, BAR'

New in Python 3: Defining functions with keyword only arguments

You can have keyword only arguments after the *args – for example, here, kwarg2 must be given as a keyword argument – not positionally:

def foo(arg, kwarg=None, *args, kwarg2=None, **kwargs): 
    return arg, kwarg, args, kwarg2, kwargs

Usage:

>>> foo(1,2,3,4,5,kwarg2='kwarg2', bar='bar', baz='baz')
(1, 2, (3, 4, 5), 'kwarg2', {'bar': 'bar', 'baz': 'baz'})

Also, * can be used by itself to indicate that keyword only arguments follow, without allowing for unlimited positional arguments.

def foo(arg, kwarg=None, *, kwarg2=None, **kwargs): 
    return arg, kwarg, kwarg2, kwargs

Here, kwarg2 again must be an explicitly named, keyword argument:

>>> foo(1,2,kwarg2='kwarg2', foo='foo', bar='bar')
(1, 2, 'kwarg2', {'foo': 'foo', 'bar': 'bar'})

And we can no longer accept unlimited positional arguments because we don’t have *args*:

>>> foo(1,2,3,4,5, kwarg2='kwarg2', foo='foo', bar='bar')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() takes from 1 to 2 positional arguments 
    but 5 positional arguments (and 1 keyword-only argument) were given

Again, more simply, here we require kwarg to be given by name, not positionally:

def bar(*, kwarg=None): 
    return kwarg

In this example, we see that if we try to pass kwarg positionally, we get an error:

>>> bar('kwarg')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: bar() takes 0 positional arguments but 1 was given

We must explicitly pass the kwarg parameter as a keyword argument.

>>> bar(kwarg='kwarg')
'kwarg'

Python 2 compatible demos

*args (typically said “star-args”) and **kwargs (stars can be implied by saying “kwargs”, but be explicit with “double-star kwargs”) are common idioms of Python for using the * and ** notation. These specific variable names aren’t required (e.g. you could use *foos and **bars), but a departure from convention is likely to enrage your fellow Python coders.

We typically use these when we don’t know what our function is going to receive or how many arguments we may be passing, and sometimes even when naming every variable separately would get very messy and redundant (but this is a case where usually explicit is better than implicit).

Example 1

The following function describes how they can be used, and demonstrates behavior. Note the named b argument will be consumed by the second positional argument before :

def foo(a, b=10, *args, **kwargs):
    '''
    this function takes required argument a, not required keyword argument b
    and any number of unknown positional arguments and keyword arguments after
    '''
    print('a is a required argument, and its value is {0}'.format(a))
    print('b not required, its default value is 10, actual value: {0}'.format(b))
    # we can inspect the unknown arguments we were passed:
    #  - args:
    print('args is of type {0} and length {1}'.format(type(args), len(args)))
    for arg in args:
        print('unknown arg: {0}'.format(arg))
    #  - kwargs:
    print('kwargs is of type {0} and length {1}'.format(type(kwargs),
                                                        len(kwargs)))
    for kw, arg in kwargs.items():
        print('unknown kwarg - kw: {0}, arg: {1}'.format(kw, arg))
    # But we don't have to know anything about them 
    # to pass them to other functions.
    print('Args or kwargs can be passed without knowing what they are.')
    # max can take two or more positional args: max(a, b, c...)
    print('e.g. max(a, b, *args) \n{0}'.format(
      max(a, b, *args))) 
    kweg = 'dict({0})'.format( # named args same as unknown kwargs
      ', '.join('{k}={v}'.format(k=k, v=v) 
                             for k, v in sorted(kwargs.items())))
    print('e.g. dict(**kwargs) (same as {kweg}) returns: \n{0}'.format(
      dict(**kwargs), kweg=kweg))

We can check the online help for the function’s signature, with help(foo), which tells us

foo(a, b=10, *args, **kwargs)

Let’s call this function with foo(1, 2, 3, 4, e=5, f=6, g=7)

which prints:

a is a required argument, and its value is 1
b not required, its default value is 10, actual value: 2
args is of type <type 'tuple'> and length 2
unknown arg: 3
unknown arg: 4
kwargs is of type <type 'dict'> and length 3
unknown kwarg - kw: e, arg: 5
unknown kwarg - kw: g, arg: 7
unknown kwarg - kw: f, arg: 6
Args or kwargs can be passed without knowing what they are.
e.g. max(a, b, *args) 
4
e.g. dict(**kwargs) (same as dict(e=5, f=6, g=7)) returns: 
{'e': 5, 'g': 7, 'f': 6}

Example 2

We can also call it using another function, into which we just provide a:

def bar(a):
    b, c, d, e, f = 2, 3, 4, 5, 6
    # dumping every local variable into foo as a keyword argument 
    # by expanding the locals dict:
    foo(**locals()) 

bar(100) prints:

a is a required argument, and its value is 100
b not required, its default value is 10, actual value: 2
args is of type <type 'tuple'> and length 0
kwargs is of type <type 'dict'> and length 4
unknown kwarg - kw: c, arg: 3
unknown kwarg - kw: e, arg: 5
unknown kwarg - kw: d, arg: 4
unknown kwarg - kw: f, arg: 6
Args or kwargs can be passed without knowing what they are.
e.g. max(a, b, *args) 
100
e.g. dict(**kwargs) (same as dict(c=3, d=4, e=5, f=6)) returns: 
{'c': 3, 'e': 5, 'd': 4, 'f': 6}

Example 3: practical usage in decorators

OK, so maybe we’re not seeing the utility yet. So imagine you have several functions with redundant code before and/or after the differentiating code. The following named functions are just pseudo-code for illustrative purposes.

def foo(a, b, c, d=0, e=100):
    # imagine this is much more code than a simple function call
    preprocess() 
    differentiating_process_foo(a,b,c,d,e)
    # imagine this is much more code than a simple function call
    postprocess()

def bar(a, b, c=None, d=0, e=100, f=None):
    preprocess()
    differentiating_process_bar(a,b,c,d,e,f)
    postprocess()

def baz(a, b, c, d, e, f):
    ... and so on

We might be able to handle this differently, but we can certainly extract the redundancy with a decorator, and so our below example demonstrates how *args and **kwargs can be very useful:

def decorator(function):
    '''function to wrap other functions with a pre- and postprocess'''
    @functools.wraps(function) # applies module, name, and docstring to wrapper
    def wrapper(*args, **kwargs):
        # again, imagine this is complicated, but we only write it once!
        preprocess()
        function(*args, **kwargs)
        postprocess()
    return wrapper

And now every wrapped function can be written much more succinctly, as we’ve factored out the redundancy:

@decorator
def foo(a, b, c, d=0, e=100):
    differentiating_process_foo(a,b,c,d,e)

@decorator
def bar(a, b, c=None, d=0, e=100, f=None):
    differentiating_process_bar(a,b,c,d,e,f)

@decorator
def baz(a, b, c=None, d=0, e=100, f=None, g=None):
    differentiating_process_baz(a,b,c,d,e,f, g)

@decorator
def quux(a, b, c=None, d=0, e=100, f=None, g=None, h=None):
    differentiating_process_quux(a,b,c,d,e,f,g,h)

And by factoring out our code, which *args and **kwargs allows us to do, we reduce lines of code, improve readability and maintainability, and have sole canonical locations for the logic in our program. If we need to change any part of this structure, we have one place in which to make each change.


回答 4

首先让我们了解什么是位置参数和关键字参数。下面是带有位置参数的函数定义的示例

def test(a,b,c):
     print(a)
     print(b)
     print(c)

test(1,2,3)
#output:
1
2
3

因此,这是带有位置参数的函数定义。您也可以使用关键字/命名参数来调用它:

def test(a,b,c):
     print(a)
     print(b)
     print(c)

test(a=1,b=2,c=3)
#output:
1
2
3

现在让我们研究一个带有关键字参数的函数定义示例:

def test(a=0,b=0,c=0):
     print(a)
     print(b)
     print(c)
     print('-------------------------')

test(a=1,b=2,c=3)
#output :
1
2
3
-------------------------

您也可以使用位置参数调用此函数:

def test(a=0,b=0,c=0):
    print(a)
    print(b)
    print(c)
    print('-------------------------')

test(1,2,3)
# output :
1
2
3
---------------------------------

因此,我们现在知道带有位置参数以及关键字参数的函数定义。

现在让我们研究“ *”运算符和“ **”运算符。

请注意,这些运算符可以在两个区域中使用:

a)函数调用

b)功能定义

函数调用中使用“ *”运算符和“ **”运算符

让我们直接看一个例子,然后讨论它。

def sum(a,b):  #receive args from function calls as sum(1,2) or sum(a=1,b=2)
    print(a+b)

my_tuple = (1,2)
my_list = [1,2]
my_dict = {'a':1,'b':2}

# Let us unpack data structure of list or tuple or dict into arguments with help of '*' operator
sum(*my_tuple)   # becomes same as sum(1,2) after unpacking my_tuple with '*'
sum(*my_list)    # becomes same as sum(1,2) after unpacking my_list with  '*'
sum(**my_dict)   # becomes same as sum(a=1,b=2) after unpacking by '**' 

# output is 3 in all three calls to sum function.

所以记住

函数调用中使用“ *”或“ **”运算符时-

‘*’运算符将列表或元组等数据结构解压缩为函数定义所需的参数。

‘**’运算符将字典分解成函数定义所需的参数。

现在让我们研究函数定义中使用’*’运算符。例:

def sum(*args): #pack the received positional args into data structure of tuple. after applying '*' - def sum((1,2,3,4))
    sum = 0
    for a in args:
        sum+=a
    print(sum)

sum(1,2,3,4)  #positional args sent to function sum
#output:
10

在函数定义中,“ *”运算符将接收到的参数打包到一个元组中。

现在让我们看一下函数定义中使用的“ **”示例:

def sum(**args): #pack keyword args into datastructure of dict after applying '**' - def sum({a:1,b:2,c:3,d:4})
    sum=0
    for k,v in args.items():
        sum+=v
    print(sum)

sum(a=1,b=2,c=3,d=4) #positional args sent to function sum

在函数定义中,“ **”运算符将接收到的参数打包到字典中。

因此请记住:

函数调用中,“ *” 将元组或列表的数据结构解压缩为位置或关键字参数,以供函数定义接收。

函数调用中,“ **” 将字典的数据结构解压缩为位置或关键字参数,以供函数定义接收。

函数定义中,“ *” 位置参数打包到元组中。

函数定义中,“ **” 关键字参数打包到字典中。

Let us first understand what are positional arguments and keyword arguments. Below is an example of function definition with Positional arguments.

def test(a,b,c):
     print(a)
     print(b)
     print(c)

test(1,2,3)
#output:
1
2
3

So this is a function definition with positional arguments. You can call it with keyword/named arguments as well:

def test(a,b,c):
     print(a)
     print(b)
     print(c)

test(a=1,b=2,c=3)
#output:
1
2
3

Now let us study an example of function definition with keyword arguments:

def test(a=0,b=0,c=0):
     print(a)
     print(b)
     print(c)
     print('-------------------------')

test(a=1,b=2,c=3)
#output :
1
2
3
-------------------------

You can call this function with positional arguments as well:

def test(a=0,b=0,c=0):
    print(a)
    print(b)
    print(c)
    print('-------------------------')

test(1,2,3)
# output :
1
2
3
---------------------------------

So we now know function definitions with positional as well as keyword arguments.

Now let us study the ‘*’ operator and ‘**’ operator.

Please note these operators can be used in 2 areas:

a) function call

b) function definition

The use of ‘*’ operator and ‘**’ operator in function call.

Let us get straight to an example and then discuss it.

def sum(a,b):  #receive args from function calls as sum(1,2) or sum(a=1,b=2)
    print(a+b)

my_tuple = (1,2)
my_list = [1,2]
my_dict = {'a':1,'b':2}

# Let us unpack data structure of list or tuple or dict into arguments with help of '*' operator
sum(*my_tuple)   # becomes same as sum(1,2) after unpacking my_tuple with '*'
sum(*my_list)    # becomes same as sum(1,2) after unpacking my_list with  '*'
sum(**my_dict)   # becomes same as sum(a=1,b=2) after unpacking by '**' 

# output is 3 in all three calls to sum function.

So remember

when the ‘*’ or ‘**’ operator is used in a function call

‘*’ operator unpacks data structure such as a list or tuple into arguments needed by function definition.

‘**’ operator unpacks a dictionary into arguments needed by function definition.

Now let us study the ‘*’ operator use in function definition. Example:

def sum(*args): #pack the received positional args into data structure of tuple. after applying '*' - def sum((1,2,3,4))
    sum = 0
    for a in args:
        sum+=a
    print(sum)

sum(1,2,3,4)  #positional args sent to function sum
#output:
10

In function definition the ‘*’ operator packs the received arguments into a tuple.

Now let us see an example of ‘**’ used in function definition:

def sum(**args): #pack keyword args into datastructure of dict after applying '**' - def sum({a:1,b:2,c:3,d:4})
    sum=0
    for k,v in args.items():
        sum+=v
    print(sum)

sum(a=1,b=2,c=3,d=4) #positional args sent to function sum

In function definition The ‘**’ operator packs the received arguments into a dictionary.

So remember:

In a function call the ‘*’ unpacks data structure of tuple or list into positional or keyword arguments to be received by function definition.

In a function call the ‘**’ unpacks data structure of dictionary into positional or keyword arguments to be received by function definition.

In a function definition the ‘*’ packs positional arguments into a tuple.

In a function definition the ‘**’ packs keyword arguments into a dictionary.


回答 5

该表非常适合在函数构造和函数调用中使用*和使用:**

            In function construction         In function call
=======================================================================
          |  def f(*args):                 |  def f(a, b):
*args     |      for arg in args:          |      return a + b
          |          print(arg)            |  args = (1, 2)
          |  f(1, 2)                       |  f(*args)
----------|--------------------------------|---------------------------
          |  def f(a, b):                  |  def f(a, b):
**kwargs  |      return a + b              |      return a + b
          |  def g(**kwargs):              |  kwargs = dict(a=1, b=2)
          |      return f(**kwargs)        |  f(**kwargs)
          |  g(a=1, b=2)                   |
-----------------------------------------------------------------------

这实际上只是用来总结Lorin Hochstein的答案,但我发现它很有帮助。

相关:在Python 3中已扩展了star / splat运算符的用法

This table is handy for using * and ** in function construction and function call:

            In function construction         In function call
=======================================================================
          |  def f(*args):                 |  def f(a, b):
*args     |      for arg in args:          |      return a + b
          |          print(arg)            |  args = (1, 2)
          |  f(1, 2)                       |  f(*args)
----------|--------------------------------|---------------------------
          |  def f(a, b):                  |  def f(a, b):
**kwargs  |      return a + b              |      return a + b
          |  def g(**kwargs):              |  kwargs = dict(a=1, b=2)
          |      return f(**kwargs)        |  f(**kwargs)
          |  g(a=1, b=2)                   |
-----------------------------------------------------------------------

This really just serves to summarize Lorin Hochstein’s answer but I find it helpful.

Relatedly: uses for the star/splat operators have been expanded in Python 3


回答 6

***在函数参数列表中有特殊用法。* 表示该参数是一个列表,并且**表示该参数是一个字典。这允许函数接受任意数量的参数

* and ** have special usage in the function argument list. * implies that the argument is a list and ** implies that the argument is a dictionary. This allows functions to take arbitrary number of arguments


回答 7

对于那些通过榜样学习的人!

  1. 的目的* 是使您能够定义一个函数,该函数可以采用以列表形式提供的任意数量的参数(例如f(*myList))。
  2. 目的**是通过提供字典(例如f(**{'x' : 1, 'y' : 2}))使您能够输入函数的参数。

就让我们一起来通过定义一个函数,它有两个正常的变量xy以及可以接受更多的论据myArgs,并能接受更多的论据myKW。稍后,我们将展示如何y使用进行订阅myArgDict

def f(x, y, *myArgs, **myKW):
    print("# x      = {}".format(x))
    print("# y      = {}".format(y))
    print("# myArgs = {}".format(myArgs))
    print("# myKW   = {}".format(myKW))
    print("# ----------------------------------------------------------------------")

# Define a list for demonstration purposes
myList    = ["Left", "Right", "Up", "Down"]
# Define a dictionary for demonstration purposes
myDict    = {"Wubba": "lubba", "Dub": "dub"}
# Define a dictionary to feed y
myArgDict = {'y': "Why?", 'y0': "Why not?", "q": "Here is a cue!"}

# The 1st elem of myList feeds y
f("myEx", *myList, **myDict)
# x      = myEx
# y      = Left
# myArgs = ('Right', 'Up', 'Down')
# myKW   = {'Wubba': 'lubba', 'Dub': 'dub'}
# ----------------------------------------------------------------------

# y is matched and fed first
# The rest of myArgDict becomes additional arguments feeding myKW
f("myEx", **myArgDict)
# x      = myEx
# y      = Why?
# myArgs = ()
# myKW   = {'y0': 'Why not?', 'q': 'Here is a cue!'}
# ----------------------------------------------------------------------

# The rest of myArgDict becomes additional arguments feeding myArgs
f("myEx", *myArgDict)
# x      = myEx
# y      = y
# myArgs = ('y0', 'q')
# myKW   = {}
# ----------------------------------------------------------------------

# Feed extra arguments manually and append even more from my list
f("myEx", 4, 42, 420, *myList, *myDict, **myDict)
# x      = myEx
# y      = 4
# myArgs = (42, 420, 'Left', 'Right', 'Up', 'Down', 'Wubba', 'Dub')
# myKW   = {'Wubba': 'lubba', 'Dub': 'dub'}
# ----------------------------------------------------------------------

# Without the stars, the entire provided list and dict become x, and y:
f(myList, myDict)
# x      = ['Left', 'Right', 'Up', 'Down']
# y      = {'Wubba': 'lubba', 'Dub': 'dub'}
# myArgs = ()
# myKW   = {}
# ----------------------------------------------------------------------

注意事项

  1. ** 专为字典保留。
  2. 非可选参数分配首先发生。
  3. 您不能两次使用非可选参数。
  4. 如果适用,**必须*始终紧随其后。

For those of you who learn by examples!

  1. The purpose of * is to give you the ability to define a function that can take an arbitrary number of arguments provided as a list (e.g. f(*myList) ).
  2. The purpose of ** is to give you the ability to feed a function’s arguments by providing a dictionary (e.g. f(**{'x' : 1, 'y' : 2}) ).

Let us show this by defining a function that takes two normal variables x, y, and can accept more arguments as myArgs, and can accept even more arguments as myKW. Later, we will show how to feed y using myArgDict.

def f(x, y, *myArgs, **myKW):
    print("# x      = {}".format(x))
    print("# y      = {}".format(y))
    print("# myArgs = {}".format(myArgs))
    print("# myKW   = {}".format(myKW))
    print("# ----------------------------------------------------------------------")

# Define a list for demonstration purposes
myList    = ["Left", "Right", "Up", "Down"]
# Define a dictionary for demonstration purposes
myDict    = {"Wubba": "lubba", "Dub": "dub"}
# Define a dictionary to feed y
myArgDict = {'y': "Why?", 'y0': "Why not?", "q": "Here is a cue!"}

# The 1st elem of myList feeds y
f("myEx", *myList, **myDict)
# x      = myEx
# y      = Left
# myArgs = ('Right', 'Up', 'Down')
# myKW   = {'Wubba': 'lubba', 'Dub': 'dub'}
# ----------------------------------------------------------------------

# y is matched and fed first
# The rest of myArgDict becomes additional arguments feeding myKW
f("myEx", **myArgDict)
# x      = myEx
# y      = Why?
# myArgs = ()
# myKW   = {'y0': 'Why not?', 'q': 'Here is a cue!'}
# ----------------------------------------------------------------------

# The rest of myArgDict becomes additional arguments feeding myArgs
f("myEx", *myArgDict)
# x      = myEx
# y      = y
# myArgs = ('y0', 'q')
# myKW   = {}
# ----------------------------------------------------------------------

# Feed extra arguments manually and append even more from my list
f("myEx", 4, 42, 420, *myList, *myDict, **myDict)
# x      = myEx
# y      = 4
# myArgs = (42, 420, 'Left', 'Right', 'Up', 'Down', 'Wubba', 'Dub')
# myKW   = {'Wubba': 'lubba', 'Dub': 'dub'}
# ----------------------------------------------------------------------

# Without the stars, the entire provided list and dict become x, and y:
f(myList, myDict)
# x      = ['Left', 'Right', 'Up', 'Down']
# y      = {'Wubba': 'lubba', 'Dub': 'dub'}
# myArgs = ()
# myKW   = {}
# ----------------------------------------------------------------------

Caveats

  1. ** is exclusively reserved for dictionaries.
  2. Non-optional argument assignment happens first.
  3. You cannot use a non-optional argument twice.
  4. If applicable, ** must come after *, always.

回答 8

从Python文档中:

如果位置参数多于形式参数槽,则将引发TypeError异常,除非存在使用语法“ * identifier”的形式参数;否则,将引发TypeError异常。在这种情况下,该形式参数会接收包含多余位置参数的元组(如果没有多余位置参数,则为空元组)。

如果任何关键字参数与形式参数名称都不对应,则除非存在使用语法“ ** identifier”的形式参数,否则将引发TypeError异常;否则,将引发TypeError异常。在这种情况下,该形式参数将接收包含多余关键字参数的字典(使用关键字作为键,并将参数值用作对应的值),或者如果没有多余的关键字参数,则接收一个(新的)空字典。

From the Python documentation:

If there are more positional arguments than there are formal parameter slots, a TypeError exception is raised, unless a formal parameter using the syntax “*identifier” is present; in this case, that formal parameter receives a tuple containing the excess positional arguments (or an empty tuple if there were no excess positional arguments).

If any keyword argument does not correspond to a formal parameter name, a TypeError exception is raised, unless a formal parameter using the syntax “**identifier” is present; in this case, that formal parameter receives a dictionary containing the excess keyword arguments (using the keywords as keys and the argument values as corresponding values), or a (new) empty dictionary if there were no excess keyword arguments.


回答 9

* 表示将可变参数作为元组接收

** 表示接收可变参数作为字典

使用方式如下:

1)单*

def foo(*args):
    for arg in args:
        print(arg)

foo("two", 3)

输出:

two
3

2)现在 **

def bar(**kwargs):
    for key in kwargs:
        print(key, kwargs[key])

bar(dic1="two", dic2=3)

输出:

dic1 two
dic2 3

* means receive variable arguments as tuple

** means receive variable arguments as dictionary

Used like the following:

1) single *

def foo(*args):
    for arg in args:
        print(arg)

foo("two", 3)

Output:

two
3

2) Now **

def bar(**kwargs):
    for key in kwargs:
        print(key, kwargs[key])

bar(dic1="two", dic2=3)

Output:

dic1 two
dic2 3

回答 10

我想举一个别人没有提到的例子

*也可以打开生成器包装

Python3文档中的一个示例

x = [1, 2, 3]
y = [4, 5, 6]

unzip_x, unzip_y = zip(*zip(x, y))

unzip_x将为[1、2、3],unzip_y将为[4、5、6]

zip()接收多个可初始化的参数,并返回一个生成器。

zip(*zip(x,y)) -> zip((1, 4), (2, 5), (3, 6))

I want to give an example which others haven’t mentioned

* can also unpack a generator

An example from Python3 Document

x = [1, 2, 3]
y = [4, 5, 6]

unzip_x, unzip_y = zip(*zip(x, y))

unzip_x will be [1, 2, 3], unzip_y will be [4, 5, 6]

The zip() receives multiple iretable args, and return a generator.

zip(*zip(x,y)) -> zip((1, 4), (2, 5), (3, 6))

回答 11

在Python 3.5,你也可以使用这个语法listdicttuple,和set显示器(有时也称为文本)。请参阅PEP 488:其他拆包概述

>>> (0, *range(1, 4), 5, *range(6, 8))
(0, 1, 2, 3, 5, 6, 7)
>>> [0, *range(1, 4), 5, *range(6, 8)]
[0, 1, 2, 3, 5, 6, 7]
>>> {0, *range(1, 4), 5, *range(6, 8)}
{0, 1, 2, 3, 5, 6, 7}
>>> d = {'one': 1, 'two': 2, 'three': 3}
>>> e = {'six': 6, 'seven': 7}
>>> {'zero': 0, **d, 'five': 5, **e}
{'five': 5, 'seven': 7, 'two': 2, 'one': 1, 'three': 3, 'six': 6, 'zero': 0}

它还允许在单个函数调用中解压缩多个可迭代对象。

>>> range(*[1, 10], *[2])
range(1, 10, 2)

(感谢mgilson的PEP链接。)

In Python 3.5, you can also use this syntax in list, dict, tuple, and set displays (also sometimes called literals). See PEP 488: Additional Unpacking Generalizations.

>>> (0, *range(1, 4), 5, *range(6, 8))
(0, 1, 2, 3, 5, 6, 7)
>>> [0, *range(1, 4), 5, *range(6, 8)]
[0, 1, 2, 3, 5, 6, 7]
>>> {0, *range(1, 4), 5, *range(6, 8)}
{0, 1, 2, 3, 5, 6, 7}
>>> d = {'one': 1, 'two': 2, 'three': 3}
>>> e = {'six': 6, 'seven': 7}
>>> {'zero': 0, **d, 'five': 5, **e}
{'five': 5, 'seven': 7, 'two': 2, 'one': 1, 'three': 3, 'six': 6, 'zero': 0}

It also allows multiple iterables to be unpacked in a single function call.

>>> range(*[1, 10], *[2])
range(1, 10, 2)

(Thanks to mgilson for the PEP link.)


回答 12

除函数调用外,* args和** kwargs在类层次结构中很有用,并且还避免了必须__init__在Python中编写方法。在类似Django代码的框架中可以看到类似的用法。

例如,

def __init__(self, *args, **kwargs):
    for attribute_name, value in zip(self._expected_attributes, args):
        setattr(self, attribute_name, value)
        if kwargs.has_key(attribute_name):
            kwargs.pop(attribute_name)

    for attribute_name in kwargs.viewkeys():
        setattr(self, attribute_name, kwargs[attribute_name])

子类可以是

class RetailItem(Item):
    _expected_attributes = Item._expected_attributes + ['name', 'price', 'category', 'country_of_origin']

class FoodItem(RetailItem):
    _expected_attributes = RetailItem._expected_attributes +  ['expiry_date']

然后将该子类实例化为

food_item = FoodItem(name = 'Jam', 
                     price = 12.0, 
                     category = 'Foods', 
                     country_of_origin = 'US', 
                     expiry_date = datetime.datetime.now())

此外,具有仅对该子类实例有意义的新属性的子类可以调用Base类__init__以卸载属性设置。这是通过* args和** kwargs完成的。主要使用kwargs,以便使用命名参数可以读取代码。例如,

class ElectronicAccessories(RetailItem):
    _expected_attributes = RetailItem._expected_attributes +  ['specifications']
    # Depend on args and kwargs to populate the data as needed.
    def __init__(self, specifications = None, *args, **kwargs):
        self.specifications = specifications  # Rest of attributes will make sense to parent class.
        super(ElectronicAccessories, self).__init__(*args, **kwargs)

可以被形容为

usb_key = ElectronicAccessories(name = 'Sandisk', 
                                price = '$6.00', 
                                category = 'Electronics',
                                country_of_origin = 'CN',
                                specifications = '4GB USB 2.0/USB 3.0')

完整的代码在这里

In addition to function calls, *args and **kwargs are useful in class hierarchies and also avoid having to write __init__ method in Python. Similar usage can seen in frameworks like Django code.

For example,

def __init__(self, *args, **kwargs):
    for attribute_name, value in zip(self._expected_attributes, args):
        setattr(self, attribute_name, value)
        if kwargs.has_key(attribute_name):
            kwargs.pop(attribute_name)

    for attribute_name in kwargs.viewkeys():
        setattr(self, attribute_name, kwargs[attribute_name])

A subclass can then be

class RetailItem(Item):
    _expected_attributes = Item._expected_attributes + ['name', 'price', 'category', 'country_of_origin']

class FoodItem(RetailItem):
    _expected_attributes = RetailItem._expected_attributes +  ['expiry_date']

The subclass then be instantiated as

food_item = FoodItem(name = 'Jam', 
                     price = 12.0, 
                     category = 'Foods', 
                     country_of_origin = 'US', 
                     expiry_date = datetime.datetime.now())

Also, a subclass with a new attribute which makes sense only to that subclass instance can call the Base class __init__ to offload the attributes setting. This is done through *args and **kwargs. kwargs mainly used so that code is readable using named arguments. For example,

class ElectronicAccessories(RetailItem):
    _expected_attributes = RetailItem._expected_attributes +  ['specifications']
    # Depend on args and kwargs to populate the data as needed.
    def __init__(self, specifications = None, *args, **kwargs):
        self.specifications = specifications  # Rest of attributes will make sense to parent class.
        super(ElectronicAccessories, self).__init__(*args, **kwargs)

which can be instatiated as

usb_key = ElectronicAccessories(name = 'Sandisk', 
                                price = '$6.00', 
                                category = 'Electronics',
                                country_of_origin = 'CN',
                                specifications = '4GB USB 2.0/USB 3.0')

The complete code is here


回答 13

建立在昵称的答案上

def foo(param1, *param2):
    print(param1)
    print(param2)


def bar(param1, **param2):
    print(param1)
    print(param2)


def three_params(param1, *param2, **param3):
    print(param1)
    print(param2)
    print(param3)


foo(1, 2, 3, 4, 5)
print("\n")
bar(1, a=2, b=3)
print("\n")
three_params(1, 2, 3, 4, s=5)

输出:

1
(2, 3, 4, 5)

1
{'a': 2, 'b': 3}

1
(2, 3, 4)
{'s': 5}

基本上,任何数量的位置参数都可以使用* args,任何命名参数(或kwargs aka关键字参数)都可以使用** kwargs。

Building on nickd’s answer

def foo(param1, *param2):
    print(param1)
    print(param2)


def bar(param1, **param2):
    print(param1)
    print(param2)


def three_params(param1, *param2, **param3):
    print(param1)
    print(param2)
    print(param3)


foo(1, 2, 3, 4, 5)
print("\n")
bar(1, a=2, b=3)
print("\n")
three_params(1, 2, 3, 4, s=5)

Output:

1
(2, 3, 4, 5)

1
{'a': 2, 'b': 3}

1
(2, 3, 4)
{'s': 5}

Basically, any number of positional arguments can use *args and any named arguments (or kwargs aka keyword arguments) can use **kwargs.


回答 14

*args**kwargs:允许您将可变数量的参数传递给函数。

*args:用于将非关键字的可变长度参数列表发送给函数:

def args(normal_arg, *argv):
    print("normal argument:", normal_arg)

    for arg in argv:
        print("Argument in list of arguments from *argv:", arg)

args('animals', 'fish', 'duck', 'bird')

将生成:

normal argument: animals
Argument in list of arguments from *argv: fish
Argument in list of arguments from *argv: duck
Argument in list of arguments from *argv: bird

**kwargs*

**kwargs允许您将关键字的可变参数长度传递给函数。**kwargs如果要处理函数中的命名参数,则应使用。

def who(**kwargs):
    if kwargs is not None:
        for key, value in kwargs.items():
            print("Your %s is %s." % (key, value))

who(name="Nikola", last_name="Tesla", birthday="7.10.1856", birthplace="Croatia")  

将生成:

Your name is Nikola.
Your last_name is Tesla.
Your birthday is 7.10.1856.
Your birthplace is Croatia.

*args and **kwargs: allow you to pass a variable number of arguments to a function.

*args: is used to send a non-keyworded variable length argument list to the function:

def args(normal_arg, *argv):
    print("normal argument:", normal_arg)

    for arg in argv:
        print("Argument in list of arguments from *argv:", arg)

args('animals', 'fish', 'duck', 'bird')

Will produce:

normal argument: animals
Argument in list of arguments from *argv: fish
Argument in list of arguments from *argv: duck
Argument in list of arguments from *argv: bird

**kwargs*

**kwargs allows you to pass keyworded variable length of arguments to a function. You should use **kwargs if you want to handle named arguments in a function.

def who(**kwargs):
    if kwargs is not None:
        for key, value in kwargs.items():
            print("Your %s is %s." % (key, value))

who(name="Nikola", last_name="Tesla", birthday="7.10.1856", birthplace="Croatia")  

Will produce:

Your name is Nikola.
Your last_name is Tesla.
Your birthday is 7.10.1856.
Your birthplace is Croatia.

回答 15

这个例子可以帮助您记住*args**kwargs甚至super可以立即在Python中继承。

class base(object):
    def __init__(self, base_param):
        self.base_param = base_param


class child1(base): # inherited from base class
    def __init__(self, child_param, *args) # *args for non-keyword args
        self.child_param = child_param
        super(child1, self).__init__(*args) # call __init__ of the base class and initialize it with a NON-KEYWORD arg

class child2(base):
    def __init__(self, child_param, **kwargs):
        self.child_param = child_param
        super(child2, self).__init__(**kwargs) # call __init__ of the base class and initialize it with a KEYWORD arg

c1 = child1(1,0)
c2 = child2(1,base_param=0)
print c1.base_param # 0
print c1.child_param # 1
print c2.base_param # 0
print c2.child_param # 1

This example would help you remember *args, **kwargs and even super and inheritance in Python at once.

class base(object):
    def __init__(self, base_param):
        self.base_param = base_param


class child1(base): # inherited from base class
    def __init__(self, child_param, *args) # *args for non-keyword args
        self.child_param = child_param
        super(child1, self).__init__(*args) # call __init__ of the base class and initialize it with a NON-KEYWORD arg

class child2(base):
    def __init__(self, child_param, **kwargs):
        self.child_param = child_param
        super(child2, self).__init__(**kwargs) # call __init__ of the base class and initialize it with a KEYWORD arg

c1 = child1(1,0)
c2 = child2(1,base_param=0)
print c1.base_param # 0
print c1.child_param # 1
print c2.base_param # 0
print c2.child_param # 1

回答 16

在函数中同时使用两者的一个很好的例子是:

>>> def foo(*arg,**kwargs):
...     print arg
...     print kwargs
>>>
>>> a = (1, 2, 3)
>>> b = {'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(*a,**b)
(1, 2, 3)
{'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(a,**b) 
((1, 2, 3),)
{'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(a,b) 
((1, 2, 3), {'aa': 11, 'bb': 22})
{}
>>>
>>>
>>> foo(a,*b)
((1, 2, 3), 'aa', 'bb')
{}

A good example of using both in a function is:

>>> def foo(*arg,**kwargs):
...     print arg
...     print kwargs
>>>
>>> a = (1, 2, 3)
>>> b = {'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(*a,**b)
(1, 2, 3)
{'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(a,**b) 
((1, 2, 3),)
{'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(a,b) 
((1, 2, 3), {'aa': 11, 'bb': 22})
{}
>>>
>>>
>>> foo(a,*b)
((1, 2, 3), 'aa', 'bb')
{}

回答 17

TL; DR

以下是6种不同的使用情况*,并**在Python编程:

  1. 要使用*args:接受任意数量的位置参数 def foo(*args): pass,此处foo接受任意数量的位置参数,即以下调用有效foo(1)foo(1, 'bar')
  2. 要使用**kwargs:接受任意数量的关键字参数 def foo(**kwargs): pass,此处的’foo’接受任意数量的关键字参数,即以下调用有效foo(name='Tom')foo(name='Tom', age=33)
  3. 要使用*args, **kwargs:接受任意数量的位置和关键字参数 def foo(*args, **kwargs): pass,此处foo接受任意数量的位置和关键字参数,即以下调用是有效的foo(1,name='Tom')foo(1, 'bar', name='Tom', age=33)
  4. 要使用*:强制使用 仅关键字参数def foo(pos1, pos2, *, kwarg1): pass,这*意味着foo仅在pos2之后接受关键字参数,因此foo(1, 2, 3)引发TypeError但foo(1, 2, kwarg1=3)可以。
  5. 要使用*_(注:仅是一种约定)对更多的位置参数不再表示兴趣 def foo(bar, baz, *_): pass:(按约定)意味着(按约定)在其工作中foo仅使用barbaz参数,而将忽略其他参数。
  6. 要使用\**_(注:仅是一种约定)对更多关键字参数不再表示兴趣 def foo(bar, baz, **_): pass:(按约定)意味着(按约定)在其工作中foo仅使用barbaz参数,而将忽略其他参数。

奖励:从python 3.8开始,可以/在函数定义中使用来强制仅位置参数。在以下示例中,参数a和b是仅位置信息,而c或d可以是位置信息或关键字,而e或f必须是关键字:

def f(a, b, /, c, d, *, e, f):
    pass

TL;DR

Below are 6 different use cases for * and ** in python programming:

  1. To accept any number of positional arguments using *args: def foo(*args): pass, here foo accepts any number of positional arguments, i. e., the following calls are valid foo(1), foo(1, 'bar')
  2. To accept any number of keyword arguments using **kwargs: def foo(**kwargs): pass, here ‘foo’ accepts any number of keyword arguments, i. e., the following calls are valid foo(name='Tom'), foo(name='Tom', age=33)
  3. To accept any number of positional and keyword arguments using *args, **kwargs: def foo(*args, **kwargs): pass, here foo accepts any number of positional and keyword arguments, i. e., the following calls are valid foo(1,name='Tom'), foo(1, 'bar', name='Tom', age=33)
  4. To enforce keyword only arguments using *: def foo(pos1, pos2, *, kwarg1): pass, here * means that foo only accept keyword arguments after pos2, hence foo(1, 2, 3) raises TypeError but foo(1, 2, kwarg1=3) is ok.
  5. To express no further interest in more positional arguments using *_ (Note: this is a convention only): def foo(bar, baz, *_): pass means (by convention) foo only uses bar and baz arguments in its working and will ignore others.
  6. To express no further interest in more keyword arguments using \**_ (Note: this is a convention only): def foo(bar, baz, **_): pass means (by convention) foo only uses bar and baz arguments in its working and will ignore others.

BONUS: From python 3.8 onward, one can use / in function definition to enforce positional only parameters. In the following example, parameters a and b are positional-only, while c or d can be positional or keyword, and e or f are required to be keywords:

def f(a, b, /, c, d, *, e, f):
    pass

回答 18

TL; DR

它包传递给函数的参数将listdict分别在函数体中。当您定义函数签名时,如下所示:

def func(*args, **kwds):
    # do stuff

可以使用任意数量的参数和关键字参数来调用它。非关键字参数打包到args函数体内调用的列表中,而关键字参数打包到kwds函数体内调用的dict中。

func("this", "is a list of", "non-keyowrd", "arguments", keyword="ligma", options=[1,2,3])

现在函数体,当函数被调用里面,有两个局部变量,args这是一个有值列表["this", "is a list of", "non-keyword", "arguments"]kwds它是一个dict具有价值{"keyword" : "ligma", "options" : [1,2,3]}


这也可以反向进行,即从呼叫方进行。例如,如果您将函数定义为:

def f(a, b, c, d=1, e=10):
    # do stuff

您可以通过解压缩调用范围中的迭代器或映射来调用它:

iterable = [1, 20, 500]
mapping = {"d" : 100, "e": 3}
f(*iterable, **mapping)
# That call is equivalent to
f(1, 20, 500, d=100, e=3)

TL;DR

It packs arguments passed to the function into list and dict respectively inside the function body. When you define a function signature like this:

def func(*args, **kwds):
    # do stuff

it can be called with any number of arguments and keyword arguments. The non-keyword arguments get packed into a list called args inside the the function body and the keyword arguments get packed into a dict called kwds inside the function body.

func("this", "is a list of", "non-keyowrd", "arguments", keyword="ligma", options=[1,2,3])

now inside the function body, when the function is called, there are two local variables, args which is a list having value ["this", "is a list of", "non-keyword", "arguments"] and kwds which is a dict having value {"keyword" : "ligma", "options" : [1,2,3]}


This also works in reverse, i.e. from the caller side. for example if you have a function defined as:

def f(a, b, c, d=1, e=10):
    # do stuff

you can call it with by unpacking iterables or mappings you have in the calling scope:

iterable = [1, 20, 500]
mapping = {"d" : 100, "e": 3}
f(*iterable, **mapping)
# That call is equivalent to
f(1, 20, 500, d=100, e=3)

回答 19

语境

  • python 3.x
  • 开箱 **
  • 与字符串格式一起使用

与字符串格式一起使用

除了此主题中的答案外,这是其他地方未提及的另一个细节。这扩展了布拉德·所罗门答案

**使用python时,使用进行解包也很有用str.format

这有点类似于您可以使用python f-strings f-string进行的操作,但是增加了声明保留变量的字典的开销(f-string不需要字典)。

快速范例

  ## init vars
  ddvars = dict()
  ddcalc = dict()
  pass
  ddvars['fname']     = 'Huomer'
  ddvars['lname']     = 'Huimpson'
  ddvars['motto']     = 'I love donuts!'
  ddvars['age']       = 33
  pass
  ddcalc['ydiff']     = 5
  ddcalc['ycalc']     = ddvars['age'] + ddcalc['ydiff']
  pass
  vdemo = []

  ## ********************
  ## single unpack supported in py 2.7
  vdemo.append('''
  Hello {fname} {lname}!

  Today you are {age} years old!

  We love your motto "{motto}" and we agree with you!
  '''.format(**ddvars)) 
  pass

  ## ********************
  ## multiple unpack supported in py 3.x
  vdemo.append('''
  Hello {fname} {lname}!

  In {ydiff} years you will be {ycalc} years old!
  '''.format(**ddvars,**ddcalc)) 
  pass

  ## ********************
  print(vdemo[-1])

Context

  • python 3.x
  • unpacking with **
  • use with string formatting

Use with string formatting

In addition to the answers in this thread, here is another detail that was not mentioned elsewhere. This expands on the answer by Brad Solomon

Unpacking with ** is also useful when using python str.format.

This is somewhat similar to what you can do with python f-strings f-string but with the added overhead of declaring a dict to hold the variables (f-string does not require a dict).

Quick Example

  ## init vars
  ddvars = dict()
  ddcalc = dict()
  pass
  ddvars['fname']     = 'Huomer'
  ddvars['lname']     = 'Huimpson'
  ddvars['motto']     = 'I love donuts!'
  ddvars['age']       = 33
  pass
  ddcalc['ydiff']     = 5
  ddcalc['ycalc']     = ddvars['age'] + ddcalc['ydiff']
  pass
  vdemo = []

  ## ********************
  ## single unpack supported in py 2.7
  vdemo.append('''
  Hello {fname} {lname}!

  Today you are {age} years old!

  We love your motto "{motto}" and we agree with you!
  '''.format(**ddvars)) 
  pass

  ## ********************
  ## multiple unpack supported in py 3.x
  vdemo.append('''
  Hello {fname} {lname}!

  In {ydiff} years you will be {ycalc} years old!
  '''.format(**ddvars,**ddcalc)) 
  pass

  ## ********************
  print(vdemo[-1])


回答 20

  • def foo(param1, *param2):是一种方法,可以接受任意数量的值*param2
  • def bar(param1, **param2): 是一种可以使用键接受任意数量的值的方法 *param2
  • param1 是一个简单的参数。

例如,在Java中实现varargs的语法如下:

accessModifier methodName(datatype arg) {
    // method body
}
  • def foo(param1, *param2): is a method can accept arbitrary number of values for *param2,
  • def bar(param1, **param2): is a method can accept arbitrary number of values with keys for *param2
  • param1 is a simple parameter.

For example, the syntax for implementing varargs in Java as follows:

accessModifier methodName(datatype… arg) {
    // method body
}

__init__.py的作用是什么?

问题:__init__.py的作用是什么?

什么是__init__.py一个Python源目录?

What is __init__.py for in a Python source directory?


回答 0

它曾经是软件包的必需部分(旧的3.3之前的“常规软件包”,而不是新的3.3+“命名空间软件包”)。

这是文档。

Python定义了两种类型的程序包,常规程序包和命名空间程序包。常规软件包是Python 3.2及更早版本中存在的传统软件包。常规软件包通常实现为包含__init__.py文件的目录。导入常规程序包时,__init__.py将隐式执行此文件,并将其定义的对象绑定到程序包命名空间中的名称。该__init__.py文件可以包含任何其他模块可以包含的相同Python代码,并且Python在导入时会向该模块添加一些其他属性。

但是只需单击链接,它就会包含一个示例,更多信息以及对命名空间包的说明,这些命名空间包不含__init__.py

It used to be a required part of a package (old, pre-3.3 “regular package”, not newer 3.3+ “namespace package”).

Here’s the documentation.

Python defines two types of packages, regular packages and namespace packages. Regular packages are traditional packages as they existed in Python 3.2 and earlier. A regular package is typically implemented as a directory containing an __init__.py file. When a regular package is imported, this __init__.py file is implicitly executed, and the objects it defines are bound to names in the package’s namespace. The __init__.py file can contain the same Python code that any other module can contain, and Python will add some additional attributes to the module when it is imported.

But just click the link, it contains an example, more information, and an explanation of namespace packages, the kind of packages without __init__.py.


回答 1

命名__init__.py的文件用于将磁盘上的目录标记为Python软件包目录。如果您有文件

mydir/spam/__init__.py
mydir/spam/module.py

并且mydir在您的路径上,您可以将代码导入module.py

import spam.module

要么

from spam import module

如果删除该__init__.py文件,Python将不再在该目录中查找子模块,因此尝试导入该模块将失败。

__init__.py文件通常为空,但可用于以更方便的名称导出包的选定部分,保留方便的功能等。给定上面的示例,可以按以下方式访问init模块的内容:

import spam

基于

Files named __init__.py are used to mark directories on disk as Python package directories. If you have the files

mydir/spam/__init__.py
mydir/spam/module.py

and mydir is on your path, you can import the code in module.py as

import spam.module

or

from spam import module

If you remove the __init__.py file, Python will no longer look for submodules inside that directory, so attempts to import the module will fail.

The __init__.py file is usually empty, but can be used to export selected portions of the package under more convenient name, hold convenience functions, etc. Given the example above, the contents of the init module can be accessed as

import spam

based on this


回答 2

除了将目录标记为Python软件包并定义之外__all__,还__init__.py允许您在软件包级别定义任何变量。如果程序包定义了将以类似于API的方式频繁导入的内容,则这样做通常很方便。这种模式促进了对Pythonic的“扁平优于嵌套”哲学的坚持。

一个例子

这是我的一个项目的示例,在该项目中,我经常导入sessionmaker被叫Session以与数据库交互。我写了一个带有一些模块的“数据库”包:

database/
    __init__.py
    schema.py
    insertions.py
    queries.py

__init__.py包含以下代码:

import os

from sqlalchemy.orm import sessionmaker
from sqlalchemy import create_engine

engine = create_engine(os.environ['DATABASE_URL'])
Session = sessionmaker(bind=engine)

既然我Session在这里定义,就可以使用以下语法开始新的会话。此代码将从“数据库”包目录的内部或外部执行相同。

from database import Session
session = Session()

当然,这是一个小方便—替代方法是Session在数据库包中的新文件(例如“ create_session.py”)中定义,然后使用以下命令启动新会话:

from database.create_session import Session
session = Session()

进一步阅读

有一个非常有趣的reddit线程,涵盖了__init__.py此处的适当用法:

http://www.reddit.com/r/Python/comments/1bbbwk/whats_your_opinion_on_what_to_include_in_init_py/

大多数人似乎认为__init__.py文件应该非常薄,以避免违反“显式优于隐式”的哲学。

In addition to labeling a directory as a Python package and defining __all__, __init__.py allows you to define any variable at the package level. Doing so is often convenient if a package defines something that will be imported frequently, in an API-like fashion. This pattern promotes adherence to the Pythonic “flat is better than nested” philosophy.

An example

Here is an example from one of my projects, in which I frequently import a sessionmaker called Session to interact with my database. I wrote a “database” package with a few modules:

database/
    __init__.py
    schema.py
    insertions.py
    queries.py

My __init__.py contains the following code:

import os

from sqlalchemy.orm import sessionmaker
from sqlalchemy import create_engine

engine = create_engine(os.environ['DATABASE_URL'])
Session = sessionmaker(bind=engine)

Since I define Session here, I can start a new session using the syntax below. This code would be the same executed from inside or outside of the “database” package directory.

from database import Session
session = Session()

Of course, this is a small convenience — the alternative would be to define Session in a new file like “create_session.py” in my database package, and start new sessions using:

from database.create_session import Session
session = Session()

Further reading

There is a pretty interesting reddit thread covering appropriate uses of __init__.py here:

http://www.reddit.com/r/Python/comments/1bbbwk/whats_your_opinion_on_what_to_include_in_init_py/

The majority opinion seems to be that __init__.py files should be very thin to avoid violating the “explicit is better than implicit” philosophy.


回答 3

有两个主要原因 __init__.py

  1. 为方便起见:其他用户将不需要知道您的函数在包层次结构中的确切位置。

    your_package/
      __init__.py
      file1.py
      file2.py
        ...
      fileN.py
    # in __init__.py
    from file1 import *
    from file2 import *
    ...
    from fileN import *
    # in file1.py
    def add():
        pass

    然后其他人可以通过以下方式调用add()

    from your_package import add

    不知道file1,例如

    from your_package.file1 import add
  2. 如果您想初始化一些东西;例如,日志记录(应放在顶层):

    import logging.config
    logging.config.dictConfig(Your_logging_config)

There are 2 main reasons for __init__.py

  1. For convenience: the other users will not need to know your functions’ exact location in your package hierarchy.

    your_package/
      __init__.py
      file1.py
      file2.py
        ...
      fileN.py
    
    # in __init__.py
    from file1 import *
    from file2 import *
    ...
    from fileN import *
    
    # in file1.py
    def add():
        pass
    

    then others can call add() by

    from your_package import add
    

    without knowing file1, like

    from your_package.file1 import add
    
  2. If you want something to be initialized; for example, logging (which should be put in the top level):

    import logging.config
    logging.config.dictConfig(Your_logging_config)
    

回答 4

__init__.py文件使Python将包含它的目录视为模块。

此外,这是要在模块中加载的第一个文件,因此您可以使用它来执行每次加载模块时要运行的代码,或指定要导出的子模块。

The __init__.py file makes Python treat directories containing it as modules.

Furthermore, this is the first file to be loaded in a module, so you can use it to execute code that you want to run each time a module is loaded, or specify the submodules to be exported.


回答 5

从Python 3.3开始,__init__.py不再需要将目录定义为可导入的Python包。

检查PEP 420:隐式命名空间包

对不需要__init__.py标记文件并且可以自动跨越多个路径段的软件包目录的本地支持(受PEP 420中所述的各种第三方方法启发,用于命名空间软件包)

这是测试:

$ mkdir -p /tmp/test_init
$ touch /tmp/test_init/module.py /tmp/test_init/__init__.py
$ tree -at /tmp/test_init
/tmp/test_init
├── module.py
└── __init__.py
$ python3

>>> import sys
>>> sys.path.insert(0, '/tmp')
>>> from test_init import module
>>> import test_init.module

$ rm -f /tmp/test_init/__init__.py
$ tree -at /tmp/test_init
/tmp/test_init
└── module.py
$ python3

>>> import sys
>>> sys.path.insert(0, '/tmp')
>>> from test_init import module
>>> import test_init.module

参考:
https
: //docs.python.org/3/whatsnew/3.3.html#pep-420-implicit-namespace-packages https://www.python.org/dev/peps/pep-0420/
是__init__。 py对于Python 3中的软件包不是必需的?

Since Python 3.3, __init__.py is no longer required to define directories as importable Python packages.

Check PEP 420: Implicit Namespace Packages:

Native support for package directories that don’t require __init__.py marker files and can automatically span multiple path segments (inspired by various third party approaches to namespace packages, as described in PEP 420)

Here’s the test:

$ mkdir -p /tmp/test_init
$ touch /tmp/test_init/module.py /tmp/test_init/__init__.py
$ tree -at /tmp/test_init
/tmp/test_init
├── module.py
└── __init__.py
$ python3

>>> import sys
>>> sys.path.insert(0, '/tmp')
>>> from test_init import module
>>> import test_init.module

$ rm -f /tmp/test_init/__init__.py
$ tree -at /tmp/test_init
/tmp/test_init
└── module.py
$ python3

>>> import sys
>>> sys.path.insert(0, '/tmp')
>>> from test_init import module
>>> import test_init.module

references:
https://docs.python.org/3/whatsnew/3.3.html#pep-420-implicit-namespace-packages
https://www.python.org/dev/peps/pep-0420/
Is __init__.py not required for packages in Python 3?


回答 6

在Python中,包的定义非常简单。像Java一样,层次结构和目录结构相同。但是您必须将__init__.py其打包。我将__init__.py用以下示例解释该文件:

package_x/
|--  __init__.py
|--    subPackage_a/
|------  __init__.py
|------  module_m1.py
|--    subPackage_b/
|------  __init__.py
|------  module_n1.py
|------  module_n2.py
|------  module_n3.py

__init__.py只要存在就可以为空。它指示该目录应视为一个包。当然__init__.py也可以设置适当的内容。

如果我们在module_n1中添加一个函数:

def function_X():
    print "function_X in module_n1"
    return

运行后:

>>>from package_x.subPackage_b.module_n1 import function_X
>>>function_X()

function_X in module_n1 

然后,我们遵循层次结构包,并将module_n1称为函数。我们可以__init__.py像这样在subPackage_b中使用:

__all__ = ['module_n2', 'module_n3']

运行后:

>>>from package_x.subPackage_b import * 
>>>module_n1.function_X()

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ImportError: No module named module_n1

因此,使用*导入,模块包受__init__.py内容的约束。

In Python the definition of package is very simple. Like Java the hierarchical structure and the directory structure are the same. But you have to have __init__.py in a package. I will explain the __init__.py file with the example below:

package_x/
|--  __init__.py
|--    subPackage_a/
|------  __init__.py
|------  module_m1.py
|--    subPackage_b/
|------  __init__.py
|------  module_n1.py
|------  module_n2.py
|------  module_n3.py

__init__.py can be empty, as long as it exists. It indicates that the directory should be regarded as a package. Of course, __init__.py can also set the appropriate content.

If we add a function in module_n1:

def function_X():
    print "function_X in module_n1"
    return

After running:

>>>from package_x.subPackage_b.module_n1 import function_X
>>>function_X()

function_X in module_n1 

Then we followed the hierarchy package and called module_n1 the function. We can use __init__.py in subPackage_b like this:

__all__ = ['module_n2', 'module_n3']

After running:

>>>from package_x.subPackage_b import * 
>>>module_n1.function_X()

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ImportError: No module named module_n1

Hence using * importing, module package is subject to __init__.py content.


回答 7

尽管Python在没有__init__.py文件的情况下仍可工作,但您仍应包含一个文件。

它指定应将程序包视为模块,因此将其包括在内(即使它为空)。

在某些情况下,您实际上可能会使用__init__.py文件:

假设您具有以下文件结构:

main_methods 
    |- methods.py

methods.py包含以下内容:

def foo():
    return 'foo'

要使用,foo()您需要以下条件之一:

from main_methods.methods import foo # Call with foo()
from main_methods import methods # Call with methods.foo()
import main_methods.methods # Call with main_methods.methods.foo()

也许您需要(或想要)保留methods.py在内部main_methods(例如,运行时/依赖项),但只想导入main_methods


如果将的名称更改为methods.py__init__.pyfoo()只需导入即可使用main_methods

import main_methods
print(main_methods.foo()) # Prints 'foo'

这是有效的,因为__init__.py它被视为包装的一部分。


一些Python软件包实际上是这样做的。以JSON为例,其中running import json实际上是__init__.pyjson包中导入的(请参阅此处的包文件结构):

源代码: Lib/json/__init__.py

Although Python works without an __init__.py file you should still include one.

It specifies a package should be treated as a module, so therefore include it (even if it is empty).

There is also a case where you may actually use an __init__.py file:

Imagine you had the following file structure:

main_methods 
    |- methods.py

And methods.py contained this:

def foo():
    return 'foo'

To use foo() you would need one of the following:

from main_methods.methods import foo # Call with foo()
from main_methods import methods # Call with methods.foo()
import main_methods.methods # Call with main_methods.methods.foo()

Maybe there you need (or want) to keep methods.py inside main_methods (runtimes/dependencies for example) but you only want to import main_methods.


If you changed the name of methods.py to __init__.py then you could use foo() by just importing main_methods:

import main_methods
print(main_methods.foo()) # Prints 'foo'

This works because __init__.py is treated as part of the package.


Some Python packages actually do this. An example is with JSON, where running import json is actually importing __init__.py from the json package (see the package file structure here):

Source code: Lib/json/__init__.py


回答 8

__init__.py 会将其所在目录视为可加载模块。

对于喜欢阅读代码的人,我在这里添加了两位炼金术士的评论。

$ find /tmp/mydir/
/tmp/mydir/
/tmp/mydir//spam
/tmp/mydir//spam/__init__.py
/tmp/mydir//spam/module.py
$ cd ~
$ python
>>> import sys
>>> sys.path.insert(0, '/tmp/mydir')
>>> from spam import module
>>> module.myfun(3)
9
>>> exit()
$ 
$ rm /tmp/mydir/spam/__init__.py*
$ 
$ python
>>> import sys
>>> sys.path.insert(0, '/tmp/mydir')
>>> from spam import module
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ImportError: No module named spam
>>> 

__init__.py will treat the directory it is in as a loadable module.

For people who prefer reading code, I put Two-Bit Alchemist’s comment here.

$ find /tmp/mydir/
/tmp/mydir/
/tmp/mydir//spam
/tmp/mydir//spam/__init__.py
/tmp/mydir//spam/module.py
$ cd ~
$ python
>>> import sys
>>> sys.path.insert(0, '/tmp/mydir')
>>> from spam import module
>>> module.myfun(3)
9
>>> exit()
$ 
$ rm /tmp/mydir/spam/__init__.py*
$ 
$ python
>>> import sys
>>> sys.path.insert(0, '/tmp/mydir')
>>> from spam import module
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ImportError: No module named spam
>>> 

回答 9

它有助于导入其他python文件。当您将此文件放置在包含其他py文件的目录中(例如,东西)时,可以执行诸如import stuff.other之类的操作。

root\
    stuff\
         other.py

    morestuff\
         another.py

如果__init__.py在目录东西中没有此内容,则无法导入other.py,因为Python不知道东西的源代码在哪里,也无法将其识别为包。

It facilitates importing other python files. When you placed this file in a directory (say stuff)containing other py files, then you can do something like import stuff.other.

root\
    stuff\
         other.py

    morestuff\
         another.py

Without this __init__.py inside the directory stuff, you couldn’t import other.py, because Python doesn’t know where the source code for stuff is and unable to recognize it as a package.


回答 10

一个__init__.py文件使得进口容易。当__init__.py包中包含an时,a()可以从文件中导入函数,b.py如下所示:

from b import a

但是,没有它,您将无法直接导入。您必须修改系统路径:

import sys
sys.path.insert(0, 'path/to/b.py')

from b import a

An __init__.py file makes imports easy. When an __init__.py is present within a package, function a() can be imported from file b.py like so:

from b import a

Without it, however, you can’t import directly. You have to amend the system path:

import sys
sys.path.insert(0, 'path/to/b.py')

from b import a

将字节转换为字符串

问题:将字节转换为字符串

我正在使用以下代码从外部程序获取标准输出:

>>> from subprocess import *
>>> command_stdout = Popen(['ls', '-l'], stdout=PIPE).communicate()[0]

communication()方法返回一个字节数组:

>>> command_stdout
b'total 0\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2\n'

但是,我想将输出作为普通的Python字符串使用。这样我就可以像这样打印它:

>>> print(command_stdout)
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2

我认为这就是binascii.b2a_qp()方法的用途,但是当我尝试使用它时,我又得到了相同的字节数组:

>>> binascii.b2a_qp(command_stdout)
b'total 0\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2\n'

如何将字节值转换回字符串?我的意思是,使用“电池”而不是手动进行操作。我希望它与Python 3兼容。

I’m using this code to get standard output from an external program:

>>> from subprocess import *
>>> command_stdout = Popen(['ls', '-l'], stdout=PIPE).communicate()[0]

The communicate() method returns an array of bytes:

>>> command_stdout
b'total 0\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2\n'

However, I’d like to work with the output as a normal Python string. So that I could print it like this:

>>> print(command_stdout)
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2

I thought that’s what the binascii.b2a_qp() method is for, but when I tried it, I got the same byte array again:

>>> binascii.b2a_qp(command_stdout)
b'total 0\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2\n'

How do I convert the bytes value back to string? I mean, using the “batteries” instead of doing it manually. And I’d like it to be OK with Python 3.


回答 0

您需要解码bytes对象以产生一个字符串:

>>> b"abcde"
b'abcde'

# utf-8 is used here because it is a very common encoding, but you
# need to use the encoding your data is actually in.
>>> b"abcde".decode("utf-8") 
'abcde'

You need to decode the bytes object to produce a string:

>>> b"abcde"
b'abcde'

# utf-8 is used here because it is a very common encoding, but you
# need to use the encoding your data is actually in.
>>> b"abcde".decode("utf-8") 
'abcde'

回答 1

您需要解码该字节字符串,然后将其转换为字符(Unicode)字符串。

在Python 2上

encoding = 'utf-8'
'hello'.decode(encoding)

要么

unicode('hello', encoding)

在Python 3上

encoding = 'utf-8'
b'hello'.decode(encoding)

要么

str(b'hello', encoding)

You need to decode the byte string and turn it in to a character (Unicode) string.

On Python 2

encoding = 'utf-8'
'hello'.decode(encoding)

or

unicode('hello', encoding)

On Python 3

encoding = 'utf-8'
b'hello'.decode(encoding)

or

str(b'hello', encoding)

回答 2

我认为这种方式很简单:

>>> bytes_data = [112, 52, 52]
>>> "".join(map(chr, bytes_data))
'p44'

I think this way is easy:

>>> bytes_data = [112, 52, 52]
>>> "".join(map(chr, bytes_data))
'p44'

回答 3

如果您不知道编码,则要以Python 3和Python 2兼容的方式将二进制输入读取为字符串,请使用古老的MS-DOS CP437编码:

PY3K = sys.version_info >= (3, 0)

lines = []
for line in stream:
    if not PY3K:
        lines.append(line)
    else:
        lines.append(line.decode('cp437'))

因为编码是未知的,所以希望将非英语符号转换为字符cp437(不会翻译英语字符,因为它们在大多数单字节编码和UTF-8中都匹配)。

将任意二进制输入解码为UTF-8是不安全的,因为您可能会得到以下信息:

>>> b'\x00\x01\xffsd'.decode('utf-8')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xff in position 2: invalid
start byte

同样适用于latin-1,这在Python 2中很流行(默认?)。请参见“ 代码页布局”中的遗漏之处-这是Python臭名昭著的地方ordinal not in range

UPDATE 20150604:有传言称Python 3具有surrogateescape错误策略,可将内容编码为二进制数据而不会导致数据丢失和崩溃,但它需要进行转换测试[binary] -> [str] -> [binary],以验证性能和可靠性。

更新20170116:感谢评论-还可以使用backslashreplace错误处理程序对所有未知字节进行斜线转义。这仅适用于Python 3,因此即使采用这种解决方法,您仍然会从不同的Python版本获得不一致的输出:

PY3K = sys.version_info >= (3, 0)

lines = []
for line in stream:
    if not PY3K:
        lines.append(line)
    else:
        lines.append(line.decode('utf-8', 'backslashreplace'))

看到 详细信息, Python的Unicode支持

更新20170119:我决定实现适用于Python 2和Python 3的斜线转义解码。它应该比cp437解决方案要慢,但是在每个Python版本上都应产生相同的结果

# --- preparation

import codecs

def slashescape(err):
    """ codecs error handler. err is UnicodeDecode instance. return
    a tuple with a replacement for the unencodable part of the input
    and a position where encoding should continue"""
    #print err, dir(err), err.start, err.end, err.object[:err.start]
    thebyte = err.object[err.start:err.end]
    repl = u'\\x'+hex(ord(thebyte))[2:]
    return (repl, err.end)

codecs.register_error('slashescape', slashescape)

# --- processing

stream = [b'\x80abc']

lines = []
for line in stream:
    lines.append(line.decode('utf-8', 'slashescape'))

If you don’t know the encoding, then to read binary input into string in Python 3 and Python 2 compatible way, use the ancient MS-DOS CP437 encoding:

PY3K = sys.version_info >= (3, 0)

lines = []
for line in stream:
    if not PY3K:
        lines.append(line)
    else:
        lines.append(line.decode('cp437'))

Because encoding is unknown, expect non-English symbols to translate to characters of cp437 (English characters are not translated, because they match in most single byte encodings and UTF-8).

Decoding arbitrary binary input to UTF-8 is unsafe, because you may get this:

>>> b'\x00\x01\xffsd'.decode('utf-8')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xff in position 2: invalid
start byte

The same applies to latin-1, which was popular (the default?) for Python 2. See the missing points in Codepage Layout – it is where Python chokes with infamous ordinal not in range.

UPDATE 20150604: There are rumors that Python 3 has the surrogateescape error strategy for encoding stuff into binary data without data loss and crashes, but it needs conversion tests, [binary] -> [str] -> [binary], to validate both performance and reliability.

UPDATE 20170116: Thanks to comment by Nearoo – there is also a possibility to slash escape all unknown bytes with backslashreplace error handler. That works only for Python 3, so even with this workaround you will still get inconsistent output from different Python versions:

PY3K = sys.version_info >= (3, 0)

lines = []
for line in stream:
    if not PY3K:
        lines.append(line)
    else:
        lines.append(line.decode('utf-8', 'backslashreplace'))

See Python’s Unicode Support for details.

UPDATE 20170119: I decided to implement slash escaping decode that works for both Python 2 and Python 3. It should be slower than the cp437 solution, but it should produce identical results on every Python version.

# --- preparation

import codecs

def slashescape(err):
    """ codecs error handler. err is UnicodeDecode instance. return
    a tuple with a replacement for the unencodable part of the input
    and a position where encoding should continue"""
    #print err, dir(err), err.start, err.end, err.object[:err.start]
    thebyte = err.object[err.start:err.end]
    repl = u'\\x'+hex(ord(thebyte))[2:]
    return (repl, err.end)

codecs.register_error('slashescape', slashescape)

# --- processing

stream = [b'\x80abc']

lines = []
for line in stream:
    lines.append(line.decode('utf-8', 'slashescape'))

回答 4

在Python 3中,默认编码为"utf-8",因此您可以直接使用:

b'hello'.decode()

相当于

b'hello'.decode(encoding="utf-8")

另一方面,在Python 2中,编码默认为默认的字符串编码。因此,您应该使用:

b'hello'.decode(encoding)

encoding您想要的编码在哪里。

注意:在Python 2.7中添加了对关键字参数的支持。

In Python 3, the default encoding is "utf-8", so you can directly use:

b'hello'.decode()

which is equivalent to

b'hello'.decode(encoding="utf-8")

On the other hand, in Python 2, encoding defaults to the default string encoding. Thus, you should use:

b'hello'.decode(encoding)

where encoding is the encoding you want.

Note: support for keyword arguments was added in Python 2.7.


回答 5

我认为您实际上想要这样:

>>> from subprocess import *
>>> command_stdout = Popen(['ls', '-l'], stdout=PIPE).communicate()[0]
>>> command_text = command_stdout.decode(encoding='windows-1252')

亚伦的答案是正确的,除了您需要知道哪个要使用编码。而且我相信Windows使用的是“ windows-1252”。仅当您的内容中包含一些不寻常的(非ASCII)字符时,这才有意义,但这将有所作为。

顺便说一句,它事实上事情的原因了Python转移到使用两种不同类型的二进制和文本数据:它不能神奇地将它们转换之间,因为它不知道编码,除非你告诉它!您唯一知道的方法是阅读Windows文档(或在此处阅读)。

I think you actually want this:

>>> from subprocess import *
>>> command_stdout = Popen(['ls', '-l'], stdout=PIPE).communicate()[0]
>>> command_text = command_stdout.decode(encoding='windows-1252')

Aaron’s answer was correct, except that you need to know which encoding to use. And I believe that Windows uses ‘windows-1252’. It will only matter if you have some unusual (non-ASCII) characters in your content, but then it will make a difference.

By the way, the fact that it does matter is the reason that Python moved to using two different types for binary and text data: it can’t convert magically between them, because it doesn’t know the encoding unless you tell it! The only way YOU would know is to read the Windows documentation (or read it here).


回答 6

将Universal_newlines设置为True,即

command_stdout = Popen(['ls', '-l'], stdout=PIPE, universal_newlines=True).communicate()[0]

Set universal_newlines to True, i.e.

command_stdout = Popen(['ls', '-l'], stdout=PIPE, universal_newlines=True).communicate()[0]

回答 7

虽然@Aaron Maenpaa的答案有效,但最近有用户

有没有更简单的方法?’fhand.read()。decode(“ ASCII”)'[…]太长了!

您可以使用:

command_stdout.decode()

decode()有一个标准参数

codecs.decode(obj, encoding='utf-8', errors='strict')

While @Aaron Maenpaa’s answer just works, a user recently asked:

Is there any more simply way? ‘fhand.read().decode(“ASCII”)’ […] It’s so long!

You can use:

command_stdout.decode()

decode() has a standard argument:

codecs.decode(obj, encoding='utf-8', errors='strict')


回答 8

要将字节序列解释为文本,您必须知道相应的字符编码:

unicode_text = bytestring.decode(character_encoding)

例:

>>> b'\xc2\xb5'.decode('utf-8')
'µ'

ls命令可能会产生无法解释为文本的输出。Unix上的文件名可以是任何字节序列,但斜杠b'/'和零 除外b'\0'

>>> open(bytes(range(0x100)).translate(None, b'\0/'), 'w').close()

尝试使用utf-8编码对此类字节汤进行解码将引发UnicodeDecodeError

可能会更糟。 如果使用错误的不兼容编码,解码可能会默默失败并产生mojibake

>>> '—'.encode('utf-8').decode('cp1252')
'—'

数据已损坏,但是您的程序仍然不知道发生了故障。

通常,要使用的字符编码不会嵌入字节序列本身。您必须带外传达此信息。一些结果比其他结果更有可能,因此chardet存在可以猜测字符编码的模块。单个Python脚本可能在不同位置使用多种字符编码。


ls可以使用os.fsdecode() 即使对于无法解码的文件名也成功的函数将输出转换为Python字符串(在Unix上使用 sys.getfilesystemencoding()surrogateescape错误处理程序):

import os
import subprocess

output = os.fsdecode(subprocess.check_output('ls'))

要获取原始字节,可以使用os.fsencode()

如果传递universal_newlines=True参数,则subprocess用于 locale.getpreferredencoding(False)解码字节,例如,它可以 cp1252在Windows上使用。

要实时解码字节流, io.TextIOWrapper() 可以使用:example

不同的命令可能对其输出使用不同的字符编码,例如,dir内部命令(cmd)可能使用cp437。要解码其输出,可以显式传递编码(Python 3.6+):

output = subprocess.check_output('dir', shell=True, encoding='cp437')

文件名可能与os.listdir()(使用Windows Unicode API)不同(例如,'\xb6'可以用'\x14'—Python的cp437编解码器映射b'\x14'代替)来控制字符U + 0014而不是U + 00B6(¶)。要支持带有任意Unicode字符的文件名,请参阅将 PowerShell输出可能包含非ASCII Unicode字符解码为Python字符串。

To interpret a byte sequence as a text, you have to know the corresponding character encoding:

unicode_text = bytestring.decode(character_encoding)

Example:

>>> b'\xc2\xb5'.decode('utf-8')
'µ'

ls command may produce output that can’t be interpreted as text. File names on Unix may be any sequence of bytes except slash b'/' and zero b'\0':

>>> open(bytes(range(0x100)).translate(None, b'\0/'), 'w').close()

Trying to decode such byte soup using utf-8 encoding raises UnicodeDecodeError.

It can be worse. The decoding may fail silently and produce mojibake if you use a wrong incompatible encoding:

>>> '—'.encode('utf-8').decode('cp1252')
'—'

The data is corrupted but your program remains unaware that a failure has occurred.

In general, what character encoding to use is not embedded in the byte sequence itself. You have to communicate this info out-of-band. Some outcomes are more likely than others and therefore chardet module exists that can guess the character encoding. A single Python script may use multiple character encodings in different places.


ls output can be converted to a Python string using os.fsdecode() function that succeeds even for undecodable filenames (it uses sys.getfilesystemencoding() and surrogateescape error handler on Unix):

import os
import subprocess

output = os.fsdecode(subprocess.check_output('ls'))

To get the original bytes, you could use os.fsencode().

If you pass universal_newlines=True parameter then subprocess uses locale.getpreferredencoding(False) to decode bytes e.g., it can be cp1252 on Windows.

To decode the byte stream on-the-fly, io.TextIOWrapper() could be used: example.

Different commands may use different character encodings for their output e.g., dir internal command (cmd) may use cp437. To decode its output, you could pass the encoding explicitly (Python 3.6+):

output = subprocess.check_output('dir', shell=True, encoding='cp437')

The filenames may differ from os.listdir() (which uses Windows Unicode API) e.g., '\xb6' can be substituted with '\x14'—Python’s cp437 codec maps b'\x14' to control character U+0014 instead of U+00B6 (¶). To support filenames with arbitrary Unicode characters, see Decode PowerShell output possibly containing non-ASCII Unicode characters into a Python string


回答 9

由于这个问题实际上是在询问subprocess输出,因此您可以使用更直接的方法,因为它Popen接受了encoding关键字(在Python 3.6+中):

>>> from subprocess import Popen, PIPE
>>> text = Popen(['ls', '-l'], stdout=PIPE, encoding='utf-8').communicate()[0]
>>> type(text)
str
>>> print(text)
total 0
-rw-r--r-- 1 wim badger 0 May 31 12:45 some_file.txt

其他用户的一般答案是将字节解码为文本:

>>> b'abcde'.decode()
'abcde'

没有参数,sys.getdefaultencoding()将被使用。如果您的数据不是sys.getdefaultencoding(),那么您必须在decode调用中显式指定编码:

>>> b'caf\xe9'.decode('cp1250')
'café'

Since this question is actually asking about subprocess output, you have a more direct approach available since Popen accepts an encoding keyword (in Python 3.6+):

>>> from subprocess import Popen, PIPE
>>> text = Popen(['ls', '-l'], stdout=PIPE, encoding='utf-8').communicate()[0]
>>> type(text)
str
>>> print(text)
total 0
-rw-r--r-- 1 wim badger 0 May 31 12:45 some_file.txt

The general answer for other users is to decode bytes to text:

>>> b'abcde'.decode()
'abcde'

With no argument, sys.getdefaultencoding() will be used. If your data is not sys.getdefaultencoding(), then you must specify the encoding explicitly in the decode call:

>>> b'caf\xe9'.decode('cp1250')
'café'

回答 10

如果您应该尝试以下操作decode()

AttributeError:“ str”对象没有属性“ decode”

您还可以直接在转换中指定编码类型:

>>> my_byte_str
b'Hello World'

>>> str(my_byte_str, 'utf-8')
'Hello World'

If you should get the following by trying decode():

AttributeError: ‘str’ object has no attribute ‘decode’

You can also specify the encoding type straight in a cast:

>>> my_byte_str
b'Hello World'

>>> str(my_byte_str, 'utf-8')
'Hello World'

回答 11

当使用Windows系统中的数据(以\r\n行结尾)时,我的答案是

String = Bytes.decode("utf-8").replace("\r\n", "\n")

为什么?尝试使用多行Input.txt:

Bytes = open("Input.txt", "rb").read()
String = Bytes.decode("utf-8")
open("Output.txt", "w").write(String)

您所有的行尾都将加倍(以 \r\r\n),从而导致多余的空行。Python的文本读取函数通常会规范行尾,因此字符串只能使用\n。如果您从Windows系统接收二进制数据,Python将没有机会这样做。从而,

Bytes = open("Input.txt", "rb").read()
String = Bytes.decode("utf-8").replace("\r\n", "\n")
open("Output.txt", "w").write(String)

将复制您的原始文件。

When working with data from Windows systems (with \r\n line endings), my answer is

String = Bytes.decode("utf-8").replace("\r\n", "\n")

Why? Try this with a multiline Input.txt:

Bytes = open("Input.txt", "rb").read()
String = Bytes.decode("utf-8")
open("Output.txt", "w").write(String)

All your line endings will be doubled (to \r\r\n), leading to extra empty lines. Python’s text-read functions usually normalize line endings so that strings use only \n. If you receive binary data from a Windows system, Python does not have a chance to do that. Thus,

Bytes = open("Input.txt", "rb").read()
String = Bytes.decode("utf-8").replace("\r\n", "\n")
open("Output.txt", "w").write(String)

will replicate your original file.


回答 12

我做了一个清理清单的功能

def cleanLists(self, lista):
    lista = [x.strip() for x in lista]
    lista = [x.replace('\n', '') for x in lista]
    lista = [x.replace('\b', '') for x in lista]
    lista = [x.encode('utf8') for x in lista]
    lista = [x.decode('utf8') for x in lista]

    return lista

I made a function to clean a list

def cleanLists(self, lista):
    lista = [x.strip() for x in lista]
    lista = [x.replace('\n', '') for x in lista]
    lista = [x.replace('\b', '') for x in lista]
    lista = [x.encode('utf8') for x in lista]
    lista = [x.decode('utf8') for x in lista]

    return lista

回答 13

对于Python 3,这是一个更安全和Python的方法来从转换bytestring

def byte_to_str(bytes_or_str):
    if isinstance(bytes_or_str, bytes): # Check if it's in bytes
        print(bytes_or_str.decode('utf-8'))
    else:
        print("Object not of byte type")

byte_to_str(b'total 0\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2\n')

输出:

total 0
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2

For Python 3, this is a much safer and Pythonic approach to convert from byte to string:

def byte_to_str(bytes_or_str):
    if isinstance(bytes_or_str, bytes): # Check if it's in bytes
        print(bytes_or_str.decode('utf-8'))
    else:
        print("Object not of byte type")

byte_to_str(b'total 0\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1\n-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2\n')

Output:

total 0
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file1
-rw-rw-r-- 1 thomas thomas 0 Mar  3 07:03 file2

回答 14

sys —系统特定的参数和功能

要从标准流写入二进制数据或从标准流读取二进制数据,请使用基础二进制缓冲区。例如,要将字节写入stdout,请使用sys.stdout.buffer.write(b'abc')

From sys — System-specific parameters and functions:

To write or read binary data from/to the standard streams, use the underlying binary buffer. For example, to write bytes to stdout, use sys.stdout.buffer.write(b'abc').


回答 15

def toString(string):    
    try:
        return v.decode("utf-8")
    except ValueError:
        return string

b = b'97.080.500'
s = '97.080.500'
print(toString(b))
print(toString(s))
def toString(string):    
    try:
        return v.decode("utf-8")
    except ValueError:
        return string

b = b'97.080.500'
s = '97.080.500'
print(toString(b))
print(toString(s))

回答 16

对于“运行shell命令并以文本而不是字节形式获取其输出” 的特定情况,在Python 3.7上,您应该使用subprocess.run并传入text=True(以及capture_output=True捕获输出)

command_result = subprocess.run(["ls", "-l"], capture_output=True, text=True)
command_result.stdout  # is a `str` containing your program's stdout

text过去称为universal_newlines,并在Python 3.7中进行了更改(很好,为别名)。如果要支持3.7之前的Python版本,请传入universal_newlines=True而不是text=True

For your specific case of “run a shell command and get its output as text instead of bytes”, on Python 3.7, you should use subprocess.run and pass in text=True (as well as capture_output=True to capture the output)

command_result = subprocess.run(["ls", "-l"], capture_output=True, text=True)
command_result.stdout  # is a `str` containing your program's stdout

text used to be called universal_newlines, and was changed (well, aliased) in Python 3.7. If you want to support Python versions before 3.7, pass in universal_newlines=True instead of text=True


回答 17

如果要转换任何字节,而不仅仅是将字符串转换为字节:

with open("bytesfile", "rb") as infile:
    str = base64.b85encode(imageFile.read())

with open("bytesfile", "rb") as infile:
    str2 = json.dumps(list(infile.read()))

但是,这不是很有效。它将2 MB的图片变成9 MB。

If you want to convert any bytes, not just string converted to bytes:

with open("bytesfile", "rb") as infile:
    str = base64.b85encode(imageFile.read())

with open("bytesfile", "rb") as infile:
    str2 = json.dumps(list(infile.read()))

This is not very efficient, however. It will turn a 2 MB picture into 9 MB.


回答 18

尝试这个

bytes.fromhex('c3a9').decode('utf-8') 

try this

bytes.fromhex('c3a9').decode('utf-8') 

如何将列表分成大小均匀的块?

问题:如何将列表分成大小均匀的块?

我有一个任意长度的列表,我需要将其分成相等大小的块并对其进行操作。有一些明显的方法可以做到这一点,例如保留一个计数器和两个列表,当第二个列表填满时,将其添加到第一个列表中,并为第二轮数据清空第二个列表,但这可能会非常昂贵。

我想知道是否有人对任何长度的列表都有很好的解决方案,例如使用生成器。

我一直在寻找有用的东西,itertools但找不到任何明显有用的东西。可能已经错过了。

相关问题:遍历大块列表的最“ pythonic”方法是什么?

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

I was looking for something useful in itertools but I couldn’t find anything obviously useful. Might’ve missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?


回答 0

这是一个生成所需块的生成器:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

如果您使用的是Python 2,则应使用xrange()而不是range()

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

同样,您可以简单地使用列表理解而不是编写函数,尽管将这样的操作封装在命名函数中是个好主意,这样您的代码更易于理解。Python 3:

[lst[i:i + n] for i in range(0, len(lst), n)]

Python 2版本:

[lst[i:i + n] for i in xrange(0, len(lst), n)]

Here’s a generator that yields the chunks you want:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

If you’re using Python 2, you should use xrange() instead of range():

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

Also you can simply use list comprehension instead of writing a function, though it’s a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:

[lst[i:i + n] for i in range(0, len(lst), n)]

Python 2 version:

[lst[i:i + n] for i in xrange(0, len(lst), n)]

回答 1

如果您想要超级简单的东西:

def chunks(l, n):
    n = max(1, n)
    return (l[i:i+n] for i in range(0, len(l), n))

在Python 2.x中使用xrange()代替range()

If you want something super simple:

def chunks(l, n):
    n = max(1, n)
    return (l[i:i+n] for i in range(0, len(l), n))

Use xrange() instead of range() in the case of Python 2.x


回答 2

直接来自(旧的)Python文档(itertools的注意事项):

from itertools import izip, chain, repeat

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

JFSebastian建议的当前版本:

#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

我猜想Guido的时间机器可以工作了,可以工作了,可以工作了,可以再次工作。

这些解决方案之所以有效,是因为[iter(iterable)]*n(或早期版本中的等效项)创建了一个迭代器,n并在列表中重复了几次。izip_longest然后有效地执行“每个”迭代器的循环;因为这是相同的迭代器,所以每次此类调用都会对其进行高级处理,从而使每个此类zip-roundrobin生成一个元组n项。

Directly from the (old) Python documentation (recipes for itertools):

from itertools import izip, chain, repeat

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

The current version, as suggested by J.F.Sebastian:

#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

I guess Guido’s time machine works—worked—will work—will have worked—was working again.

These solutions work because [iter(iterable)]*n (or the equivalent in the earlier version) creates one iterator, repeated n times in the list. izip_longest then effectively performs a round-robin of “each” iterator; because this is the same iterator, it is advanced by each such call, resulting in each such zip-roundrobin generating one tuple of n items.


回答 3

我知道这有点陈旧,但没有人提到numpy.array_split

import numpy as np

lst = range(50)
np.array_split(lst, 5)
# [array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
#  array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),
#  array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29]),
#  array([30, 31, 32, 33, 34, 35, 36, 37, 38, 39]),
#  array([40, 41, 42, 43, 44, 45, 46, 47, 48, 49])]

I know this is kind of old but nobody yet mentioned numpy.array_split:

import numpy as np

lst = range(50)
np.array_split(lst, 5)
# [array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
#  array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),
#  array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29]),
#  array([30, 31, 32, 33, 34, 35, 36, 37, 38, 39]),
#  array([40, 41, 42, 43, 44, 45, 46, 47, 48, 49])]

回答 4

令我惊讶的是,没有人想到使用iter二元形式

from itertools import islice

def chunk(it, size):
    it = iter(it)
    return iter(lambda: tuple(islice(it, size)), ())

演示:

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]

这可以与任何迭代一起工作,并产生延迟输出。它返回元组而不是迭代器,但是我认为它仍然具有一定的优雅。它也不会填充;如果您想进行填充,则只需对上述内容进行简单的修改即可:

from itertools import islice, chain, repeat

def chunk_pad(it, size, padval=None):
    it = chain(iter(it), repeat(padval))
    return iter(lambda: tuple(islice(it, size)), (padval,) * size)

演示:

>>> list(chunk_pad(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk_pad(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]

izip_longest基于解决方案的解决方案一样,以上内容总是可以解决的。据我所知,没有可选的填充函数的单行或两行itertools配方。通过结合以上两种方法,这一方法非常接近:

_no_padding = object()

def chunk(it, size, padval=_no_padding):
    if padval == _no_padding:
        it = iter(it)
        sentinel = ()
    else:
        it = chain(iter(it), repeat(padval))
        sentinel = (padval,) * size
    return iter(lambda: tuple(islice(it, size)), sentinel)

演示:

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
>>> list(chunk(range(14), 3, None))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]

我认为这是最短的分块器,建议提供可选的填充。

饰演Tomasz Gandor 观察到的,如果两个填充分块器遇到一长串填充值,它们将意外停止。这是一个可以合理解决该问题的最终变体:

_no_padding = object()
def chunk(it, size, padval=_no_padding):
    it = iter(it)
    chunker = iter(lambda: tuple(islice(it, size)), ())
    if padval == _no_padding:
        yield from chunker
    else:
        for ch in chunker:
            yield ch if len(ch) == size else ch + (padval,) * (size - len(ch))

演示:

>>> list(chunk([1, 2, (), (), 5], 2))
[(1, 2), ((), ()), (5,)]
>>> list(chunk([1, 2, None, None, 5], 2, None))
[(1, 2), (None, None), (5, None)]

I’m surprised nobody has thought of using iter‘s two-argument form:

from itertools import islice

def chunk(it, size):
    it = iter(it)
    return iter(lambda: tuple(islice(it, size)), ())

Demo:

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]

This works with any iterable and produces output lazily. It returns tuples rather than iterators, but I think it has a certain elegance nonetheless. It also doesn’t pad; if you want padding, a simple variation on the above will suffice:

from itertools import islice, chain, repeat

def chunk_pad(it, size, padval=None):
    it = chain(iter(it), repeat(padval))
    return iter(lambda: tuple(islice(it, size)), (padval,) * size)

Demo:

>>> list(chunk_pad(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk_pad(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]

Like the izip_longest-based solutions, the above always pads. As far as I know, there’s no one- or two-line itertools recipe for a function that optionally pads. By combining the above two approaches, this one comes pretty close:

_no_padding = object()

def chunk(it, size, padval=_no_padding):
    if padval == _no_padding:
        it = iter(it)
        sentinel = ()
    else:
        it = chain(iter(it), repeat(padval))
        sentinel = (padval,) * size
    return iter(lambda: tuple(islice(it, size)), sentinel)

Demo:

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
>>> list(chunk(range(14), 3, None))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]

I believe this is the shortest chunker proposed that offers optional padding.

As Tomasz Gandor observed, the two padding chunkers will stop unexpectedly if they encounter a long sequence of pad values. Here’s a final variation that works around that problem in a reasonable way:

_no_padding = object()
def chunk(it, size, padval=_no_padding):
    it = iter(it)
    chunker = iter(lambda: tuple(islice(it, size)), ())
    if padval == _no_padding:
        yield from chunker
    else:
        for ch in chunker:
            yield ch if len(ch) == size else ch + (padval,) * (size - len(ch))

Demo:

>>> list(chunk([1, 2, (), (), 5], 2))
[(1, 2), ((), ()), (5,)]
>>> list(chunk([1, 2, None, None, 5], 2, None))
[(1, 2), (None, None), (5, None)]

回答 5

这是处理任意可迭代对象的生成器:

def split_seq(iterable, size):
    it = iter(iterable)
    item = list(itertools.islice(it, size))
    while item:
        yield item
        item = list(itertools.islice(it, size))

例:

>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

Here is a generator that work on arbitrary iterables:

def split_seq(iterable, size):
    it = iter(iterable)
    item = list(itertools.islice(it, size))
    while item:
        yield item
        item = list(itertools.islice(it, size))

Example:

>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

回答 6

def chunk(input, size):
    return map(None, *([iter(input)] * size))
def chunk(input, size):
    return map(None, *([iter(input)] * size))

回答 7

简单而优雅

l = range(1, 1000)
print [l[x:x+10] for x in xrange(0, len(l), 10)]

或者,如果您喜欢:

def chunks(l, n): return [l[x: x+n] for x in xrange(0, len(l), n)]
chunks(l, 10)

Simple yet elegant

l = range(1, 1000)
print [l[x:x+10] for x in xrange(0, len(l), 10)]

or if you prefer:

def chunks(l, n): return [l[x: x+n] for x in xrange(0, len(l), n)]
chunks(l, 10)

回答 8

批判其他答案在这里:

这些答案都不是均匀大小的块,它们都在末尾留下欠缺的块,因此它们并不完全平衡。如果您使用这些功能来分配工作,那么您就建立了一个前景可能比其他事情早完成的前景,因此在其他人继续努力的同时,它什么也没做。

例如,当前的最佳答案以:

[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]

我只是讨厌最后那个矮子!

其他人,例如list(grouper(3, xrange(7)))和,chunk(xrange(7), 3)都返回:[(0, 1, 2), (3, 4, 5), (6, None, None)]。的None只是填充,在我看来相当不雅。他们没有将可迭代对象均匀地分块。

为什么我们不能更好地划分这些?

我的解决方案

这是一个平衡的解决方案,它是根据我在生产环境中使用过的函数改编而成的(Python 3中的Note替换xrangerange):

def baskets_from(items, maxbaskets=25):
    baskets = [[] for _ in xrange(maxbaskets)] # in Python 3 use range
    for i, item in enumerate(items):
        baskets[i % maxbaskets].append(item)
    return filter(None, baskets) 

我创建了一个生成器,如果将其放入列表中,它的功能也相同:

def iter_baskets_from(items, maxbaskets=3):
    '''generates evenly balanced baskets from indexable iterable'''
    item_count = len(items)
    baskets = min(item_count, maxbaskets)
    for x_i in xrange(baskets):
        yield [items[y_i] for y_i in xrange(x_i, item_count, baskets)]

最后,由于我看到上述所有函数均按连续顺序返回元素(如给出的那样):

def iter_baskets_contiguous(items, maxbaskets=3, item_count=None):
    '''
    generates balanced baskets from iterable, contiguous contents
    provide item_count if providing a iterator that doesn't support len()
    '''
    item_count = item_count or len(items)
    baskets = min(item_count, maxbaskets)
    items = iter(items)
    floor = item_count // baskets 
    ceiling = floor + 1
    stepdown = item_count % baskets
    for x_i in xrange(baskets):
        length = ceiling if x_i < stepdown else floor
        yield [items.next() for _ in xrange(length)]

输出量

要测试它们:

print(baskets_from(xrange(6), 8))
print(list(iter_baskets_from(xrange(6), 8)))
print(list(iter_baskets_contiguous(xrange(6), 8)))
print(baskets_from(xrange(22), 8))
print(list(iter_baskets_from(xrange(22), 8)))
print(list(iter_baskets_contiguous(xrange(22), 8)))
print(baskets_from('ABCDEFG', 3))
print(list(iter_baskets_from('ABCDEFG', 3)))
print(list(iter_baskets_contiguous('ABCDEFG', 3)))
print(baskets_from(xrange(26), 5))
print(list(iter_baskets_from(xrange(26), 5)))
print(list(iter_baskets_contiguous(xrange(26), 5)))

打印出:

[[0], [1], [2], [3], [4], [5]]
[[0], [1], [2], [3], [4], [5]]
[[0], [1], [2], [3], [4], [5]]
[[0, 8, 16], [1, 9, 17], [2, 10, 18], [3, 11, 19], [4, 12, 20], [5, 13, 21], [6, 14], [7, 15]]
[[0, 8, 16], [1, 9, 17], [2, 10, 18], [3, 11, 19], [4, 12, 20], [5, 13, 21], [6, 14], [7, 15]]
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13, 14], [15, 16, 17], [18, 19], [20, 21]]
[['A', 'D', 'G'], ['B', 'E'], ['C', 'F']]
[['A', 'D', 'G'], ['B', 'E'], ['C', 'F']]
[['A', 'B', 'C'], ['D', 'E'], ['F', 'G']]
[[0, 5, 10, 15, 20, 25], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]
[[0, 5, 10, 15, 20, 25], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]
[[0, 1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]]

请注意,连续生成器以与其他两个相同的长度模式提供块,但是所有项都是有序的,并且它们被均匀地划分为一个可以划分一列离散元素的形式。

Critique of other answers here:

None of these answers are evenly sized chunks, they all leave a runt chunk at the end, so they’re not completely balanced. If you were using these functions to distribute work, you’ve built-in the prospect of one likely finishing well before the others, so it would sit around doing nothing while the others continued working hard.

For example, the current top answer ends with:

[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]

I just hate that runt at the end!

Others, like list(grouper(3, xrange(7))), and chunk(xrange(7), 3) both return: [(0, 1, 2), (3, 4, 5), (6, None, None)]. The None‘s are just padding, and rather inelegant in my opinion. They are NOT evenly chunking the iterables.

Why can’t we divide these better?

My Solution(s)

Here’s a balanced solution, adapted from a function I’ve used in production (Note in Python 3 to replace xrange with range):

def baskets_from(items, maxbaskets=25):
    baskets = [[] for _ in xrange(maxbaskets)] # in Python 3 use range
    for i, item in enumerate(items):
        baskets[i % maxbaskets].append(item)
    return filter(None, baskets) 

And I created a generator that does the same if you put it into a list:

def iter_baskets_from(items, maxbaskets=3):
    '''generates evenly balanced baskets from indexable iterable'''
    item_count = len(items)
    baskets = min(item_count, maxbaskets)
    for x_i in xrange(baskets):
        yield [items[y_i] for y_i in xrange(x_i, item_count, baskets)]

And finally, since I see that all of the above functions return elements in a contiguous order (as they were given):

def iter_baskets_contiguous(items, maxbaskets=3, item_count=None):
    '''
    generates balanced baskets from iterable, contiguous contents
    provide item_count if providing a iterator that doesn't support len()
    '''
    item_count = item_count or len(items)
    baskets = min(item_count, maxbaskets)
    items = iter(items)
    floor = item_count // baskets 
    ceiling = floor + 1
    stepdown = item_count % baskets
    for x_i in xrange(baskets):
        length = ceiling if x_i < stepdown else floor
        yield [items.next() for _ in xrange(length)]

Output

To test them out:

print(baskets_from(xrange(6), 8))
print(list(iter_baskets_from(xrange(6), 8)))
print(list(iter_baskets_contiguous(xrange(6), 8)))
print(baskets_from(xrange(22), 8))
print(list(iter_baskets_from(xrange(22), 8)))
print(list(iter_baskets_contiguous(xrange(22), 8)))
print(baskets_from('ABCDEFG', 3))
print(list(iter_baskets_from('ABCDEFG', 3)))
print(list(iter_baskets_contiguous('ABCDEFG', 3)))
print(baskets_from(xrange(26), 5))
print(list(iter_baskets_from(xrange(26), 5)))
print(list(iter_baskets_contiguous(xrange(26), 5)))

Which prints out:

[[0], [1], [2], [3], [4], [5]]
[[0], [1], [2], [3], [4], [5]]
[[0], [1], [2], [3], [4], [5]]
[[0, 8, 16], [1, 9, 17], [2, 10, 18], [3, 11, 19], [4, 12, 20], [5, 13, 21], [6, 14], [7, 15]]
[[0, 8, 16], [1, 9, 17], [2, 10, 18], [3, 11, 19], [4, 12, 20], [5, 13, 21], [6, 14], [7, 15]]
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13, 14], [15, 16, 17], [18, 19], [20, 21]]
[['A', 'D', 'G'], ['B', 'E'], ['C', 'F']]
[['A', 'D', 'G'], ['B', 'E'], ['C', 'F']]
[['A', 'B', 'C'], ['D', 'E'], ['F', 'G']]
[[0, 5, 10, 15, 20, 25], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]
[[0, 5, 10, 15, 20, 25], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]
[[0, 1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]]

Notice that the contiguous generator provide chunks in the same length patterns as the other two, but the items are all in order, and they are as evenly divided as one may divide a list of discrete elements.


回答 9

我在其中看到了最棒的Python式答案 这个问题重复部分中,:

from itertools import zip_longest

a = range(1, 16)
i = iter(a)
r = list(zip_longest(i, i, i))
>>> print(r)
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)]

您可以为任何n个创建n个元组。如果为a = range(1, 15),则结果将为:

[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, None)]

如果列表平均分配,则可以替换zip_longestzip,否则三元组(13, 14, None)将丢失。上面使用了Python 3。对于Python 2,请使用izip_longest

I saw the most awesome Python-ish answer in a duplicate of this question:

from itertools import zip_longest

a = range(1, 16)
i = iter(a)
r = list(zip_longest(i, i, i))
>>> print(r)
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)]

You can create n-tuple for any n. If a = range(1, 15), then the result will be:

[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, None)]

If the list is divided evenly, then you can replace zip_longest with zip, otherwise the triplet (13, 14, None) would be lost. Python 3 is used above. For Python 2, use izip_longest.


回答 10

如果您知道列表大小:

def SplitList(mylist, chunk_size):
    return [mylist[offs:offs+chunk_size] for offs in range(0, len(mylist), chunk_size)]

如果不这样做(一个迭代器):

def IterChunks(sequence, chunk_size):
    res = []
    for item in sequence:
        res.append(item)
        if len(res) >= chunk_size:
            yield res
            res = []
    if res:
        yield res  # yield the last, incomplete, portion

在后一种情况下,如果可以确定序列始终包含给定大小的所有块(即不存在不完整的最后一个块),则可以用更漂亮的方式来重新措词。

If you know list size:

def SplitList(mylist, chunk_size):
    return [mylist[offs:offs+chunk_size] for offs in range(0, len(mylist), chunk_size)]

If you don’t (an iterator):

def IterChunks(sequence, chunk_size):
    res = []
    for item in sequence:
        res.append(item)
        if len(res) >= chunk_size:
            yield res
            res = []
    if res:
        yield res  # yield the last, incomplete, portion

In the latter case, it can be rephrased in a more beautiful way if you can be sure that the sequence always contains a whole number of chunks of given size (i.e. there is no incomplete last chunk).


回答 11

图尔茨库具有partition此功能:

from toolz.itertoolz.core import partition

list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]

The toolz library has the partition function for this:

from toolz.itertoolz.core import partition

list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]

回答 12

例如,如果块大小为3,则可以执行以下操作:

zip(*[iterable[i::3] for i in range(3)]) 

来源:http//code.activestate.com/recipes/303060-group-a-list-into-sequential-n-tuples/

当我可以输入的块大小为固定数字(例如“ 3”)并且永远不会更改时,我将使用此选项。

If you had a chunk size of 3 for example, you could do:

zip(*[iterable[i::3] for i in range(3)]) 

source: http://code.activestate.com/recipes/303060-group-a-list-into-sequential-n-tuples/

I would use this when my chunk size is fixed number I can type, e.g. ‘3’, and would never change.


回答 13

我非常喜欢tzot和JFSebastian提出的Python文档版本,但是它有两个缺点:

  • 这不是很明确
  • 我通常不希望在最后一块填充值

我在代码中经常使用此代码:

from itertools import islice

def chunks(n, iterable):
    iterable = iter(iterable)
    while True:
        yield tuple(islice(iterable, n)) or iterable.next()

更新:惰性块版本:

from itertools import chain, islice

def chunks(n, iterable):
   iterable = iter(iterable)
   while True:
       yield chain([next(iterable)], islice(iterable, n-1))

I like the Python doc’s version proposed by tzot and J.F.Sebastian a lot, but it has two shortcomings:

  • it is not very explicit
  • I usually don’t want a fill value in the last chunk

I’m using this one a lot in my code:

from itertools import islice

def chunks(n, iterable):
    iterable = iter(iterable)
    while True:
        yield tuple(islice(iterable, n)) or iterable.next()

UPDATE: A lazy chunks version:

from itertools import chain, islice

def chunks(n, iterable):
   iterable = iter(iterable)
   while True:
       yield chain([next(iterable)], islice(iterable, n-1))

回答 14

[AA[i:i+SS] for i in range(len(AA))[::SS]]

其中AA是数组,SS是块大小。例如:

>>> AA=range(10,21);SS=3
>>> [AA[i:i+SS] for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]
# or [range(10, 13), range(13, 16), range(16, 19), range(19, 21)] in py3
[AA[i:i+SS] for i in range(len(AA))[::SS]]

Where AA is array, SS is chunk size. For example:

>>> AA=range(10,21);SS=3
>>> [AA[i:i+SS] for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]
# or [range(10, 13), range(13, 16), range(16, 19), range(19, 21)] in py3

回答 15

我很好奇不同方法的性能,这里是:

在Python 3.5.1上测试

import time
batch_size = 7
arr_len = 298937

#---------slice-------------

print("\r\nslice")
start = time.time()
arr = [i for i in range(0, arr_len)]
while True:
    if not arr:
        break

    tmp = arr[0:batch_size]
    arr = arr[batch_size:-1]
print(time.time() - start)

#-----------index-----------

print("\r\nindex")
arr = [i for i in range(0, arr_len)]
start = time.time()
for i in range(0, round(len(arr) / batch_size + 1)):
    tmp = arr[batch_size * i : batch_size * (i + 1)]
print(time.time() - start)

#----------batches 1------------

def batch(iterable, n=1):
    l = len(iterable)
    for ndx in range(0, l, n):
        yield iterable[ndx:min(ndx + n, l)]

print("\r\nbatches 1")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
    tmp = x
print(time.time() - start)

#----------batches 2------------

from itertools import islice, chain

def batch(iterable, size):
    sourceiter = iter(iterable)
    while True:
        batchiter = islice(sourceiter, size)
        yield chain([next(batchiter)], batchiter)


print("\r\nbatches 2")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
    tmp = x
print(time.time() - start)

#---------chunks-------------
def chunks(l, n):
    """Yield successive n-sized chunks from l."""
    for i in range(0, len(l), n):
        yield l[i:i + n]
print("\r\nchunks")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in chunks(arr, batch_size):
    tmp = x
print(time.time() - start)

#-----------grouper-----------

from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

def grouper(iterable, n, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

arr = [i for i in range(0, arr_len)]
print("\r\ngrouper")
start = time.time()
for x in grouper(arr, batch_size):
    tmp = x
print(time.time() - start)

结果:

slice
31.18285083770752

index
0.02184295654296875

batches 1
0.03503894805908203

batches 2
0.22681021690368652

chunks
0.019841909408569336

grouper
0.006506919860839844

I was curious about the performance of different approaches and here it is:

Tested on Python 3.5.1

import time
batch_size = 7
arr_len = 298937

#---------slice-------------

print("\r\nslice")
start = time.time()
arr = [i for i in range(0, arr_len)]
while True:
    if not arr:
        break

    tmp = arr[0:batch_size]
    arr = arr[batch_size:-1]
print(time.time() - start)

#-----------index-----------

print("\r\nindex")
arr = [i for i in range(0, arr_len)]
start = time.time()
for i in range(0, round(len(arr) / batch_size + 1)):
    tmp = arr[batch_size * i : batch_size * (i + 1)]
print(time.time() - start)

#----------batches 1------------

def batch(iterable, n=1):
    l = len(iterable)
    for ndx in range(0, l, n):
        yield iterable[ndx:min(ndx + n, l)]

print("\r\nbatches 1")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
    tmp = x
print(time.time() - start)

#----------batches 2------------

from itertools import islice, chain

def batch(iterable, size):
    sourceiter = iter(iterable)
    while True:
        batchiter = islice(sourceiter, size)
        yield chain([next(batchiter)], batchiter)


print("\r\nbatches 2")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
    tmp = x
print(time.time() - start)

#---------chunks-------------
def chunks(l, n):
    """Yield successive n-sized chunks from l."""
    for i in range(0, len(l), n):
        yield l[i:i + n]
print("\r\nchunks")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in chunks(arr, batch_size):
    tmp = x
print(time.time() - start)

#-----------grouper-----------

from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

def grouper(iterable, n, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

arr = [i for i in range(0, arr_len)]
print("\r\ngrouper")
start = time.time()
for x in grouper(arr, batch_size):
    tmp = x
print(time.time() - start)

Results:

slice
31.18285083770752

index
0.02184295654296875

batches 1
0.03503894805908203

batches 2
0.22681021690368652

chunks
0.019841909408569336

grouper
0.006506919860839844

回答 16

码:

def split_list(the_list, chunk_size):
    result_list = []
    while the_list:
        result_list.append(the_list[:chunk_size])
        the_list = the_list[chunk_size:]
    return result_list

a_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

print split_list(a_list, 3)

结果:

[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]

code:

def split_list(the_list, chunk_size):
    result_list = []
    while the_list:
        result_list.append(the_list[:chunk_size])
        the_list = the_list[chunk_size:]
    return result_list

a_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

print split_list(a_list, 3)

result:

[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]

回答 17

您还可以get_chunksutilspie库函数用作:

>>> from utilspie import iterutils
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]

>>> list(iterutils.get_chunks(a, 5))
[[1, 2, 3, 4, 5], [6, 7, 8, 9]]

您可以utilspie通过pip 安装:

sudo pip install utilspie

免责声明:我是utilspie library 的创建者

You may also use get_chunks function of utilspie library as:

>>> from utilspie import iterutils
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]

>>> list(iterutils.get_chunks(a, 5))
[[1, 2, 3, 4, 5], [6, 7, 8, 9]]

You can install utilspie via pip:

sudo pip install utilspie

Disclaimer: I am the creator of utilspie library.


回答 18

在这一点上,我认为我们需要一个递归生成器,以防万一…

在python 2:

def chunks(li, n):
    if li == []:
        return
    yield li[:n]
    for e in chunks(li[n:], n):
        yield e

在python 3:

def chunks(li, n):
    if li == []:
        return
    yield li[:n]
    yield from chunks(li[n:], n)

同样,在外星人大规模入侵的情况下,经过修饰的递归生成器可能会派上用场:

def dec(gen):
    def new_gen(li, n):
        for e in gen(li, n):
            if e == []:
                return
            yield e
    return new_gen

@dec
def chunks(li, n):
    yield li[:n]
    for e in chunks(li[n:], n):
        yield e

At this point, I think we need a recursive generator, just in case…

In python 2:

def chunks(li, n):
    if li == []:
        return
    yield li[:n]
    for e in chunks(li[n:], n):
        yield e

In python 3:

def chunks(li, n):
    if li == []:
        return
    yield li[:n]
    yield from chunks(li[n:], n)

Also, in case of massive Alien invasion, a decorated recursive generator might become handy:

def dec(gen):
    def new_gen(li, n):
        for e in gen(li, n):
            if e == []:
                return
            yield e
    return new_gen

@dec
def chunks(li, n):
    yield li[:n]
    for e in chunks(li[n:], n):
        yield e

回答 19

使用Python 3.8中的赋值表达式,它变得非常不错:

import itertools

def batch(iterable, size):
    it = iter(iterable)
    while item := list(itertools.islice(it, size)):
        yield item

这适用于任意迭代,而不仅仅是列表。

>>> import pprint
>>> pprint.pprint(list(batch(range(75), 10)))
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

With Assignment Expressions in Python 3.8 it becomes quite nice:

import itertools

def batch(iterable, size):
    it = iter(iterable)
    while item := list(itertools.islice(it, size)):
        yield item

This works on an arbitrary iterable, not just a list.

>>> import pprint
>>> pprint.pprint(list(batch(range(75), 10)))
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

回答 20

呵呵,单行版

In [48]: chunk = lambda ulist, step:  map(lambda i: ulist[i:i+step],  xrange(0, len(ulist), step))

In [49]: chunk(range(1,100), 10)
Out[49]: 
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
 [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
 [21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
 [31, 32, 33, 34, 35, 36, 37, 38, 39, 40],
 [41, 42, 43, 44, 45, 46, 47, 48, 49, 50],
 [51, 52, 53, 54, 55, 56, 57, 58, 59, 60],
 [61, 62, 63, 64, 65, 66, 67, 68, 69, 70],
 [71, 72, 73, 74, 75, 76, 77, 78, 79, 80],
 [81, 82, 83, 84, 85, 86, 87, 88, 89, 90],
 [91, 92, 93, 94, 95, 96, 97, 98, 99]]

heh, one line version

In [48]: chunk = lambda ulist, step:  map(lambda i: ulist[i:i+step],  xrange(0, len(ulist), step))

In [49]: chunk(range(1,100), 10)
Out[49]: 
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
 [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
 [21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
 [31, 32, 33, 34, 35, 36, 37, 38, 39, 40],
 [41, 42, 43, 44, 45, 46, 47, 48, 49, 50],
 [51, 52, 53, 54, 55, 56, 57, 58, 59, 60],
 [61, 62, 63, 64, 65, 66, 67, 68, 69, 70],
 [71, 72, 73, 74, 75, 76, 77, 78, 79, 80],
 [81, 82, 83, 84, 85, 86, 87, 88, 89, 90],
 [91, 92, 93, 94, 95, 96, 97, 98, 99]]

回答 21

def split_seq(seq, num_pieces):
    start = 0
    for i in xrange(num_pieces):
        stop = start + len(seq[i::num_pieces])
        yield seq[start:stop]
        start = stop

用法:

seq = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

for seq in split_seq(seq, 3):
    print seq
def split_seq(seq, num_pieces):
    start = 0
    for i in xrange(num_pieces):
        stop = start + len(seq[i::num_pieces])
        yield seq[start:stop]
        start = stop

usage:

seq = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

for seq in split_seq(seq, 3):
    print seq

回答 22

另一个更明确的版本。

def chunkList(initialList, chunkSize):
    """
    This function chunks a list into sub lists 
    that have a length equals to chunkSize.

    Example:
    lst = [3, 4, 9, 7, 1, 1, 2, 3]
    print(chunkList(lst, 3)) 
    returns
    [[3, 4, 9], [7, 1, 1], [2, 3]]
    """
    finalList = []
    for i in range(0, len(initialList), chunkSize):
        finalList.append(initialList[i:i+chunkSize])
    return finalList

Another more explicit version.

def chunkList(initialList, chunkSize):
    """
    This function chunks a list into sub lists 
    that have a length equals to chunkSize.

    Example:
    lst = [3, 4, 9, 7, 1, 1, 2, 3]
    print(chunkList(lst, 3)) 
    returns
    [[3, 4, 9], [7, 1, 1], [2, 3]]
    """
    finalList = []
    for i in range(0, len(initialList), chunkSize):
        finalList.append(initialList[i:i+chunkSize])
    return finalList

回答 23

在不调用len()的情况下,该方法非常适合大型列表:

def splitter(l, n):
    i = 0
    chunk = l[:n]
    while chunk:
        yield chunk
        i += n
        chunk = l[i:i+n]

这是针对可迭代对象的:

def isplitter(l, n):
    l = iter(l)
    chunk = list(islice(l, n))
    while chunk:
        yield chunk
        chunk = list(islice(l, n))

以上功能的味道:

def isplitter2(l, n):
    return takewhile(bool,
                     (tuple(islice(start, n))
                            for start in repeat(iter(l))))

要么:

def chunks_gen_sentinel(n, seq):
    continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
    return iter(imap(tuple, continuous_slices).next,())

要么:

def chunks_gen_filter(n, seq):
    continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
    return takewhile(bool,imap(tuple, continuous_slices))

Without calling len() which is good for large lists:

def splitter(l, n):
    i = 0
    chunk = l[:n]
    while chunk:
        yield chunk
        i += n
        chunk = l[i:i+n]

And this is for iterables:

def isplitter(l, n):
    l = iter(l)
    chunk = list(islice(l, n))
    while chunk:
        yield chunk
        chunk = list(islice(l, n))

The functional flavour of the above:

def isplitter2(l, n):
    return takewhile(bool,
                     (tuple(islice(start, n))
                            for start in repeat(iter(l))))

OR:

def chunks_gen_sentinel(n, seq):
    continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
    return iter(imap(tuple, continuous_slices).next,())

OR:

def chunks_gen_filter(n, seq):
    continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
    return takewhile(bool,imap(tuple, continuous_slices))

回答 24

以下是其他方法的列表:

给定

import itertools as it
import collections as ct

import more_itertools as mit


iterable = range(11)
n = 3

标准图书馆

list(it.zip_longest(*[iter(iterable)] * n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

d = {}
for i, x in enumerate(iterable):
    d.setdefault(i//n, []).append(x)

list(d.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

dd = ct.defaultdict(list)
for i, x in enumerate(iterable):
    dd[i//n].append(x)

list(dd.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

more_itertools+

list(mit.chunked(iterable, n))
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

list(mit.sliced(iterable, n))
# [range(0, 3), range(3, 6), range(6, 9), range(9, 11)]

list(mit.grouper(n, iterable))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

list(mit.windowed(iterable, len(iterable)//n, step=n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

参考文献

+一个实现itertools配方等的第三方库。> pip install more_itertools

Here is a list of additional approaches:

Given

import itertools as it
import collections as ct

import more_itertools as mit


iterable = range(11)
n = 3

Code

The Standard Library

list(it.zip_longest(*[iter(iterable)] * n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

d = {}
for i, x in enumerate(iterable):
    d.setdefault(i//n, []).append(x)

list(d.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

dd = ct.defaultdict(list)
for i, x in enumerate(iterable):
    dd[i//n].append(x)

list(dd.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

more_itertools+

list(mit.chunked(iterable, n))
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

list(mit.sliced(iterable, n))
# [range(0, 3), range(3, 6), range(6, 9), range(9, 11)]

list(mit.grouper(n, iterable))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

list(mit.windowed(iterable, len(iterable)//n, step=n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

References

+ A third-party library that implements itertools recipes and more. > pip install more_itertools


回答 25

看到这个参考

>>> orange = range(1, 1001)
>>> otuples = list( zip(*[iter(orange)]*10))
>>> print(otuples)
[(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), ... (991, 992, 993, 994, 995, 996, 997, 998, 999, 1000)]
>>> olist = [list(i) for i in otuples]
>>> print(olist)
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], ..., [991, 992, 993, 994, 995, 996, 997, 998, 999, 1000]]
>>> 

Python3

See this reference

>>> orange = range(1, 1001)
>>> otuples = list( zip(*[iter(orange)]*10))
>>> print(otuples)
[(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), ... (991, 992, 993, 994, 995, 996, 997, 998, 999, 1000)]
>>> olist = [list(i) for i in otuples]
>>> print(olist)
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], ..., [991, 992, 993, 994, 995, 996, 997, 998, 999, 1000]]
>>> 

Python3


回答 26

由于这里的每个人都在谈论迭代器。boltons为此有一个完美的方法,称为iterutils.chunked_iter

from boltons import iterutils

list(iterutils.chunked_iter(list(range(50)), 11))

输出:

[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
 [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21],
 [22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32],
 [33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43],
 [44, 45, 46, 47, 48, 49]]

但是,如果您不想对内存存留怜悯,可以使用old-way并通过将数据存储list在第一位iterutils.chunked

Since everybody here talking about iterators. boltons has perfect method for that, called iterutils.chunked_iter.

from boltons import iterutils

list(iterutils.chunked_iter(list(range(50)), 11))

Output:

[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
 [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21],
 [22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32],
 [33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43],
 [44, 45, 46, 47, 48, 49]]

But if you don’t want to be mercy on memory, you can use old-way and store the full list in the first place with iterutils.chunked.


回答 27

另一种解决方案

def make_chunks(data, chunk_size): 
    while data:
        chunk, data = data[:chunk_size], data[chunk_size:]
        yield chunk

>>> for chunk in make_chunks([1, 2, 3, 4, 5, 6, 7], 2):
...     print chunk
... 
[1, 2]
[3, 4]
[5, 6]
[7]
>>> 

One more solution

def make_chunks(data, chunk_size): 
    while data:
        chunk, data = data[:chunk_size], data[chunk_size:]
        yield chunk

>>> for chunk in make_chunks([1, 2, 3, 4, 5, 6, 7], 2):
...     print chunk
... 
[1, 2]
[3, 4]
[5, 6]
[7]
>>> 

回答 28

def chunks(iterable,n):
    """assumes n is an integer>0
    """
    iterable=iter(iterable)
    while True:
        result=[]
        for i in range(n):
            try:
                a=next(iterable)
            except StopIteration:
                break
            else:
                result.append(a)
        if result:
            yield result
        else:
            break

g1=(i*i for i in range(10))
g2=chunks(g1,3)
print g2
'<generator object chunks at 0x0337B9B8>'
print list(g2)
'[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81]]'
def chunks(iterable,n):
    """assumes n is an integer>0
    """
    iterable=iter(iterable)
    while True:
        result=[]
        for i in range(n):
            try:
                a=next(iterable)
            except StopIteration:
                break
            else:
                result.append(a)
        if result:
            yield result
        else:
            break

g1=(i*i for i in range(10))
g2=chunks(g1,3)
print g2
'<generator object chunks at 0x0337B9B8>'
print list(g2)
'[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81]]'

回答 29

考虑使用matplotlib.cbook片段

例如:

import matplotlib.cbook as cbook
segments = cbook.pieces(np.arange(20), 3)
for s in segments:
     print s

Consider using matplotlib.cbook pieces

for example:

import matplotlib.cbook as cbook
segments = cbook.pieces(np.arange(20), 3)
for s in segments:
     print s

在Python中手动引发(抛出)异常

问题:在Python中手动引发(抛出)异常

如何在Python中引发异常,以便以后可以通过except块将其捕获?

How can I raise an exception in Python so that it can later be caught via an except block?


回答 0

如何在Python中手动引发/引发异常?

使用在语义上适合您的问题的最特定的Exception构造函数

在您的消息中要具体,例如:

raise ValueError('A very specific bad thing happened.')

不要引发通用异常

避免提出泛型Exception。要捕获它,您必须捕获将其子类化的所有其他更具体的异常。

问题1:隐藏错误

raise Exception('I know Python!') # Don't! If you catch, likely to hide bugs.

例如:

def demo_bad_catch():
    try:
        raise ValueError('Represents a hidden bug, do not catch this')
        raise Exception('This is the exception you expect to handle')
    except Exception as error:
        print('Caught this error: ' + repr(error))

>>> demo_bad_catch()
Caught this error: ValueError('Represents a hidden bug, do not catch this',)

问题2:无法抓住

而且更具体的捕获不会捕获一般异常:

def demo_no_catch():
    try:
        raise Exception('general exceptions not caught by specific handling')
    except ValueError as e:
        print('we will not catch exception: Exception')


>>> demo_no_catch()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in demo_no_catch
Exception: general exceptions not caught by specific handling

最佳做法:raise声明

而是使用在语义上适合您的issue的最特定的Exception构造函数

raise ValueError('A very specific bad thing happened')

这也方便地允许将任意数量的参数传递给构造函数:

raise ValueError('A very specific bad thing happened', 'foo', 'bar', 'baz') 

这些参数由对象args上的属性访问Exception。例如:

try:
    some_code_that_may_raise_our_value_error()
except ValueError as err:
    print(err.args)

版画

('message', 'foo', 'bar', 'baz')    

在Python 2.5中,message添加了一个实际属性,以BaseException鼓励用户继承Exceptions子类并停止使用args,但是args 的引入message和最初的弃用已被收回

最佳做法:except条款

例如,在except子句中时,您可能想要记录发生了特定类型的错误,然后重新引发。保留堆栈跟踪时执行此操作的最佳方法是使用裸机抬高语句。例如:

logger = logging.getLogger(__name__)

try:
    do_something_in_app_that_breaks_easily()
except AppError as error:
    logger.error(error)
    raise                 # just this!
    # raise AppError      # Don't do this, you'll lose the stack trace!

不要修改您的错误…但是如果您坚持的话。

您可以使用来保留stacktrace(和错误值)sys.exc_info(),但这更容易出错,并且在Python 2和3之间存在兼容性问题,建议使用裸机raise重新引发。

解释- sys.exc_info()返回类型,值和回溯。

type, value, traceback = sys.exc_info()

这是Python 2中的语法-请注意,这与Python 3不兼容:

    raise AppError, error, sys.exc_info()[2] # avoid this.
    # Equivalently, as error *is* the second object:
    raise sys.exc_info()[0], sys.exc_info()[1], sys.exc_info()[2]

如果愿意,您可以修改新加薪后的情况-例如args,为实例设置新值:

def error():
    raise ValueError('oops!')

def catch_error_modify_message():
    try:
        error()
    except ValueError:
        error_type, error_instance, traceback = sys.exc_info()
        error_instance.args = (error_instance.args[0] + ' <modification>',)
        raise error_type, error_instance, traceback

并且我们在修改args时保留了整个回溯。请注意,这不是最佳做法,并且在Python 3中是无效的语法(使得保持兼容性变得更加困难)。

>>> catch_error_modify_message()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in catch_error_modify_message
  File "<stdin>", line 2, in error
ValueError: oops! <modification>

Python 3中

    raise error.with_traceback(sys.exc_info()[2])

同样:避免手动操作回溯。它效率较低,更容易出错。而且,如果您正在使用线程,sys.exc_info甚至可能会得到错误的回溯(特别是如果您对控制流使用异常处理,我个人倾向于避免这种情况。)

Python 3,异常链接

在Python 3中,您可以链接异常,以保留回溯:

    raise RuntimeError('specific message') from error

意识到:

  • 确实允许更改引发的错误类型,并且
  • 这与Python 2 兼容。

不推荐使用的方法:

这些可以轻松隐藏甚至进入生产代码。您想提出一个exceptions,而这样做会引发一个exceptions,但不是一个预期的exceptions

在Python 2中有效,但在Python 3中无效

raise ValueError, 'message' # Don't do this, it's deprecated!

在更旧的Python版本(2.4及更低版本)中有效,您仍然可以看到有人在引发字符串:

raise 'message' # really really wrong. don't do this.

在所有现代版本中,这实际上会引发一个TypeError,因为您没有引发一个BaseException类型。如果您没有检查正确的exceptions情况,并且没有知道此问题的审阅者,那么它可能会投入生产。

用法示例

我提出异常以警告使用者如果我的API使用不正确:

def api_func(foo):
    '''foo should be either 'baz' or 'bar'. returns something very useful.'''
    if foo not in _ALLOWED_ARGS:
        raise ValueError('{foo} wrong, use "baz" or "bar"'.format(foo=repr(foo)))

适当时创建自己的错误类型

“我想故意犯一个错误,以便将其排除在外”

您可以创建自己的错误类型,如果要指示应用程序存在某些特定的错误,只需在异常层次结构中将适当的点子类化:

class MyAppLookupError(LookupError):
    '''raise this when there's a lookup error for my app'''

和用法:

if important_key not in resource_dict and not ok_to_be_missing:
    raise MyAppLookupError('resource is missing, and that is not ok.')

How do I manually throw/raise an exception in Python?

Use the most specific Exception constructor that semantically fits your issue.

Be specific in your message, e.g.:

raise ValueError('A very specific bad thing happened.')

Don’t raise generic exceptions

Avoid raising a generic Exception. To catch it, you’ll have to catch all other more specific exceptions that subclass it.

Problem 1: Hiding bugs

raise Exception('I know Python!') # Don't! If you catch, likely to hide bugs.

For example:

def demo_bad_catch():
    try:
        raise ValueError('Represents a hidden bug, do not catch this')
        raise Exception('This is the exception you expect to handle')
    except Exception as error:
        print('Caught this error: ' + repr(error))

>>> demo_bad_catch()
Caught this error: ValueError('Represents a hidden bug, do not catch this',)

Problem 2: Won’t catch

And more specific catches won’t catch the general exception:

def demo_no_catch():
    try:
        raise Exception('general exceptions not caught by specific handling')
    except ValueError as e:
        print('we will not catch exception: Exception')


>>> demo_no_catch()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in demo_no_catch
Exception: general exceptions not caught by specific handling

Best Practices: raise statement

Instead, use the most specific Exception constructor that semantically fits your issue.

raise ValueError('A very specific bad thing happened')

which also handily allows an arbitrary number of arguments to be passed to the constructor:

raise ValueError('A very specific bad thing happened', 'foo', 'bar', 'baz') 

These arguments are accessed by the args attribute on the Exception object. For example:

try:
    some_code_that_may_raise_our_value_error()
except ValueError as err:
    print(err.args)

prints

('message', 'foo', 'bar', 'baz')    

In Python 2.5, an actual message attribute was added to BaseException in favor of encouraging users to subclass Exceptions and stop using args, but the introduction of message and the original deprecation of args has been retracted.

Best Practices: except clause

When inside an except clause, you might want to, for example, log that a specific type of error happened, and then re-raise. The best way to do this while preserving the stack trace is to use a bare raise statement. For example:

logger = logging.getLogger(__name__)

try:
    do_something_in_app_that_breaks_easily()
except AppError as error:
    logger.error(error)
    raise                 # just this!
    # raise AppError      # Don't do this, you'll lose the stack trace!

Don’t modify your errors… but if you insist.

You can preserve the stacktrace (and error value) with sys.exc_info(), but this is way more error prone and has compatibility problems between Python 2 and 3, prefer to use a bare raise to re-raise.

To explain – the sys.exc_info() returns the type, value, and traceback.

type, value, traceback = sys.exc_info()

This is the syntax in Python 2 – note this is not compatible with Python 3:

    raise AppError, error, sys.exc_info()[2] # avoid this.
    # Equivalently, as error *is* the second object:
    raise sys.exc_info()[0], sys.exc_info()[1], sys.exc_info()[2]

If you want to, you can modify what happens with your new raise – e.g. setting new args for the instance:

def error():
    raise ValueError('oops!')

def catch_error_modify_message():
    try:
        error()
    except ValueError:
        error_type, error_instance, traceback = sys.exc_info()
        error_instance.args = (error_instance.args[0] + ' <modification>',)
        raise error_type, error_instance, traceback

And we have preserved the whole traceback while modifying the args. Note that this is not a best practice and it is invalid syntax in Python 3 (making keeping compatibility much harder to work around).

>>> catch_error_modify_message()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in catch_error_modify_message
  File "<stdin>", line 2, in error
ValueError: oops! <modification>

In Python 3:

    raise error.with_traceback(sys.exc_info()[2])

Again: avoid manually manipulating tracebacks. It’s less efficient and more error prone. And if you’re using threading and sys.exc_info you may even get the wrong traceback (especially if you’re using exception handling for control flow – which I’d personally tend to avoid.)

Python 3, Exception chaining

In Python 3, you can chain Exceptions, which preserve tracebacks:

    raise RuntimeError('specific message') from error

Be aware:

  • this does allow changing the error type raised, and
  • this is not compatible with Python 2.

Deprecated Methods:

These can easily hide and even get into production code. You want to raise an exception, and doing them will raise an exception, but not the one intended!

Valid in Python 2, but not in Python 3 is the following:

raise ValueError, 'message' # Don't do this, it's deprecated!

Only valid in much older versions of Python (2.4 and lower), you may still see people raising strings:

raise 'message' # really really wrong. don't do this.

In all modern versions, this will actually raise a TypeError, because you’re not raising a BaseException type. If you’re not checking for the right exception and don’t have a reviewer that’s aware of the issue, it could get into production.

Example Usage

I raise Exceptions to warn consumers of my API if they’re using it incorrectly:

def api_func(foo):
    '''foo should be either 'baz' or 'bar'. returns something very useful.'''
    if foo not in _ALLOWED_ARGS:
        raise ValueError('{foo} wrong, use "baz" or "bar"'.format(foo=repr(foo)))

Create your own error types when apropos

“I want to make an error on purpose, so that it would go into the except”

You can create your own error types, if you want to indicate something specific is wrong with your application, just subclass the appropriate point in the exception hierarchy:

class MyAppLookupError(LookupError):
    '''raise this when there's a lookup error for my app'''

and usage:

if important_key not in resource_dict and not ok_to_be_missing:
    raise MyAppLookupError('resource is missing, and that is not ok.')

回答 1

不要这样做。赤身裸体Exception绝对不是正确的选择。请参阅亚伦·霍尔(Aaron Hall)的出色答案

不能得到比这更多的pythonic:

raise Exception("I know python!")

如果您需要更多信息,请参阅python 的凸起语句文档

DON’T DO THIS. Raising a bare Exception is absolutely not the right thing to do; see Aaron Hall’s excellent answer instead.

Can’t get much more pythonic than this:

raise Exception("I know python!")

See the raise statement docs for python if you’d like more info.


回答 2

在Python3中,有四种用于引发异常的语法:

1. raise exception 
2. raise exception (args) 
3. raise
4. raise exception (args) from original_exception

1.引发异常vs. 2.引发异常(参数)

如果raise exception (args) 用于引发异常,则在 args打印异常对象时将打印出-如下例所示。

  #raise exception (args)
    try:
        raise ValueError("I have raised an Exception")
    except ValueError as exp:
        print ("Error", exp)     # Output -> Error I have raised an Exception 



  #raise execption 
    try:
        raise ValueError
    except ValueError as exp:
        print ("Error", exp)     # Output -> Error 

3.提高

raise不带任何参数的语句重新引发最后一个异常。如果您需要在捕获异常后执行一些操作然后重新引发它,这将很有用。但是,如果以前没有异常,则raise语句引发 TypeErrorException。

def somefunction():
    print("some cleaning")

a=10
b=0 
result=None

try:
    result=a/b
    print(result)

except Exception:            #Output ->
    somefunction()           #some cleaning
    raise                    #Traceback (most recent call last):
                             #File "python", line 8, in <module>
                             #ZeroDivisionError: division by zero

4.从original_exception引发异常(参数)

该语句用于创建异常链接,其中响应另一个异常而引发的异常可以包含原始异常的详细信息-如下例所示。

class MyCustomException(Exception):
pass

a=10
b=0 
reuslt=None
try:
    try:
        result=a/b

    except ZeroDivisionError as exp:
        print("ZeroDivisionError -- ",exp)
        raise MyCustomException("Zero Division ") from exp

except MyCustomException as exp:
        print("MyException",exp)
        print(exp.__cause__)

输出:

ZeroDivisionError --  division by zero
MyException Zero Division 
division by zero

In Python3 there are 4 different syntaxes for rasing exceptions:

1. raise exception 
2. raise exception (args) 
3. raise
4. raise exception (args) from original_exception

1. raise exception vs. 2. raise exception (args)

If you use raise exception (args) to raise an exception then the args will be printed when you print the exception object – as shown in the example below.

  #raise exception (args)
    try:
        raise ValueError("I have raised an Exception")
    except ValueError as exp:
        print ("Error", exp)     # Output -> Error I have raised an Exception 



  #raise execption 
    try:
        raise ValueError
    except ValueError as exp:
        print ("Error", exp)     # Output -> Error 

3.raise

raise statement without any arguments re-raises the last exception. This is useful if you need to perform some actions after catching the exception and then want to re-raise it. But if there was no exception before, raise statement raises TypeError Exception.

def somefunction():
    print("some cleaning")

a=10
b=0 
result=None

try:
    result=a/b
    print(result)

except Exception:            #Output ->
    somefunction()           #some cleaning
    raise                    #Traceback (most recent call last):
                             #File "python", line 8, in <module>
                             #ZeroDivisionError: division by zero

4. raise exception (args) from original_exception

This statement is used to create exception chaining in which an exception that is raised in response to another exception can contain the details of the original exception – as shown in the example below.

class MyCustomException(Exception):
pass

a=10
b=0 
reuslt=None
try:
    try:
        result=a/b

    except ZeroDivisionError as exp:
        print("ZeroDivisionError -- ",exp)
        raise MyCustomException("Zero Division ") from exp

except MyCustomException as exp:
        print("MyException",exp)
        print(exp.__cause__)

Output:

ZeroDivisionError --  division by zero
MyException Zero Division 
division by zero

回答 3

对于常见的情况,您需要针对某些意外情况抛出异常,并且您从不打算抓住它,而只是快速失败以使您能够从那里进行调试(如果发生的话)—最合乎逻辑的是AssertionError

if 0 < distance <= RADIUS:
    #Do something.
elif RADIUS < distance:
    #Do something.
else:
    raise AssertionError("Unexpected value of 'distance'!", distance)

For the common case where you need to throw an exception in response to some unexpected conditions, and that you never intend to catch, but simply to fail fast to enable you to debug from there if it ever happens — the most logical one seems to be AssertionError:

if 0 < distance <= RADIUS:
    #Do something.
elif RADIUS < distance:
    #Do something.
else:
    raise AssertionError("Unexpected value of 'distance'!", distance)

回答 4

首先阅读现有的答案,这只是一个附录。

请注意,可以带或不带参数引发异常。

例:

raise SystemExit

退出程序,但是您可能想知道发生了什么。因此可以使用它。

raise SystemExit("program exited")

这将在关闭程序之前将“程序退出”打印到stderr。

Read the existing answers first, this is just an addendum.

Notice that you can raise exceptions with or without arguments.

Example:

raise SystemExit

exits the program but you might want to know what happened.So you can use this.

raise SystemExit("program exited")

this will print “program exited” to stderr before closing the program.


回答 5

抛出异常的另一种方法是assert。您可以使用assert来验证是否满足条件,否则条件将上升AssertionError。有关更多详细信息,请在此处查看

def avg(marks):
    assert len(marks) != 0,"List is empty."
    return sum(marks)/len(marks)

mark2 = [55,88,78,90,79]
print("Average of mark2:",avg(mark2))

mark1 = []
print("Average of mark1:",avg(mark1))

Another way to throw an exceptions is assert. You can use assert to verify a condition is being fulfilled if not then it will raise AssertionError. For more details have a look here.

def avg(marks):
    assert len(marks) != 0,"List is empty."
    return sum(marks)/len(marks)

mark2 = [55,88,78,90,79]
print("Average of mark2:",avg(mark2))

mark1 = []
print("Average of mark1:",avg(mark1))

回答 6

只是要注意:有时候您确实想处理通用异常。如果要处理大量文件并记录错误,则可能要捕获文件发生的任何错误,将其记录下来,然后继续处理其余文件。在这种情况下,

try:
    foo() 
except Exception as e:
    print(str(e)) # Print out handled error

阻止这样做的好方法。您仍然需要raise特定的异常,以便您了解异常的含义。

Just to note: there are times when you DO want to handle generic exceptions. If you’re processing a bunch of files and logging your errors, you might want to catch any error that occurs for a file, log it, and continue processing the rest of the files. In that case, a

try:
    foo() 
except Exception as e:
    print(str(e)) # Print out handled error

block a good way to do it. You’ll still want to raise specific exceptions so you know what they mean, though.


回答 7

您应该为此学习python的凸起语句。它应该保存在try块中。范例-

try:
    raise TypeError            #remove TypeError by any other error if you want
except TypeError:
    print('TypeError raised')

You should learn the raise statement of python for that. It should be kept inside the try block. Example –

try:
    raise TypeError            #remove TypeError by any other error if you want
except TypeError:
    print('TypeError raised')

如何将字符串解析为float或int?

问题:如何将字符串解析为float或int?

在Python中,如何解析类似于"545.2222"其对应的float值的数字字符串545.2222?还是将字符串解析为"31"整数31

我只是想知道如何分析一个浮动 strfloat,和(单独)的INT strint

In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 545.2222? Or parse the string "31" to an integer, 31?

I just want to know how to parse a float str to a float, and (separately) an int str to an int.


回答 0

>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545
>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545

回答 1

def num(s):
    try:
        return int(s)
    except ValueError:
        return float(s)
def num(s):
    try:
        return int(s)
    except ValueError:
        return float(s)

回答 2

检查字符串是否为浮点数的Python方法:

def is_float(value):
  try:
    float(value)
    return True
  except:
    return False

此功能的更长更准确的名称可能是: is_convertible_to_float(value)

什么是Python中的浮点数,哪些不是浮点数,可能会让您感到惊讶:

val                   is_float(val) Note
--------------------  ----------   --------------------------------
""                    False        Blank string
"127"                 True         Passed string
True                  True         Pure sweet Truth
"True"                False        Vile contemptible lie
False                 True         So false it becomes true
"123.456"             True         Decimal
"      -127    "      True         Spaces trimmed
"\t\n12\r\n"          True         whitespace ignored
"NaN"                 True         Not a number
"NaNanananaBATMAN"    False        I am Batman
"-iNF"                True         Negative infinity
"123.E4"              True         Exponential notation
".1"                  True         mantissa only
"1,234"               False        Commas gtfo
u'\x30'               True         Unicode is fine.
"NULL"                False        Null is not special
0x3fade               True         Hexadecimal
"6e7777777777777"     True         Shrunk to infinity
"1.797693e+308"       True         This is max value
"infinity"            True         Same as inf
"infinityandBEYOND"   False        Extra characters wreck it
"12.34.56"            False        Only one dot allowed
u'四'                 False        Japanese '4' is not a float.
"#56"                 False        Pound sign
"56%"                 False        Percent of what?
"0E0"                 True         Exponential, move dot 0 places
0**0                  True         0___0  Exponentiation
"-5e-5"               True         Raise to a negative number
"+1e1"                True         Plus is OK with exponent
"+1e1^5"              False        Fancy exponent not interpreted
"+1e1.3"              False        No decimals in exponent
"-+1"                 False        Make up your mind
"(1)"                 False        Parenthesis is bad

您以为知道什么数字?你不像你想的那样好!并不奇怪。

不要在对生命至关重要的软件上使用此代码!

用这种方式捕获广泛的异常,杀死金丝雀和吞噬异常会产生很小的机会,即有效的float字符串将返回false。该float(...)行代码可以失败的任何什么都没有做的字符串的内容一千个理由。但是,如果您使用Python这样的鸭子式原型语言来编写至关重要的软件,那么您将遇到更大的问题。

Python method to check if a string is a float:

def is_float(value):
  try:
    float(value)
    return True
  except:
    return False

A longer and more accurate name for this function could be: is_convertible_to_float(value)

What is, and is not a float in Python may surprise you:

val                   is_float(val) Note
--------------------  ----------   --------------------------------
""                    False        Blank string
"127"                 True         Passed string
True                  True         Pure sweet Truth
"True"                False        Vile contemptible lie
False                 True         So false it becomes true
"123.456"             True         Decimal
"      -127    "      True         Spaces trimmed
"\t\n12\r\n"          True         whitespace ignored
"NaN"                 True         Not a number
"NaNanananaBATMAN"    False        I am Batman
"-iNF"                True         Negative infinity
"123.E4"              True         Exponential notation
".1"                  True         mantissa only
"1,234"               False        Commas gtfo
u'\x30'               True         Unicode is fine.
"NULL"                False        Null is not special
0x3fade               True         Hexadecimal
"6e7777777777777"     True         Shrunk to infinity
"1.797693e+308"       True         This is max value
"infinity"            True         Same as inf
"infinityandBEYOND"   False        Extra characters wreck it
"12.34.56"            False        Only one dot allowed
u'四'                 False        Japanese '4' is not a float.
"#56"                 False        Pound sign
"56%"                 False        Percent of what?
"0E0"                 True         Exponential, move dot 0 places
0**0                  True         0___0  Exponentiation
"-5e-5"               True         Raise to a negative number
"+1e1"                True         Plus is OK with exponent
"+1e1^5"              False        Fancy exponent not interpreted
"+1e1.3"              False        No decimals in exponent
"-+1"                 False        Make up your mind
"(1)"                 False        Parenthesis is bad

You think you know what numbers are? You are not so good as you think! Not big surprise.

Don’t use this code on life-critical software!

Catching broad exceptions this way, killing canaries and gobbling the exception creates a tiny chance that a valid float as string will return false. The float(...) line of code can failed for any of a thousand reasons that have nothing to do with the contents of the string. But if you’re writing life-critical software in a duck-typing prototype language like Python, then you’ve got much larger problems.


回答 3

这是另一个值得一提的方法ast.literal_eval

这可用于安全地评估包含来自不受信任来源的Python表达式的字符串,而无需自己解析值。

也就是说,一个安全的“评估”

>>> import ast
>>> ast.literal_eval("545.2222")
545.2222
>>> ast.literal_eval("31")
31

This is another method which deserves to be mentioned here, ast.literal_eval:

This can be used for safely evaluating strings containing Python expressions from untrusted sources without the need to parse the values oneself.

That is, a safe ‘eval’

>>> import ast
>>> ast.literal_eval("545.2222")
545.2222
>>> ast.literal_eval("31")
31

回答 4

float(x) if '.' in x else int(x)
float(x) if '.' in x else int(x)

回答 5

本地化和逗号

您应该考虑数字的字符串表示形式中可能出现逗号的情况,例如 float("545,545.2222")抛出异常的情况。而是使用in locale中的方法将字符串转换为数字并正确解释逗号。locale.atof一旦为所需的数字约定设置了语言环境,该方法便会一步转换为浮点数。

示例1-美国数字约定

在美国和英国,逗号可以用作千位分隔符。在具有美国语言环境的此示例中,逗号作为分隔符正确处理:

>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>

示例2-欧洲数字约定

在世界上大多数国家/地区,逗号用于小数点而不是句点。在此使用法语语言环境的示例中,逗号被正确处理为小数点:

>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222

该方法locale.atoi也可用,但参数应为整数。

Localization and commas

You should consider the possibility of commas in the string representation of a number, for cases like float("545,545.2222") which throws an exception. Instead, use methods in locale to convert the strings to numbers and interpret commas correctly. The locale.atof method converts to a float in one step once the locale has been set for the desired number convention.

Example 1 — United States number conventions

In the United States and the UK, commas can be used as a thousands separator. In this example with American locale, the comma is handled properly as a separator:

>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>

Example 2 — European number conventions

In the majority of countries of the world, commas are used for decimal marks instead of periods. In this example with French locale, the comma is correctly handled as a decimal mark:

>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222

The method locale.atoi is also available, but the argument should be an integer.


回答 6

如果您不喜欢第三方模块,则可以签出fastnumbers模块。它提供了一个名为fast_real的函数,该函数可以完全满足此问题的要求,并且比纯Python实现要快:

>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int

If you aren’t averse to third-party modules, you could check out the fastnumbers module. It provides a function called fast_real that does exactly what this question is asking for and does it faster than a pure-Python implementation:

>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int

回答 7

用户codelogicharley是正确的,但是请记住,如果您知道字符串是整数(例如545),则可以调用int(“ 545”)而不先进行浮点运算。

如果您的字符串在列表中,则也可以使用map函数。

>>> x = ["545.0", "545.6", "999.2"]
>>> map(float, x)
[545.0, 545.60000000000002, 999.20000000000005]
>>>

只有它们都是相同的类型才是好的。

Users codelogic and harley are correct, but keep in mind if you know the string is an integer (for example, 545) you can call int(“545”) without first casting to float.

If your strings are in a list, you could use the map function as well.

>>> x = ["545.0", "545.6", "999.2"]
>>> map(float, x)
[545.0, 545.60000000000002, 999.20000000000005]
>>>

It is only good if they’re all the same type.


回答 8

在Python中,如何将“ 545.2222”之类的数字字符串解析为其对应的浮点值542.2222?还是将字符串“ 31”解析为整数31? 我只想知道如何将float字符串解析为float,以及将int字符串分别解析为int。

您最好单独进行这些操作。如果您要混合使用它们,则可能会在以后遇到问题。简单的答案是:

"545.2222" 漂浮:

>>> float("545.2222")
545.2222

"31" 到一个整数:

>>> int("31")
31

其他与字符串和文字之间的转换,整数转换:

来自各种基准的转换,您应该事先知道基准(默认值为10)。请注意,您可以为它们加上Python期望的字面量(请参见下文)或删除前缀:

>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31

如果您不预先知道基础,但是您知道它们将具有正确的前缀,那么如果您通过0作为基础,Python可以为您推断出这个前缀:

>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31

其他基数的非十进制(即整数)文字

但是,如果您的动机是让自己的代码清楚地表示硬编码的特定值,则可能不需要从基数进行转换-您可以让Python使用正确的语法自动为您完成。

您可以使用apropos前缀自动转换为具有以下文字的整数。这些对Python 2和3有效:

二进制前缀 0b

>>> 0b11111
31

八进制,前缀 0o

>>> 0o37
31

十六进制,前缀 0x

>>> 0x1f
31

当描述二进制标志,代码中的文件许可权或颜色的十六进制值时,这很有用-例如,请注意不要使用引号:

>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215

使模棱两可的Python 2八进制与Python 3兼容

如果您在Python 2中看到一个以0开头的整数,则这是(不建议使用的)八进制语法。

>>> 037
31

这很糟糕,因为看起来值应该是37。因此,在Python 3中,它现在引发了SyntaxError

>>> 037
  File "<stdin>", line 1
    037
      ^
SyntaxError: invalid token

使用0o前缀将Python 2八进制转换为在2和3中均可使用的八进制:

>>> 0o37
31

In Python, how can I parse a numeric string like “545.2222” to its corresponding float value, 542.2222? Or parse the string “31” to an integer, 31? I just want to know how to parse a float string to a float, and (separately) an int string to an int.

It’s good that you ask to do these separately. If you’re mixing them, you may be setting yourself up for problems later. The simple answer is:

"545.2222" to float:

>>> float("545.2222")
545.2222

"31" to an integer:

>>> int("31")
31

Other conversions, ints to and from strings and literals:

Conversions from various bases, and you should know the base in advance (10 is the default). Note you can prefix them with what Python expects for its literals (see below) or remove the prefix:

>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31

If you don’t know the base in advance, but you do know they will have the correct prefix, Python can infer this for you if you pass 0 as the base:

>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31

Non-Decimal (i.e. Integer) Literals from other Bases

If your motivation is to have your own code clearly represent hard-coded specific values, however, you may not need to convert from the bases – you can let Python do it for you automatically with the correct syntax.

You can use the apropos prefixes to get automatic conversion to integers with the following literals. These are valid for Python 2 and 3:

Binary, prefix 0b

>>> 0b11111
31

Octal, prefix 0o

>>> 0o37
31

Hexadecimal, prefix 0x

>>> 0x1f
31

This can be useful when describing binary flags, file permissions in code, or hex values for colors – for example, note no quotes:

>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215

Making ambiguous Python 2 octals compatible with Python 3

If you see an integer that starts with a 0, in Python 2, this is (deprecated) octal syntax.

>>> 037
31

It is bad because it looks like the value should be 37. So in Python 3, it now raises a SyntaxError:

>>> 037
  File "<stdin>", line 1
    037
      ^
SyntaxError: invalid token

Convert your Python 2 octals to octals that work in both 2 and 3 with the 0o prefix:

>>> 0o37
31

回答 9

这个问题似乎有点老了。但是让我建议一个函数parseStr,它的功能类似,即返回整数或浮点数,并且如果无法将给定的ASCII字符串转换为其中的任何一个,则它将返回原样。当然,可以将代码调整为仅执行所需的操作:

   >>> import string
   >>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
   ...                      int(x) or x.isalnum() and x or \
   ...                      len(set(string.punctuation).intersection(x)) == 1 and \
   ...                      x.count('.') == 1 and float(x) or x
   >>> parseStr('123')
   123
   >>> parseStr('123.3')
   123.3
   >>> parseStr('3HC1')
   '3HC1'
   >>> parseStr('12.e5')
   1200000.0
   >>> parseStr('12$5')
   '12$5'
   >>> parseStr('12.2.2')
   '12.2.2'

The question seems a little bit old. But let me suggest a function, parseStr, which makes something similar, that is, returns integer or float and if a given ASCII string cannot be converted to none of them it returns it untouched. The code of course might be adjusted to do only what you want:

   >>> import string
   >>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
   ...                      int(x) or x.isalnum() and x or \
   ...                      len(set(string.punctuation).intersection(x)) == 1 and \
   ...                      x.count('.') == 1 and float(x) or x
   >>> parseStr('123')
   123
   >>> parseStr('123.3')
   123.3
   >>> parseStr('3HC1')
   '3HC1'
   >>> parseStr('12.e5')
   1200000.0
   >>> parseStr('12$5')
   '12$5'
   >>> parseStr('12.2.2')
   '12.2.2'

回答 10

float("545.2222")int(float("545.2222"))

float("545.2222") and int(float("545.2222"))


回答 11

我为此使用此功能

import ast

def parse_str(s):
   try:
      return ast.literal_eval(str(s))
   except:
      return

它将字符串转换为其类型

value = parse_str('1')  # Returns Integer
value = parse_str('1.5')  # Returns Float

I use this function for that

import ast

def parse_str(s):
   try:
      return ast.literal_eval(str(s))
   except:
      return

It will convert the string to its type

value = parse_str('1')  # Returns Integer
value = parse_str('1.5')  # Returns Float

回答 12

YAML解析器可以帮助你找出你的数据类型的字符串是什么。使用yaml.load(),然后可以使用type(result)测试类型:

>>> import yaml

>>> a = "545.2222"
>>> result = yaml.load(a)
>>> result
545.22220000000004
>>> type(result)
<type 'float'>

>>> b = "31"
>>> result = yaml.load(b)
>>> result
31
>>> type(result)
<type 'int'>

>>> c = "HI"
>>> result = yaml.load(c)
>>> result
'HI'
>>> type(result)
<type 'str'>

The YAML parser can help you figure out what datatype your string is. Use yaml.load(), and then you can use type(result) to test for type:

>>> import yaml

>>> a = "545.2222"
>>> result = yaml.load(a)
>>> result
545.22220000000004
>>> type(result)
<type 'float'>

>>> b = "31"
>>> result = yaml.load(b)
>>> result
31
>>> type(result)
<type 'int'>

>>> c = "HI"
>>> result = yaml.load(c)
>>> result
'HI'
>>> type(result)
<type 'str'>

回答 13

def get_int_or_float(v):
    number_as_float = float(v)
    number_as_int = int(number_as_float)
    return number_as_int if number_as_float == number_as_int else number_as_float
def get_int_or_float(v):
    number_as_float = float(v)
    number_as_int = int(number_as_float)
    return number_as_int if number_as_float == number_as_int else number_as_float

回答 14

def num(s):
    """num(s)
    num(3),num(3.7)-->3
    num('3')-->3, num('3.7')-->3.7
    num('3,700')-->ValueError
    num('3a'),num('a3'),-->ValueError
    num('3e4') --> 30000.0
    """
    try:
        return int(s)
    except ValueError:
        try:
            return float(s)
        except ValueError:
            raise ValueError('argument is not a string of number')
def num(s):
    """num(s)
    num(3),num(3.7)-->3
    num('3')-->3, num('3.7')-->3.7
    num('3,700')-->ValueError
    num('3a'),num('a3'),-->ValueError
    num('3e4') --> 30000.0
    """
    try:
        return int(s)
    except ValueError:
        try:
            return float(s)
        except ValueError:
            raise ValueError('argument is not a string of number')

回答 15

您需要考虑到四舍五入才能正确执行此操作。

即int(5.1)=> 5 int(5.6)=> 5-错误,应该为6所以我们做int(5.6 + 0.5)=> 6

def convert(n):
    try:
        return int(n)
    except ValueError:
        return float(n + 0.5)

You need to take into account rounding to do this properly.

I.e. int(5.1) => 5 int(5.6) => 5 — wrong, should be 6 so we do int(5.6 + 0.5) => 6

def convert(n):
    try:
        return int(n)
    except ValueError:
        return float(n + 0.5)

回答 16

我很惊讶没有人提到正则表达式,因为有时必须在转换为数字之前准备好字符串并对其进行规范化

import re
def parseNumber(value, as_int=False):
    try:
        number = float(re.sub('[^.\-\d]', '', value))
        if as_int:
            return int(number + 0.5)
        else:
            return number
    except ValueError:
        return float('nan')  # or None if you wish

用法:

parseNumber('13,345')
> 13345.0

parseNumber('- 123 000')
> -123000.0

parseNumber('99999\n')
> 99999.0

顺便说一句,以验证您有一个数字:

import numbers
def is_number(value):
    return isinstance(value, numbers.Number)
    # will work with int, float, long, Decimal

I am surprised nobody mentioned regex because sometimes string must be prepared and normalized before casting to number

import re
def parseNumber(value, as_int=False):
    try:
        number = float(re.sub('[^.\-\d]', '', value))
        if as_int:
            return int(number + 0.5)
        else:
            return number
    except ValueError:
        return float('nan')  # or None if you wish

usage:

parseNumber('13,345')
> 13345.0

parseNumber('- 123 000')
> -123000.0

parseNumber('99999\n')
> 99999.0

and by the way, something to verify you have a number:

import numbers
def is_number(value):
    return isinstance(value, numbers.Number)
    # will work with int, float, long, Decimal

回答 17

要在python中进行类型转换,请使用该类型的构造函数,并将字符串(或您尝试投射的任何值)作为参数传递。

例如:

>>>float("23.333")
   23.333

在后台,python正在调用objects __float__方法,该方法应该返回参数的float表示形式。这是特别强大的功能,因为您可以使用__float__方法定义自己的类型(使用类),以便可以使用float(myobject)将其转换为float。

To typecast in python use the constructor funtions of the type, passing the string (or whatever value you are trying to cast) as a parameter.

For example:

>>>float("23.333")
   23.333

Behind the scenes, python is calling the objects __float__ method, which should return a float representation of the parameter. This is especially powerful, as you can define your own types (using classes) with a __float__ method so that it can be casted into a float using float(myobject).


回答 18

这是一个正确版本https://stackoverflow.com/a/33017514/5973334

这将尝试解析一个字符串并返回一个intfloat取决于该字符串表示什么。它可能会引发解析异常或具有某些意外行为

  def get_int_or_float(v):
        number_as_float = float(v)
        number_as_int = int(number_as_float)
        return number_as_int if number_as_float == number_as_int else 
        number_as_float

This is a corrected version of https://stackoverflow.com/a/33017514/5973334

This will try to parse a string and return either int or float depending on what the string represents. It might rise parsing exceptions or have some unexpected behaviour.

  def get_int_or_float(v):
        number_as_float = float(v)
        number_as_int = int(number_as_float)
        return number_as_int if number_as_float == number_as_int else 
        number_as_float

回答 19

将您的字符串传递给此函数:

def string_to_number(str):
  if("." in str):
    try:
      res = float(str)
    except:
      res = str  
  elif(str.isdigit()):
    res = int(str)
  else:
    res = str
  return(res)

根据所传递的内容,它将返回int,float或string。

一个int字符串

print(type(string_to_number("124")))
<class 'int'>

浮点数的字符串

print(type(string_to_number("12.4")))
<class 'float'>

字符串即字符串

print(type(string_to_number("hello")))
<class 'str'>

看起来像个浮点数的字符串

print(type(string_to_number("hel.lo")))
<class 'str'>

Pass your string to this function:

def string_to_number(str):
  if("." in str):
    try:
      res = float(str)
    except:
      res = str  
  elif(str.isdigit()):
    res = int(str)
  else:
    res = str
  return(res)

It will return int, float or string depending on what was passed.

string that is an int

print(type(string_to_number("124")))
<class 'int'>

string that is a float

print(type(string_to_number("12.4")))
<class 'float'>

string that is a string

print(type(string_to_number("hello")))
<class 'str'>

string that looks like a float

print(type(string_to_number("hel.lo")))
<class 'str'>

回答 20

采用:

def num(s):
    try:
        for each in s:
            yield int(each)
    except ValueError:
        yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()

这是我想出的最Python化的方式。

Use:

def num(s):
    try:
        for each in s:
            yield int(each)
    except ValueError:
        yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()

This is the most Pythonic way I could come up with.


回答 21

处理十六进制,八进制,二进制,十进制和浮点数

该解决方案将处理数字的所有字符串约定(我所知道的全部)。

def to_number(n):
    ''' Convert any number representation to a number 
    This covers: float, decimal, hex, and octal numbers.
    '''

    try:
        return int(str(n), 0)
    except:
        try:
            # python 3 doesn't accept "010" as a valid octal.  You must use the
            # '0o' prefix
            return int('0o' + n, 0)
        except:
            return float(n)

该测试用例输出说明了我在说什么。

======================== CAPTURED OUTPUT =========================
to_number(3735928559)   = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0")        =          0 ==          0
to_number(100)          =        100 ==        100
to_number("42")         =         42 ==         42
to_number(8)            =          8 ==          8
to_number("0o20")       =         16 ==         16
to_number("020")        =         16 ==         16
to_number(3.14)         =       3.14 ==       3.14
to_number("2.72")       =       2.72 ==       2.72
to_number("1e3")        =     1000.0 ==       1000
to_number(0.001)        =      0.001 ==      0.001
to_number("0xA")        =         10 ==         10
to_number("012")        =         10 ==         10
to_number("0o12")       =         10 ==         10
to_number("0b01010")    =         10 ==         10
to_number("10")         =         10 ==         10
to_number("10.0")       =       10.0 ==         10
to_number("1e1")        =       10.0 ==         10

这是测试:

class test_to_number(unittest.TestCase):

    def test_hex(self):
        # All of the following should be converted to an integer
        #
        values = [

                 #          HEX
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (0xDEADBEEF  , 3735928559), # Hex
                ("0xFEEDFACE", 4277009102), # Hex
                ("0x0"       ,          0), # Hex

                 #        Decimals
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (100         ,        100), # Decimal
                ("42"        ,         42), # Decimal
            ]



        values += [
                 #        Octals
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (0o10        ,          8), # Octal
                ("0o20"      ,         16), # Octal
                ("020"       ,         16), # Octal
            ]


        values += [
                 #        Floats
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (3.14        ,       3.14), # Float
                ("2.72"      ,       2.72), # Float
                ("1e3"       ,       1000), # Float
                (1e-3        ,      0.001), # Float
            ]

        values += [
                 #        All ints
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                ("0xA"       ,         10), 
                ("012"       ,         10), 
                ("0o12"      ,         10), 
                ("0b01010"   ,         10), 
                ("10"        ,         10), 
                ("10.0"      ,         10), 
                ("1e1"       ,         10), 
            ]

        for _input, expected in values:
            value = to_number(_input)

            if isinstance(_input, str):
                cmd = 'to_number("{}")'.format(_input)
            else:
                cmd = 'to_number({})'.format(_input)

            print("{:23} = {:10} == {:10}".format(cmd, value, expected))
            self.assertEqual(value, expected)

Handles hex, octal, binary, decimal, and float

This solution will handle all of the string conventions for numbers (all that I know about).

def to_number(n):
    ''' Convert any number representation to a number 
    This covers: float, decimal, hex, and octal numbers.
    '''

    try:
        return int(str(n), 0)
    except:
        try:
            # python 3 doesn't accept "010" as a valid octal.  You must use the
            # '0o' prefix
            return int('0o' + n, 0)
        except:
            return float(n)

This test case output illustrates what I’m talking about.

======================== CAPTURED OUTPUT =========================
to_number(3735928559)   = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0")        =          0 ==          0
to_number(100)          =        100 ==        100
to_number("42")         =         42 ==         42
to_number(8)            =          8 ==          8
to_number("0o20")       =         16 ==         16
to_number("020")        =         16 ==         16
to_number(3.14)         =       3.14 ==       3.14
to_number("2.72")       =       2.72 ==       2.72
to_number("1e3")        =     1000.0 ==       1000
to_number(0.001)        =      0.001 ==      0.001
to_number("0xA")        =         10 ==         10
to_number("012")        =         10 ==         10
to_number("0o12")       =         10 ==         10
to_number("0b01010")    =         10 ==         10
to_number("10")         =         10 ==         10
to_number("10.0")       =       10.0 ==         10
to_number("1e1")        =       10.0 ==         10

Here is the test:

class test_to_number(unittest.TestCase):

    def test_hex(self):
        # All of the following should be converted to an integer
        #
        values = [

                 #          HEX
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (0xDEADBEEF  , 3735928559), # Hex
                ("0xFEEDFACE", 4277009102), # Hex
                ("0x0"       ,          0), # Hex

                 #        Decimals
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (100         ,        100), # Decimal
                ("42"        ,         42), # Decimal
            ]



        values += [
                 #        Octals
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (0o10        ,          8), # Octal
                ("0o20"      ,         16), # Octal
                ("020"       ,         16), # Octal
            ]


        values += [
                 #        Floats
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (3.14        ,       3.14), # Float
                ("2.72"      ,       2.72), # Float
                ("1e3"       ,       1000), # Float
                (1e-3        ,      0.001), # Float
            ]

        values += [
                 #        All ints
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                ("0xA"       ,         10), 
                ("012"       ,         10), 
                ("0o12"      ,         10), 
                ("0b01010"   ,         10), 
                ("10"        ,         10), 
                ("10.0"      ,         10), 
                ("1e1"       ,         10), 
            ]

        for _input, expected in values:
            value = to_number(_input)

            if isinstance(_input, str):
                cmd = 'to_number("{}")'.format(_input)
            else:
                cmd = 'to_number({})'.format(_input)

            print("{:23} = {:10} == {:10}".format(cmd, value, expected))
            self.assertEqual(value, expected)

回答 22

采用:

>>> str_float = "545.2222"
>>> float(str_float)
545.2222
>>> type(_) # Check its type
<type 'float'>

>>> str_int = "31"
>>> int(str_int)
31
>>> type(_) # Check its type
<type 'int'>

Use:

>>> str_float = "545.2222"
>>> float(str_float)
545.2222
>>> type(_) # Check its type
<type 'float'>

>>> str_int = "31"
>>> int(str_int)
31
>>> type(_) # Check its type
<type 'int'>

回答 23

这是将转换任何一个函数object(不只是str)到intfloat方法,依据实际的字符串提供模样 intfloat。此外,如果它是同时具有__float__int__方法的对象,则默认使用__float__

def conv_to_num(x, num_type='asis'):
    '''Converts an object to a number if possible.
    num_type: int, float, 'asis'
    Defaults to floating point in case of ambiguity.
    '''
    import numbers

    is_num, is_str, is_other = [False]*3

    if isinstance(x, numbers.Number):
        is_num = True
    elif isinstance(x, str):
        is_str = True

    is_other = not any([is_num, is_str])

    if is_num:
        res = x
    elif is_str:
        is_float, is_int, is_char = [False]*3
        try:
            res = float(x)
            if '.' in x:
                is_float = True
            else:
                is_int = True
        except ValueError:
            res = x
            is_char = True

    else:
        if num_type == 'asis':
            funcs = [int, float]
        else:
            funcs = [num_type]

        for func in funcs:
            try:
                res = func(x)
                break
            except TypeError:
                continue
        else:
            res = x

This is a function which will convert any object (not just str) to int or float, based on if the actual string supplied looks like int or float. Further if it’s an object which has both __float and __int__ methods, it defaults to using __float__

def conv_to_num(x, num_type='asis'):
    '''Converts an object to a number if possible.
    num_type: int, float, 'asis'
    Defaults to floating point in case of ambiguity.
    '''
    import numbers

    is_num, is_str, is_other = [False]*3

    if isinstance(x, numbers.Number):
        is_num = True
    elif isinstance(x, str):
        is_str = True

    is_other = not any([is_num, is_str])

    if is_num:
        res = x
    elif is_str:
        is_float, is_int, is_char = [False]*3
        try:
            res = float(x)
            if '.' in x:
                is_float = True
            else:
                is_int = True
        except ValueError:
            res = x
            is_char = True

    else:
        if num_type == 'asis':
            funcs = [int, float]
        else:
            funcs = [num_type]

        for func in funcs:
            try:
                res = func(x)
                break
            except TypeError:
                continue
        else:
            res = x

回答 24

通过使用int和float方法,我们可以将字符串转换为整数和浮点数。

s="45.8"
print(float(s))

y='67'
print(int(y))

By using int and float methods we can convert a string to integer and floats.

s="45.8"
print(float(s))

y='67'
print(int(y))

回答 25

eval()是这个问题的很好解决方案。它不需要检查数字是int还是float,它只给出相应的等价物。如果需要其他方法,请尝试

if '.' in string:
    print(float(string))
else:
    print(int(string))

try-except也可以用作替代方法。尝试在try块中将字符串转换为int。如果该字符串是一个浮点值,它将抛出一个错误,该错误将在except块中捕获,像这样

try:
    print(int(string))
except:
    print(float(string))

eval() is a very good solution to this question. It doesn’t need to check if the number is int or float, it just gives the corresponding equivalent. If other methods are required, try

if '.' in string:
    print(float(string))
else:
    print(int(string))

try-except can also be used as an alternative. Try converting string to int inside the try block. If the string would be a float value, it will throw an error which will be catched in the except block, like this

try:
    print(int(string))
except:
    print(float(string))

回答 26

这是您问题的另一种解释(提示:含糊)。您可能正在寻找这样的东西:

def parseIntOrFloat( aString ):
    return eval( aString )

它是这样的…

>>> parseIntOrFloat("545.2222")
545.22220000000004
>>> parseIntOrFloat("545")
545

从理论上讲,存在注入漏洞。字符串可以是例如"import os; os.abort()"。但是,由于没有关于字符串来自何处的任何背景,因此可能是理论上的推测。由于问题很模糊,因此尚不清楚此漏洞是否确实存在。

Here’s another interpretation of your question (hint: it’s vague). It’s possible you’re looking for something like this:

def parseIntOrFloat( aString ):
    return eval( aString )

It works like this…

>>> parseIntOrFloat("545.2222")
545.22220000000004
>>> parseIntOrFloat("545")
545

Theoretically, there’s an injection vulnerability. The string could, for example be "import os; os.abort()". Without any background on where the string comes from, however, the possibility is theoretical speculation. Since the question is vague, it’s not at all clear if this vulnerability actually exists or not.


将字符串转换为日期时间

问题:将字符串转换为日期时间

我有大量的日期时间列表,例如字符串:

Jun 1 2005  1:33PM
Aug 28 1999 12:00AM

我将把它们推回到数据库中正确的日期时间字段中,因此我需要将它们魔术化为实际的日期时间对象。

这是通过Django的ORM进行的,因此我无法使用SQL进行插入时的转换。

I’ve got a huge list of date-times like this as strings:

Jun 1 2005  1:33PM
Aug 28 1999 12:00AM

I’m going to be shoving these back into proper datetime fields in a database so I need to magic them into real datetime objects.

This is going through Django’s ORM so I can’t use SQL to do the conversion on insert.


回答 0

datetime.strptime是将字符串解析为日期时间的主要例程。它可以处理各种格式,格式由您为其指定的格式字符串确定:

from datetime import datetime

datetime_object = datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

生成的datetime对象是时区未使用的。

链接:

笔记:

  • strptime =“字符串解析时间”
  • strftime =“字符串格式时间”
  • 今天大声发音,您将在6个月内无需再次搜索。

datetime.strptime is the main routine for parsing strings into datetimes. It can handle all sorts of formats, with the format determined by a format string you give it:

from datetime import datetime

datetime_object = datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

The resulting datetime object is timezone-naive.

Links:

Notes:

  • strptime = “string parse time”
  • strftime = “string format time”
  • Pronounce it out loud today & you won’t have to search for it again in 6 months.

回答 1

使用第三方dateutil库:

from dateutil import parser
parser.parse("Aug 28 1999 12:00AM")  # datetime.datetime(1999, 8, 28, 0, 0)

它可以处理大多数日期格式,包括您需要解析的格式。它比strptime大多数时候都可以猜测正确的格式要方便得多。

这对于编写测试非常有用,在测试中,可读性比性能更重要。

您可以使用以下方法安装它:

pip install python-dateutil

Use the third party dateutil library:

from dateutil import parser
parser.parse("Aug 28 1999 12:00AM")  # datetime.datetime(1999, 8, 28, 0, 0)

It can handle most date formats, including the one you need to parse. It’s more convenient than strptime as it can guess the correct format most of the time.

It’s very useful for writing tests, where readability is more important than performance.

You can install it with:

pip install python-dateutil

回答 2

时间模块中strptime。它与strftime相反。

$ python
>>> import time
>>> my_time = time.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')
time.struct_time(tm_year=2005, tm_mon=6, tm_mday=1,
                 tm_hour=13, tm_min=33, tm_sec=0,
                 tm_wday=2, tm_yday=152, tm_isdst=-1)

timestamp = time.mktime(my_time)
# convert time object to datetime
from datetime import datetime
my_datetime = datetime.fromtimestamp(timestamp)
# convert time object to date
from datetime import date
my_date = date.fromtimestamp(timestamp)

Check out strptime in the time module. It is the inverse of strftime.

$ python
>>> import time
>>> my_time = time.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')
time.struct_time(tm_year=2005, tm_mon=6, tm_mday=1,
                 tm_hour=13, tm_min=33, tm_sec=0,
                 tm_wday=2, tm_yday=152, tm_isdst=-1)

timestamp = time.mktime(my_time)
# convert time object to datetime
from datetime import datetime
my_datetime = datetime.fromtimestamp(timestamp)
# convert time object to date
from datetime import date
my_date = date.fromtimestamp(timestamp)

回答 3

我整理了一个可以转换一些真正简洁的表达式的项目。查看时间字符串

以下是一些示例:

pip install timestring
>>> import timestring
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm')
<timestring.Date 2015-08-15 20:40:00 4491909392>
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm').date
datetime.datetime(2015, 8, 15, 20, 40)
>>> timestring.Range('next week')
<timestring.Range From 03/10/14 00:00:00 to 03/03/14 00:00:00 4496004880>
>>> (timestring.Range('next week').start.date, timestring.Range('next week').end.date)
(datetime.datetime(2014, 3, 10, 0, 0), datetime.datetime(2014, 3, 14, 0, 0))

I have put together a project that can convert some really neat expressions. Check out timestring.

Here are some examples below:

pip install timestring
>>> import timestring
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm')
<timestring.Date 2015-08-15 20:40:00 4491909392>
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm').date
datetime.datetime(2015, 8, 15, 20, 40)
>>> timestring.Range('next week')
<timestring.Range From 03/10/14 00:00:00 to 03/03/14 00:00:00 4496004880>
>>> (timestring.Range('next week').start.date, timestring.Range('next week').end.date)
(datetime.datetime(2014, 3, 10, 0, 0), datetime.datetime(2014, 3, 14, 0, 0))

回答 4

记住这一点,您无需再次对日期时间转换感到困惑。

日期时间对象的字符串= strptime

datetime对象为其他格式= strftime

Jun 1 2005 1:33PM

等于

%b %d %Y %I:%M%p

%b月作为语言环境的缩写名称(六月)

%d月中的一天,以零填充的十进制数字(1)

%Y以世纪为十进制数字的年份(2015)

%I小时(12小时制),为零填充的十进制数字(01)

%M分钟,为零填充的十进制数字(33)

等同于AM或PM(PM)的%p语言环境

所以你需要strptime即转换string

>>> dates = []
>>> dates.append('Jun 1 2005  1:33PM')
>>> dates.append('Aug 28 1999 12:00AM')
>>> from datetime import datetime
>>> for d in dates:
...     date = datetime.strptime(d, '%b %d %Y %I:%M%p')
...     print type(date)
...     print date
... 

输出量

<type 'datetime.datetime'>
2005-06-01 13:33:00
<type 'datetime.datetime'>
1999-08-28 00:00:00

如果日期格式不同,可以使用panda或dateutil.parse怎么办?

>>> import dateutil
>>> dates = []
>>> dates.append('12 1 2017')
>>> dates.append('1 1 2017')
>>> dates.append('1 12 2017')
>>> dates.append('June 1 2017 1:30:00AM')
>>> [parser.parse(x) for x in dates]

输出

[datetime.datetime(2017, 12, 1, 0, 0), datetime.datetime(2017, 1, 1, 0, 0), datetime.datetime(2017, 1, 12, 0, 0), datetime.datetime(2017, 6, 1, 1, 30)]

Remember this and you didn’t need to get confused in datetime conversion again.

String to datetime object = strptime

datetime object to other formats = strftime

Jun 1 2005 1:33PM

is equals to

%b %d %Y %I:%M%p

%b Month as locale’s abbreviated name(Jun)

%d Day of the month as a zero-padded decimal number(1)

%Y Year with century as a decimal number(2015)

%I Hour (12-hour clock) as a zero-padded decimal number(01)

%M Minute as a zero-padded decimal number(33)

%p Locale’s equivalent of either AM or PM(PM)

so you need strptime i-e converting string to

>>> dates = []
>>> dates.append('Jun 1 2005  1:33PM')
>>> dates.append('Aug 28 1999 12:00AM')
>>> from datetime import datetime
>>> for d in dates:
...     date = datetime.strptime(d, '%b %d %Y %I:%M%p')
...     print type(date)
...     print date
... 

Output

<type 'datetime.datetime'>
2005-06-01 13:33:00
<type 'datetime.datetime'>
1999-08-28 00:00:00

What if you have different format of dates you can use panda or dateutil.parse

>>> import dateutil
>>> dates = []
>>> dates.append('12 1 2017')
>>> dates.append('1 1 2017')
>>> dates.append('1 12 2017')
>>> dates.append('June 1 2017 1:30:00AM')
>>> [parser.parse(x) for x in dates]

OutPut

[datetime.datetime(2017, 12, 1, 0, 0), datetime.datetime(2017, 1, 1, 0, 0), datetime.datetime(2017, 1, 12, 0, 0), datetime.datetime(2017, 6, 1, 1, 30)]

回答 5

在Python> = 3.7.0中,

转换YYYY-MM-DD字符串DateTime对象datetime.fromisoformat都可以使用。

>>> from datetime import datetime

>>> date_string = "2012-12-12 10:10:10"
>>> print (datetime.fromisoformat(date_string))
>>> 2012-12-12 10:10:10

In Python >= 3.7.0,

to convert YYYY-MM-DD string to datetime object, datetime.fromisoformat could be used.

>>> from datetime import datetime

>>> date_string = "2012-12-12 10:10:10"
>>> print (datetime.fromisoformat(date_string))
>>> 2012-12-12 10:10:10

回答 6

许多时间戳都有一个隐含的时区。为了确保您的代码在每个时区都能工作,您应该在内部使用UTC,并在每次异物进入系统时都附加一个时区。

Python 3.2+:

>>> datetime.datetime.strptime(
...     "March 5, 2014, 20:13:50", "%B %d, %Y, %H:%M:%S"
... ).replace(tzinfo=datetime.timezone(datetime.timedelta(hours=-3)))

Many timestamps have an implied timezone. To ensure that your code will work in every timezone, you should use UTC internally and attach a timezone each time a foreign object enters the system.

Python 3.2+:

>>> datetime.datetime.strptime(
...     "March 5, 2014, 20:13:50", "%B %d, %Y, %H:%M:%S"
... ).replace(tzinfo=datetime.timezone(datetime.timedelta(hours=-3)))

回答 7

这是两个使用Pandas将格式为字符串的日期转换为datetime.date对象的解决方案。

import pandas as pd

dates = ['2015-12-25', '2015-12-26']

# 1) Use a list comprehension.
>>> [d.date() for d in pd.to_datetime(dates)]
[datetime.date(2015, 12, 25), datetime.date(2015, 12, 26)]

# 2) Convert the dates to a DatetimeIndex and extract the python dates.
>>> pd.DatetimeIndex(dates).date.tolist()
[datetime.date(2015, 12, 25), datetime.date(2015, 12, 26)]

时机

dates = pd.DatetimeIndex(start='2000-1-1', end='2010-1-1', freq='d').date.tolist()

>>> %timeit [d.date() for d in pd.to_datetime(dates)]
# 100 loops, best of 3: 3.11 ms per loop

>>> %timeit pd.DatetimeIndex(dates).date.tolist()
# 100 loops, best of 3: 6.85 ms per loop

这是如何转换OP的原始日期时间示例:

datetimes = ['Jun 1 2005  1:33PM', 'Aug 28 1999 12:00AM']

>>> pd.to_datetime(datetimes).to_pydatetime().tolist()
[datetime.datetime(2005, 6, 1, 13, 33), 
 datetime.datetime(1999, 8, 28, 0, 0)]

使用可以从字符串转换为Pandas Timestamps有很多选项to_datetime,因此请检查文档如果需要任何特殊。

同样,时间戳除了具有许多可访问的属性和方法外,.date

Here are two solutions using Pandas to convert dates formatted as strings into datetime.date objects.

import pandas as pd

dates = ['2015-12-25', '2015-12-26']

# 1) Use a list comprehension.
>>> [d.date() for d in pd.to_datetime(dates)]
[datetime.date(2015, 12, 25), datetime.date(2015, 12, 26)]

# 2) Convert the dates to a DatetimeIndex and extract the python dates.
>>> pd.DatetimeIndex(dates).date.tolist()
[datetime.date(2015, 12, 25), datetime.date(2015, 12, 26)]

Timings

dates = pd.DatetimeIndex(start='2000-1-1', end='2010-1-1', freq='d').date.tolist()

>>> %timeit [d.date() for d in pd.to_datetime(dates)]
# 100 loops, best of 3: 3.11 ms per loop

>>> %timeit pd.DatetimeIndex(dates).date.tolist()
# 100 loops, best of 3: 6.85 ms per loop

And here is how to convert the OP’s original date-time examples:

datetimes = ['Jun 1 2005  1:33PM', 'Aug 28 1999 12:00AM']

>>> pd.to_datetime(datetimes).to_pydatetime().tolist()
[datetime.datetime(2005, 6, 1, 13, 33), 
 datetime.datetime(1999, 8, 28, 0, 0)]

There are many options for converting from the strings to Pandas Timestamps using to_datetime, so check the docs if you need anything special.

Likewise, Timestamps have many properties and methods that can be accessed in addition to .date


回答 8

我个人喜欢使用parser模块的解决方案,这是该问题的第二个答案,而且很漂亮,因为您不必构造任何字符串文字即可使其工作。但是,缺点是它比接受的答案慢90%strptime

from dateutil import parser
from datetime import datetime
import timeit

def dt():
    dt = parser.parse("Jun 1 2005  1:33PM")
def strptime():
    datetime_object = datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

print(timeit.timeit(stmt=dt, number=10**5))
print(timeit.timeit(stmt=strptime, number=10**5))
>10.70296801342902
>1.3627995655316933

只要你是不是做这个一百万一遍又一遍的时间,我还是觉得parser方法是更方便,会自动处理大部分的时间格式。

I personally like the solution using the parser module, which is the second Answer to this question and is beautiful, as you don’t have to construct any string literals to get it working. BUT, one downside is that it is 90% slower than the accepted answer with strptime.

from dateutil import parser
from datetime import datetime
import timeit

def dt():
    dt = parser.parse("Jun 1 2005  1:33PM")
def strptime():
    datetime_object = datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

print(timeit.timeit(stmt=dt, number=10**5))
print(timeit.timeit(stmt=strptime, number=10**5))
>10.70296801342902
>1.3627995655316933

As long as you are not doing this a million times over and over again, I still think the parser method is more convenient and will handle most of the time formats automatically.


回答 9

这里没有提到但有用的东西:在一天中添加一个后缀。我解耦了后缀逻辑,以便您可以将其用于任何您喜欢的数字,而不仅仅是日期。

import time

def num_suffix(n):
    '''
    Returns the suffix for any given int
    '''
    suf = ('th','st', 'nd', 'rd')
    n = abs(n) # wise guy
    tens = int(str(n)[-2:])
    units = n % 10
    if tens > 10 and tens < 20:
        return suf[0] # teens with 'th'
    elif units <= 3:
        return suf[units]
    else:
        return suf[0] # 'th'

def day_suffix(t):
    '''
    Returns the suffix of the given struct_time day
    '''
    return num_suffix(t.tm_mday)

# Examples
print num_suffix(123)
print num_suffix(3431)
print num_suffix(1234)
print ''
print day_suffix(time.strptime("1 Dec 00", "%d %b %y"))
print day_suffix(time.strptime("2 Nov 01", "%d %b %y"))
print day_suffix(time.strptime("3 Oct 02", "%d %b %y"))
print day_suffix(time.strptime("4 Sep 03", "%d %b %y"))
print day_suffix(time.strptime("13 Nov 90", "%d %b %y"))
print day_suffix(time.strptime("14 Oct 10", "%d %b %y"))​​​​​​​

Something that isn’t mentioned here and is useful: adding a suffix to the day. I decoupled the suffix logic so you can use it for any number you like, not just dates.

import time

def num_suffix(n):
    '''
    Returns the suffix for any given int
    '''
    suf = ('th','st', 'nd', 'rd')
    n = abs(n) # wise guy
    tens = int(str(n)[-2:])
    units = n % 10
    if tens > 10 and tens < 20:
        return suf[0] # teens with 'th'
    elif units <= 3:
        return suf[units]
    else:
        return suf[0] # 'th'

def day_suffix(t):
    '''
    Returns the suffix of the given struct_time day
    '''
    return num_suffix(t.tm_mday)

# Examples
print num_suffix(123)
print num_suffix(3431)
print num_suffix(1234)
print ''
print day_suffix(time.strptime("1 Dec 00", "%d %b %y"))
print day_suffix(time.strptime("2 Nov 01", "%d %b %y"))
print day_suffix(time.strptime("3 Oct 02", "%d %b %y"))
print day_suffix(time.strptime("4 Sep 03", "%d %b %y"))
print day_suffix(time.strptime("13 Nov 90", "%d %b %y"))
print day_suffix(time.strptime("14 Oct 10", "%d %b %y"))​​​​​​​

回答 10

In [34]: import datetime

In [35]: _now = datetime.datetime.now()

In [36]: _now
Out[36]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)

In [37]: print _now
2016-01-19 09:47:00.432000

In [38]: _parsed = datetime.datetime.strptime(str(_now),"%Y-%m-%d %H:%M:%S.%f")

In [39]: _parsed
Out[39]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)

In [40]: assert _now == _parsed
In [34]: import datetime

In [35]: _now = datetime.datetime.now()

In [36]: _now
Out[36]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)

In [37]: print _now
2016-01-19 09:47:00.432000

In [38]: _parsed = datetime.datetime.strptime(str(_now),"%Y-%m-%d %H:%M:%S.%f")

In [39]: _parsed
Out[39]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)

In [40]: assert _now == _parsed

回答 11

Django时区感知日期时间对象示例。

import datetime
from django.utils.timezone import get_current_timezone
tz = get_current_timezone()

format = '%b %d %Y %I:%M%p'
date_object = datetime.datetime.strptime('Jun 1 2005  1:33PM', format)
date_obj = tz.localize(date_object)

具备USE_TZ = True以下条件时,此转换对Django和Python非常重要:

RuntimeWarning: DateTimeField MyModel.created received a naive datetime (2016-03-04 00:00:00) while time zone support is active.

Django Timezone aware datetime object example.

import datetime
from django.utils.timezone import get_current_timezone
tz = get_current_timezone()

format = '%b %d %Y %I:%M%p'
date_object = datetime.datetime.strptime('Jun 1 2005  1:33PM', format)
date_obj = tz.localize(date_object)

This conversion is very important for Django and Python when you have USE_TZ = True:

RuntimeWarning: DateTimeField MyModel.created received a naive datetime (2016-03-04 00:00:00) while time zone support is active.

回答 12

创建一个小的实用程序函数,例如:

def date(datestr="", format="%Y-%m-%d"):
    from datetime import datetime
    if not datestr:
        return datetime.today().date()
    return datetime.strptime(datestr, format).date()

这足够通用:

  • 如果您不传递任何参数,它将返回今天的日期。
  • 有一种默认的日期格式可以覆盖。
  • 您可以轻松地对其进行修改以返回日期时间。

Create a small utility function like:

def date(datestr="", format="%Y-%m-%d"):
    from datetime import datetime
    if not datestr:
        return datetime.today().date()
    return datetime.strptime(datestr, format).date()

This is versatile enough:

  • If you don’t pass any arguments it will return today’s date.
  • There’s a date format as default that you can override.
  • You can easily modify it to return a datetime.

回答 13

它将有助于将字符串转换为日期时间以及时区

def convert_string_to_time(date_string, timezone):
    from datetime import datetime
    import pytz
    date_time_obj = datetime.strptime(date_string[:26], '%Y-%m-%d %H:%M:%S.%f')
    date_time_obj_timezone = pytz.timezone(timezone).localize(date_time_obj)

    return date_time_obj_timezone

date = '2018-08-14 13:09:24.543953+00:00'
TIME_ZONE = 'UTC'
date_time_obj_timezone = convert_string_to_time(date, TIME_ZONE)

It would do the helpful for converting string to datetime and also with time zone

def convert_string_to_time(date_string, timezone):
    from datetime import datetime
    import pytz
    date_time_obj = datetime.strptime(date_string[:26], '%Y-%m-%d %H:%M:%S.%f')
    date_time_obj_timezone = pytz.timezone(timezone).localize(date_time_obj)

    return date_time_obj_timezone

date = '2018-08-14 13:09:24.543953+00:00'
TIME_ZONE = 'UTC'
date_time_obj_timezone = convert_string_to_time(date, TIME_ZONE)

回答 14

arrow提供了许多有用的日期和时间功能。这段代码提供了对该问题的答案,并表明箭头还能够轻松格式化日期并显示其他语言环境的信息。

>>> import arrow
>>> dateStrings = [ 'Jun 1  2005 1:33PM', 'Aug 28 1999 12:00AM' ]
>>> for dateString in dateStrings:
...     dateString
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').datetime
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').format('ddd, Do MMM YYYY HH:mm')
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').humanize(locale='de')
...
'Jun 1  2005 1:33PM'
datetime.datetime(2005, 6, 1, 13, 33, tzinfo=tzutc())
'Wed, 1st Jun 2005 13:33'
'vor 11 Jahren'
'Aug 28 1999 12:00AM'
datetime.datetime(1999, 8, 28, 0, 0, tzinfo=tzutc())
'Sat, 28th Aug 1999 00:00'
'vor 17 Jahren'

有关更多信息,请参见http://arrow.readthedocs.io/en/latest/

arrow offers many useful functions for dates and times. This bit of code provides an answer to the question and shows that arrow is also capable of formatting dates easily and displaying information for other locales.

>>> import arrow
>>> dateStrings = [ 'Jun 1  2005 1:33PM', 'Aug 28 1999 12:00AM' ]
>>> for dateString in dateStrings:
...     dateString
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').datetime
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').format('ddd, Do MMM YYYY HH:mm')
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').humanize(locale='de')
...
'Jun 1  2005 1:33PM'
datetime.datetime(2005, 6, 1, 13, 33, tzinfo=tzutc())
'Wed, 1st Jun 2005 13:33'
'vor 11 Jahren'
'Aug 28 1999 12:00AM'
datetime.datetime(1999, 8, 28, 0, 0, tzinfo=tzutc())
'Sat, 28th Aug 1999 00:00'
'vor 17 Jahren'

See http://arrow.readthedocs.io/en/latest/ for more.


回答 15

您可以使用easy_date使其变得容易:

import date_converter
converted_date = date_converter.string_to_datetime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

You can use easy_date to make it easy:

import date_converter
converted_date = date_converter.string_to_datetime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

回答 16

如果只需要日期格式,则可以通过传递各个字段来手动将其转换,例如:

>>> import datetime
>>> date = datetime.date(int('2017'),int('12'),int('21'))
>>> date
datetime.date(2017, 12, 21)
>>> type(date)
<type 'datetime.date'>

您可以传递拆分的字符串值以将其转换为日期类型,例如:

selected_month_rec = '2017-09-01'
date_formate = datetime.date(int(selected_month_rec.split('-')[0]),int(selected_month_rec.split('-')[1]),int(selected_month_rec.split('-')[2]))

您将获得日期格式的结果值。

If you want only date format then you can manually convert it by passing your individual fields like:

>>> import datetime
>>> date = datetime.date(int('2017'),int('12'),int('21'))
>>> date
datetime.date(2017, 12, 21)
>>> type(date)
<type 'datetime.date'>

You can pass your split string values to convert it into date type like:

selected_month_rec = '2017-09-01'
date_formate = datetime.date(int(selected_month_rec.split('-')[0]),int(selected_month_rec.split('-')[1]),int(selected_month_rec.split('-')[2]))

You will get the resulting value in date format.


回答 17

您也可以退房 dateparser

dateparser 提供的模块可轻松解析几乎任何网页上常见的字符串格式的本地化日期。

安装:

$ pip install dateparser

我认为,这是解析日期的最简单方法。

最直接的方法是使用dateparser.parse功能,该功能包装了模块中的大多数功能。

样例代码:

import dateparser

t1 = 'Jun 1 2005  1:33PM'
t2 = 'Aug 28 1999 12:00AM'

dt1 = dateparser.parse(t1)
dt2 = dateparser.parse(t2)

print(dt1)
print(dt2)

输出:

2005-06-01 13:33:00
1999-08-28 00:00:00

You can also check out dateparser

dateparser provides modules to easily parse localized dates in almost any string formats commonly found on web pages.

Install:

$ pip install dateparser

This is, I think, the easiest way you can parse dates.

The most straightforward way is to use the dateparser.parse function, that wraps around most of the functionality in the module.

Sample Code:

import dateparser

t1 = 'Jun 1 2005  1:33PM'
t2 = 'Aug 28 1999 12:00AM'

dt1 = dateparser.parse(t1)
dt2 = dateparser.parse(t2)

print(dt1)
print(dt2)

Output:

2005-06-01 13:33:00
1999-08-28 00:00:00

回答 18

我的回答

在现实世界的数据中,这是一个实际的问题:多种,不匹配,不完整,不一致以及多语言/区域日期格式,通常在一个数据集中自由地混合使用。生产代码失败是不可能的,更不用说像狐狸一样的异常快乐了。

我们需要尝试…捕获多种日期时间格式fmt1,fmt2,…,fmtn,并strptime()为所有不匹配的对象抑制/处理(来自的)异常(尤其是避免使用yukky n缩进的try梯形图) ..catch子句)。从我的解决方案

def try_strptime(s, fmts=['%d-%b-%y','%m/%d/%Y']):
    for fmt in fmts:
        try:
            return datetime.strptime(s, fmt)
        except:
            continue

    return None # or reraise the ValueError if no format matched, if you prefer

See my answer.

In real-world data this is a real problem: multiple, mismatched, incomplete, inconsistent and multilanguage/region date formats, often mixed freely in one dataset. It’s not ok for production code to fail, let alone go exception-happy like a fox.

We need to try…catch multiple datetime formats fmt1,fmt2,…,fmtn and suppress/handle the exceptions (from strptime()) for all those that mismatch (and in particular, avoid needing a yukky n-deep indented ladder of try..catch clauses). From my solution

def try_strptime(s, fmts=['%d-%b-%y','%m/%d/%Y']):
    for fmt in fmts:
        try:
            return datetime.strptime(s, fmt)
        except:
            continue

    return None # or reraise the ValueError if no format matched, if you prefer

回答 19

emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv")
emp.info()

它显示“开始日期时间”列和“上次登录时间”在数据框中均为“对象=字符串”

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name           933 non-null object
Gender               855 non-null object
Start Date           1000 non-null object

Last Login Time      1000 non-null object
Salary               1000 non-null int64
Bonus %              1000 non-null float64
Senior Management    933 non-null object
Team                 957 non-null object
dtypes: float64(1), int64(1), object(6)
memory usage: 62.6+ KB

通过使用parse_dates选项,read_csv您可以将字符串datetime转换为pandas datetime格式。

emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv", parse_dates=["Start Date", "Last Login Time"])
emp.info()


<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name           933 non-null object
Gender               855 non-null object
Start Date           1000 non-null datetime64[ns]
Last Login Time      1000 non-null datetime64[ns]
Salary               1000 non-null int64
Bonus %              1000 non-null float64
Senior Management    933 non-null object
Team                 957 non-null object
dtypes: datetime64[ns](2), float64(1), int64(1), object(4)
memory usage: 62.6+ KB
emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv")
emp.info()

it shows “Start Date Time” Column and “Last Login Time” both are “object = strings” in data-frame

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name           933 non-null object
Gender               855 non-null object
Start Date           1000 non-null object

Last Login Time      1000 non-null object
Salary               1000 non-null int64
Bonus %              1000 non-null float64
Senior Management    933 non-null object
Team                 957 non-null object
dtypes: float64(1), int64(1), object(6)
memory usage: 62.6+ KB

By using parse_dates option in read_csv mention you can convert your string datetime into pandas datetime format.

emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv", parse_dates=["Start Date", "Last Login Time"])
emp.info()


<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name           933 non-null object
Gender               855 non-null object
Start Date           1000 non-null datetime64[ns]
Last Login Time      1000 non-null datetime64[ns]
Salary               1000 non-null int64
Bonus %              1000 non-null float64
Senior Management    933 non-null object
Team                 957 non-null object
dtypes: datetime64[ns](2), float64(1), int64(1), object(4)
memory usage: 62.6+ KB

如何删除文件或文件夹?

问题:如何删除文件或文件夹?

如何在Python中删除文件或文件夹?

How to delete a file or folder in Python?


回答 0


PathPython 3.4+ pathlib模块中的对象还公开了这些实例方法:


Path objects from the Python 3.4+ pathlib module also expose these instance methods:


回答 1

Python语法删除文件

import os
os.remove("/tmp/<file_name>.txt")

要么

import os
os.unlink("/tmp/<file_name>.txt")

要么

适用于Python版本> 3.5的pathlib

file_to_rem = pathlib.Path("/tmp/<file_name>.txt")
file_to_rem.unlink()

Path.unlink(missing_ok = False)

Unlink方法用于删除文件或符号链接。

如果missing_ok为false(默认值),则在路径不存在时引发FileNotFoundError。
如果missing_ok为true,则将忽略FileNotFoundError异常(与POSIX rm -f命令相同的行为)。
在版本3.8中更改:添加了missing_ok参数。

最佳实践

  1. 首先,检查文件或文件夹是否存在,然后仅删除该文件。这可以通过两种方式实现:
    一。os.path.isfile("/path/to/file")
    b。采用exception handling.

实例os.path.isfile

#!/usr/bin/python
import os
myfile="/tmp/foo.txt"

## If file exists, delete it ##
if os.path.isfile(myfile):
    os.remove(myfile)
else:    ## Show an error ##
    print("Error: %s file not found" % myfile)

异常处理

#!/usr/bin/python
import os

## Get input ##
myfile= raw_input("Enter file name to delete: ")

## Try to delete the file ##
try:
    os.remove(myfile)
except OSError as e:  ## if failed, report it back to the user ##
    print ("Error: %s - %s." % (e.filename, e.strerror))

预期输出

输入要删除的文件名:demo.txt
错误:demo.txt-没有这样的文件或目录。

输入要删除的文件名:rrr.txt
错误:rrr.txt-不允许操作。

输入要删除的文件名:foo.txt

删除文件夹的Python语法

shutil.rmtree()

范例 shutil.rmtree()

#!/usr/bin/python
import os
import sys
import shutil

# Get directory name
mydir= raw_input("Enter directory name: ")

## Try to remove tree; if failed show an error using try...except on screen
try:
    shutil.rmtree(mydir)
except OSError as e:
    print ("Error: %s - %s." % (e.filename, e.strerror))

Python syntax to delete a file

import os
os.remove("/tmp/<file_name>.txt")

Or

import os
os.unlink("/tmp/<file_name>.txt")

Or

pathlib Library for Python version > 3.5

file_to_rem = pathlib.Path("/tmp/<file_name>.txt")
file_to_rem.unlink()

Path.unlink(missing_ok=False)

Unlink method used to remove the file or the symbolik link.

If missing_ok is false (the default), FileNotFoundError is raised if the path does not exist.
If missing_ok is true, FileNotFoundError exceptions will be ignored (same behavior as the POSIX rm -f command).
Changed in version 3.8: The missing_ok parameter was added.

Best practice

  1. First, check whether the file or folder exists or not then only delete that file. This can be achieved in two ways :
    a. os.path.isfile("/path/to/file")
    b. Use exception handling.

EXAMPLE for os.path.isfile

#!/usr/bin/python
import os
myfile="/tmp/foo.txt"

## If file exists, delete it ##
if os.path.isfile(myfile):
    os.remove(myfile)
else:    ## Show an error ##
    print("Error: %s file not found" % myfile)

Exception Handling

#!/usr/bin/python
import os

## Get input ##
myfile= raw_input("Enter file name to delete: ")

## Try to delete the file ##
try:
    os.remove(myfile)
except OSError as e:  ## if failed, report it back to the user ##
    print ("Error: %s - %s." % (e.filename, e.strerror))

RESPECTIVE OUTPUT

Enter file name to delete : demo.txt
Error: demo.txt - No such file or directory.

Enter file name to delete : rrr.txt
Error: rrr.txt - Operation not permitted.

Enter file name to delete : foo.txt

Python syntax to delete a folder

shutil.rmtree()

Example for shutil.rmtree()

#!/usr/bin/python
import os
import sys
import shutil

# Get directory name
mydir= raw_input("Enter directory name: ")

## Try to remove tree; if failed show an error using try...except on screen
try:
    shutil.rmtree(mydir)
except OSError as e:
    print ("Error: %s - %s." % (e.filename, e.strerror))

回答 2

采用

shutil.rmtree(path[, ignore_errors[, onerror]])

(请参阅关于shutil的完整文档)和/或

os.remove

os.rmdir

(关于os的完整文档。)

Use

shutil.rmtree(path[, ignore_errors[, onerror]])

(See complete documentation on shutil) and/or

os.remove

and

os.rmdir

(Complete documentation on os.)


回答 3

这是同时使用os.remove和的强大功能shutil.rmtree

def remove(path):
    """ param <path> could either be relative or absolute. """
    if os.path.isfile(path) or os.path.islink(path):
        os.remove(path)  # remove the file
    elif os.path.isdir(path):
        shutil.rmtree(path)  # remove dir and all contains
    else:
        raise ValueError("file {} is not a file or dir.".format(path))

Here is a robust function that uses both os.remove and shutil.rmtree:

def remove(path):
    """ param <path> could either be relative or absolute. """
    if os.path.isfile(path) or os.path.islink(path):
        os.remove(path)  # remove the file
    elif os.path.isdir(path):
        shutil.rmtree(path)  # remove dir and all contains
    else:
        raise ValueError("file {} is not a file or dir.".format(path))

回答 4

您可以使用内置的pathlib模块(需要Python 3.4+,但也有旧版本PyPI上的反向移植:pathlibpathlib2)。

要删除文件,可以使用以下unlink方法:

import pathlib
path = pathlib.Path(name_of_file)
path.unlink()

rmdir删除文件夹的方法:

import pathlib
path = pathlib.Path(name_of_folder)
path.rmdir()

You can use the built-in pathlib module (requires Python 3.4+, but there are backports for older versions on PyPI: pathlib, pathlib2).

To remove a file there is the unlink method:

import pathlib
path = pathlib.Path(name_of_file)
path.unlink()

Or the rmdir method to remove an empty folder:

import pathlib
path = pathlib.Path(name_of_folder)
path.rmdir()

回答 5

如何在Python中删除文件或文件夹?

对于Python 3,要分别删除文件和目录,请分别使用unlink和对象方法:rmdir Path

from pathlib import Path
dir_path = Path.home() / 'directory' 
file_path = dir_path / 'file'

file_path.unlink() # remove file

dir_path.rmdir()   # remove directory

请注意,您还可以将相对路径与Path对象一起使用,并且可以使用来检查当前的工作目录Path.cwd

要在Python 2中删除单个文件和目录,请参见下面标记的部分。

要删除包含目录的目录,请使用shutil.rmtree,请注意,该目录在Python 2和3中可用:

from shutil import rmtree

rmtree(dir_path)

示范

Path对象是Python 3.4中的新增功能。

让我们用一个目录和文件来演示用法。请注意,我们使用/来连接路径的各个部分,这解决了操作系统之间的问题以及Windows上使用反斜杠(在其中您需要将反斜杠加倍,\\或者使用原始字符串,如r"foo\bar")引起的问题:

from pathlib import Path

# .home() is new in 3.5, otherwise use os.path.expanduser('~')
directory_path = Path.home() / 'directory'
directory_path.mkdir()

file_path = directory_path / 'file'
file_path.touch()

现在:

>>> file_path.is_file()
True

现在让我们删除它们。首先文件:

>>> file_path.unlink()     # remove file
>>> file_path.is_file()
False
>>> file_path.exists()
False

我们可以使用通配符删除多个文件-首先,我们为此创建一些文件:

>>> (directory_path / 'foo.my').touch()
>>> (directory_path / 'bar.my').touch()

然后只需遍历全局模式:

>>> for each_file_path in directory_path.glob('*.my'):
...     print(f'removing {each_file_path}')
...     each_file_path.unlink()
... 
removing ~/directory/foo.my
removing ~/directory/bar.my

现在,演示删除目录:

>>> directory_path.rmdir() # remove directory
>>> directory_path.is_dir()
False
>>> directory_path.exists()
False

如果我们要删除目录及其中的所有内容怎么办?对于此用例,请使用shutil.rmtree

让我们重新创建目录和文件:

file_path.parent.mkdir()
file_path.touch()

并注意rmdir除非它为空,否则它将失败,这就是rmtree如此方便的原因:

>>> directory_path.rmdir()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "~/anaconda3/lib/python3.6/pathlib.py", line 1270, in rmdir
    self._accessor.rmdir(self)
  File "~/anaconda3/lib/python3.6/pathlib.py", line 387, in wrapped
    return strfunc(str(pathobj), *args)
OSError: [Errno 39] Directory not empty: '/home/username/directory'

现在,导入rmtree并将目录传递给该功能:

from shutil import rmtree
rmtree(directory_path)      # remove everything 

我们可以看到整个内容已被删除:

>>> directory_path.exists()
False

Python 2

如果您使用的是Python 2,则有一个名为pathlib2的pathlib模块的反向端口,可以使用pip进行安装:

$ pip install pathlib2

然后您可以将库别名为 pathlib

import pathlib2 as pathlib

或者直接导入Path对象(如此处所示):

from pathlib2 import Path

如果太多,您可以使用删除文件os.removeos.unlink

from os import unlink, remove
from os.path import join, expanduser

remove(join(expanduser('~'), 'directory/file'))

要么

unlink(join(expanduser('~'), 'directory/file'))

您可以使用以下命令删除目录os.rmdir

from os import rmdir

rmdir(join(expanduser('~'), 'directory'))

请注意,还有一个os.removedirs-它仅以递归方式删除空目录,但它可能适合您的用例。

How do I delete a file or folder in Python?

For Python 3, to remove the file and directory individually, use the unlink and rmdir Path object methods respectively:

from pathlib import Path
dir_path = Path.home() / 'directory' 
file_path = dir_path / 'file'

file_path.unlink() # remove file

dir_path.rmdir()   # remove directory

Note that you can also use relative paths with Path objects, and you can check your current working directory with Path.cwd.

For removing individual files and directories in Python 2, see the section so labeled below.

To remove a directory with contents, use shutil.rmtree, and note that this is available in Python 2 and 3:

from shutil import rmtree

rmtree(dir_path)

Demonstration

New in Python 3.4 is the Path object.

Let’s use one to create a directory and file to demonstrate usage. Note that we use the / to join the parts of the path, this works around issues between operating systems and issues from using backslashes on Windows (where you’d need to either double up your backslashes like \\ or use raw strings, like r"foo\bar"):

from pathlib import Path

# .home() is new in 3.5, otherwise use os.path.expanduser('~')
directory_path = Path.home() / 'directory'
directory_path.mkdir()

file_path = directory_path / 'file'
file_path.touch()

and now:

>>> file_path.is_file()
True

Now let’s delete them. First the file:

>>> file_path.unlink()     # remove file
>>> file_path.is_file()
False
>>> file_path.exists()
False

We can use globbing to remove multiple files – first let’s create a few files for this:

>>> (directory_path / 'foo.my').touch()
>>> (directory_path / 'bar.my').touch()

Then just iterate over the glob pattern:

>>> for each_file_path in directory_path.glob('*.my'):
...     print(f'removing {each_file_path}')
...     each_file_path.unlink()
... 
removing ~/directory/foo.my
removing ~/directory/bar.my

Now, demonstrating removing the directory:

>>> directory_path.rmdir() # remove directory
>>> directory_path.is_dir()
False
>>> directory_path.exists()
False

What if we want to remove a directory and everything in it? For this use-case, use shutil.rmtree

Let’s recreate our directory and file:

file_path.parent.mkdir()
file_path.touch()

and note that rmdir fails unless it’s empty, which is why rmtree is so convenient:

>>> directory_path.rmdir()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "~/anaconda3/lib/python3.6/pathlib.py", line 1270, in rmdir
    self._accessor.rmdir(self)
  File "~/anaconda3/lib/python3.6/pathlib.py", line 387, in wrapped
    return strfunc(str(pathobj), *args)
OSError: [Errno 39] Directory not empty: '/home/username/directory'

Now, import rmtree and pass the directory to the funtion:

from shutil import rmtree
rmtree(directory_path)      # remove everything 

and we can see the whole thing has been removed:

>>> directory_path.exists()
False

Python 2

If you’re on Python 2, there’s a backport of the pathlib module called pathlib2, which can be installed with pip:

$ pip install pathlib2

And then you can alias the library to pathlib

import pathlib2 as pathlib

Or just directly import the Path object (as demonstrated here):

from pathlib2 import Path

If that’s too much, you can remove files with os.remove or os.unlink

from os import unlink, remove
from os.path import join, expanduser

remove(join(expanduser('~'), 'directory/file'))

or

unlink(join(expanduser('~'), 'directory/file'))

and you can remove directories with os.rmdir:

from os import rmdir

rmdir(join(expanduser('~'), 'directory'))

Note that there is also a os.removedirs – it only removes empty directories recursively, but it may suit your use-case.


回答 6

import os

folder = '/Path/to/yourDir/'
fileList = os.listdir(folder)

for f in fileList:
    filePath = folder + '/'+f

    if os.path.isfile(filePath):
        os.remove(filePath)

    elif os.path.isdir(filePath):
        newFileList = os.listdir(filePath)
        for f1 in newFileList:
            insideFilePath = filePath + '/' + f1

            if os.path.isfile(insideFilePath):
                os.remove(insideFilePath)
import os

folder = '/Path/to/yourDir/'
fileList = os.listdir(folder)

for f in fileList:
    filePath = folder + '/'+f

    if os.path.isfile(filePath):
        os.remove(filePath)

    elif os.path.isdir(filePath):
        newFileList = os.listdir(filePath)
        for f1 in newFileList:
            insideFilePath = filePath + '/' + f1

            if os.path.isfile(insideFilePath):
                os.remove(insideFilePath)

回答 7

shutil.rmtree是异步函数,因此,如果要检查它是否完成,可以使用while … loop

import os
import shutil

shutil.rmtree(path)

while os.path.exists(path):
  pass

print('done')

shutil.rmtree is the asynchronous function, so if you want to check when it complete, you can use while…loop

import os
import shutil

shutil.rmtree(path)

while os.path.exists(path):
  pass

print('done')

回答 8

删除文件:

os.unlink(path, *, dir_fd=None)

要么

os.remove(path, *, dir_fd=None)

这两个功能在语义上是相同的。此功能删除(删除)文件路径。如果path不是文件,而是目录,则会引发异常。

删除文件夹:

shutil.rmtree(path, ignore_errors=False, onerror=None)

要么

os.rmdir(path, *, dir_fd=None)

为了删除整个目录树,shutil.rmtree()可以使用。os.rmdir仅在目录为空且存在时才起作用。

要递归删除父文件夹:

os.removedirs(name)

它用self删除每个空的父目录,直到有一些内容的父目录为止

例如 os.removedirs(’abc / xyz / pqr’)如果目录为空,则会按顺序abc / xyz / pqr,abc / xyz和abc删除目录。

欲了解更多信息检查官方文档:os.unlinkos.removeos.rmdirshutil.rmtreeos.removedirs

For deleting files:

os.unlink(path, *, dir_fd=None)

or

os.remove(path, *, dir_fd=None)

Both functions are semantically same. This functions removes (deletes) the file path. If path is not a file and it is directory, then exception is raised.

For deleting folders:

shutil.rmtree(path, ignore_errors=False, onerror=None)

or

os.rmdir(path, *, dir_fd=None)

In order to remove whole directory trees, shutil.rmtree() can be used. os.rmdir only works when the directory is empty and exists.

For deleting folders recursively towards parent:

os.removedirs(name)

It remove every empty parent directory with self until parent which has some content

ex. os.removedirs(‘abc/xyz/pqr’) will remove the directories by order ‘abc/xyz/pqr’, ‘abc/xyz’ and ‘abc’ if they are empty.

For more info check official doc: os.unlink , os.remove, os.rmdir , shutil.rmtree, os.removedirs


回答 9

删除文件夹中的所有文件

import os
import glob

files = glob.glob(os.path.join('path/to/folder/*'))
files = glob.glob(os.path.join('path/to/folder/*.csv')) // It will give all csv files in folder
for file in files:
    os.remove(file)

删除目录中的所有文件夹

from shutil import rmtree
import os

// os.path.join()  # current working directory.

for dirct in os.listdir(os.path.join('path/to/folder')):
    rmtree(os.path.join('path/to/folder',dirct))

To remove all files in folder

import os
import glob

files = glob.glob(os.path.join('path/to/folder/*'))
files = glob.glob(os.path.join('path/to/folder/*.csv')) // It will give all csv files in folder
for file in files:
    os.remove(file)

To remove all folders in a directory

from shutil import rmtree
import os

// os.path.join()  # current working directory.

for dirct in os.listdir(os.path.join('path/to/folder')):
    rmtree(os.path.join('path/to/folder',dirct))

回答 10

为了避免ÉricAraujo 的注释突出显示的TOCTOU问题,您可以捕获异常以调用正确的方法:

def remove_file_or_dir(path: str) -> None:
    """ Remove a file or directory """
    try:
        shutil.rmtree(path)
    except NotADirectoryError:
        os.remove(path)

因为shutil.rmtree()将仅删除目录,os.remove()或者os.unlink()仅将删除文件。

To avoid the TOCTOU issue highlighted by Éric Araujo’s comment, you can catch an exception to call the correct method:

def remove_file_or_dir(path: str) -> None:
    """ Remove a file or directory """
    try:
        shutil.rmtree(path)
    except NotADirectoryError:
        os.remove(path)

Since shutil.rmtree() will only remove directories and os.remove() or os.unlink() will only remove files.


回答 11

subprocess如果您喜欢编写漂亮且易读的代码,那么我建议您使用:

import subprocess
subprocess.Popen("rm -r my_dir", shell=True)

而且,如果您不是软件工程师,那么可以考虑使用Jupyter。您可以简单地输入bash命令:

!rm -r my_dir

传统上,您使用shutil

import shutil
shutil.rmtree(my_dir) 

I recommend using subprocess if writing a beautiful and readable code is your cup of tea:

import subprocess
subprocess.Popen("rm -r my_dir", shell=True)

And if you are not a software engineer, then maybe consider using Jupyter; you can simply type bash commands:

!rm -r my_dir

Traditionally, you use shutil:

import shutil
shutil.rmtree(my_dir) 

如何在Python中获取字符串的子字符串?

问题:如何在Python中获取字符串的子字符串?

有没有一种方法可以在Python中为字符串加上字符串,以从第三个字符到字符串的末尾获取新的字符串?

也许喜欢myString[2:end]吗?

如果离开第二部分意味着“直到最后”,而如果离开第一部分,它是否从头开始?

Is there a way to substring a string in Python, to get a new string from the third character to the end of the string?

Maybe like myString[2:end]?

If leaving the second part means ’till the end’, and if you leave the first part, does it start from the start?


回答 0

>>> x = "Hello World!"
>>> x[2:]
'llo World!'
>>> x[:2]
'He'
>>> x[:-2]
'Hello Worl'
>>> x[-2:]
'd!'
>>> x[2:-2]
'llo Worl'

Python称这个概念为“切片”,它不仅适用于字符串,还适用于更多的领域。看看这里的一个全面的介绍。

>>> x = "Hello World!"
>>> x[2:]
'llo World!'
>>> x[:2]
'He'
>>> x[:-2]
'Hello Worl'
>>> x[-2:]
'd!'
>>> x[2:-2]
'llo Worl'

Python calls this concept “slicing” and it works on more than just strings. Take a look here for a comprehensive introduction.


回答 1

只是为了完整性,没有其他人提到过它。数组切片的第三个参数是一个步骤。因此,反转字符串很简单:

some_string[::-1]

或选择其他字符为:

"H-e-l-l-o- -W-o-r-l-d"[::2] # outputs "Hello World"

在字符串中前进和后退的能力保持了从头到尾排列切片的一致性。

Just for completeness as nobody else has mentioned it. The third parameter to an array slice is a step. So reversing a string is as simple as:

some_string[::-1]

Or selecting alternate characters would be:

"H-e-l-l-o- -W-o-r-l-d"[::2] # outputs "Hello World"

The ability to step forwards and backwards through the string maintains consistency with being able to array slice from the start or end.


回答 2

Substr()通常(即PHP和Perl)以这种方式工作:

s = Substr(s, beginning, LENGTH)

因此参数为beginningLENGTH

但是Python的行为是不同的。它期望从开始到结束(!)。初学者很难发现这一点。因此,正确替换Substr(s,Beginning,LENGTH)是

s = s[ beginning : beginning + LENGTH]

Substr() normally (i.e. PHP and Perl) works this way:

s = Substr(s, beginning, LENGTH)

So the parameters are beginning and LENGTH.

But Python’s behaviour is different; it expects beginning and one after END (!). This is difficult to spot by beginners. So the correct replacement for Substr(s, beginning, LENGTH) is

s = s[ beginning : beginning + LENGTH]

回答 3

实现此目的的一种常见方法是通过字符串切片。

MyString[a:b] 给您一个从索引a到(b-1)的子字符串。

A common way to achieve this is by string slicing.

MyString[a:b] gives you a substring from index a to (b – 1).


回答 4

这里似乎缺少一个示例:完整(浅)副本。

>>> x = "Hello World!"
>>> x
'Hello World!'
>>> x[:]
'Hello World!'
>>> x==x[:]
True
>>>

这是用于创建序列类型(而非插入字符串)的副本的常见用法[:]。浅表复制列表,请参阅无明显原因的Python列表切片语法

One example seems to be missing here: full (shallow) copy.

>>> x = "Hello World!"
>>> x
'Hello World!'
>>> x[:]
'Hello World!'
>>> x==x[:]
True
>>>

This is a common idiom for creating a copy of sequence types (not of interned strings), [:]. Shallow copies a list, see Python list slice syntax used for no obvious reason.


回答 5

有没有一种方法可以在Python中为字符串加上字符串,以从第3个字符到字符串的末尾获取新的字符串?

也许喜欢myString[2:end]吗?

是的,如果您将名称()分配或绑定end到常量单例,这实际上是可行的None

>>> end = None
>>> myString = '1234567890'
>>> myString[2:end]
'34567890'

切片符号具有3个重要参数:

  • 开始

如果未指定,则默认为None-但我们可以显式传递它们:

>>> stop = step = None
>>> start = 2
>>> myString[start:stop:step]
'34567890'

如果离开第二部分意味着“直到最后”,那么如果离开第一部分,它是否从头开始?

是的,例如:

>>> start = None
>>> stop = 2
>>> myString[start:stop:step]
'12'

请注意,我们在切片中包括了开始,但是我们仅上至(不包括)停止。

当step为时None,默认情况下切片将1用于该步骤。如果您使用负整数执行操作,则Python足够聪明,可以从头到尾进行操作。

>>> myString[::-1]
'0987654321'

我在对“解释切片符号问题”的回答中会详细解释切片符号。

Is there a way to substring a string in Python, to get a new string from the 3rd character to the end of the string?

Maybe like myString[2:end]?

Yes, this actually works if you assign, or bind, the name,end, to constant singleton, None:

>>> end = None
>>> myString = '1234567890'
>>> myString[2:end]
'34567890'

Slice notation has 3 important arguments:

  • start
  • stop
  • step

Their defaults when not given are None – but we can pass them explicitly:

>>> stop = step = None
>>> start = 2
>>> myString[start:stop:step]
'34567890'

If leaving the second part means ’till the end’, if you leave the first part, does it start from the start?

Yes, for example:

>>> start = None
>>> stop = 2
>>> myString[start:stop:step]
'12'

Note that we include start in the slice, but we only go up to, and not including, stop.

When step is None, by default the slice uses 1 for the step. If you step with a negative integer, Python is smart enough to go from the end to the beginning.

>>> myString[::-1]
'0987654321'

I explain slice notation in great detail in my answer to Explain slice notation Question.


回答 6

除了“结束”之外,您已经准备就绪。这称为切片符号。您的示例应为:

new_sub_string = myString[2:]

如果省略第二个参数,则它隐式为字符串的结尾。

You’ve got it right there except for “end”. It’s called slice notation. Your example should read:

new_sub_string = myString[2:]

If you leave out the second parameter it is implicitly the end of the string.


回答 7

我想在讨论中添加两点:

  1. 您可以None改为在空白处使用“从头开始”或“到末尾”来指定:

    'abcde'[2:None] == 'abcde'[2:] == 'cde'

    这在不能提供空格作为参数的函数中特别有用:

    def substring(s, start, end):
        """Remove `start` characters from the beginning and `end` 
        characters from the end of string `s`.
    
        Examples
        --------
        >>> substring('abcde', 0, 3)
        'abc'
        >>> substring('abcde', 1, None)
        'bcde'
        """
        return s[start:end]
  2. Python具有切片对象:

    idx = slice(2, None)
    'abcde'[idx] == 'abcde'[2:] == 'cde'

I would like to add two points to the discussion:

  1. You can use None instead on an empty space to specify “from the start” or “to the end”:

    'abcde'[2:None] == 'abcde'[2:] == 'cde'
    

    This is particularly helpful in functions, where you can’t provide an empty space as an argument:

    def substring(s, start, end):
        """Remove `start` characters from the beginning and `end` 
        characters from the end of string `s`.
    
        Examples
        --------
        >>> substring('abcde', 0, 3)
        'abc'
        >>> substring('abcde', 1, None)
        'bcde'
        """
        return s[start:end]
    
  2. Python has slice objects:

    idx = slice(2, None)
    'abcde'[idx] == 'abcde'[2:] == 'cde'
    

回答 8

如果myString包含以偏移量6开始且长度为9的帐号,则可以通过以下方式提取该帐号: acct = myString[6:][:9]

如果OP接受,他们可能想尝试一下,

myString[2:][:999999]

它可以正常工作-不会引发任何错误,也不会发生默认的“字符串填充”。

If myString contains an account number that begins at offset 6 and has length 9, then you can extract the account number this way: acct = myString[6:][:9].

If the OP accepts that, they might want to try, in an experimental fashion,

myString[2:][:999999]

It works – no error is raised, and no default ‘string padding’ occurs.


回答 9

也许我错过了,但是在此页面上找不到原始问题的完整答案,因为这里没有进一步讨论变量。所以我不得不继续寻找。

由于尚未允许我发表评论,因此让我在这里添加我的结论。我确定访问此页面时,我不是唯一对此感兴趣的人:

 >>>myString = 'Hello World'
 >>>end = 5

 >>>myString[2:end]
 'llo'

如果您离开第一部分,您会得到

 >>>myString[:end]
 'Hello' 

如果在中间也留下了:,则会得到最简单的子字符串,它是第5个字符(计数从0开始,因此在这种情况下为空白):

 >>>myString[end]
 ' '

Maybe I missed it, but I couldn’t find a complete answer on this page to the original question(s) because variables are not further discussed here. So I had to go on searching.

Since I’m not yet allowed to comment, let me add my conclusion here. I’m sure I was not the only one interested in it when accessing this page:

 >>>myString = 'Hello World'
 >>>end = 5

 >>>myString[2:end]
 'llo'

If you leave the first part, you get

 >>>myString[:end]
 'Hello' 

And if you left the : in the middle as well you got the simplest substring, which would be the 5th character (count starting with 0, so it’s the blank in this case):

 >>>myString[end]
 ' '

回答 10

好吧,我遇到了需要将PHP脚本转换为Python的情况,并且它有许多用法substr(string, beginning, LENGTH)
如果选择Python,string[beginning:end]则必须计算大量的结束索引,因此更简单的方法是使用string[beginning:][:length],这为我省去了很多麻烦。

Well, I got a situation where I needed to translate a PHP script to Python, and it had many usages of substr(string, beginning, LENGTH).
If I chose Python’s string[beginning:end] I’d have to calculate a lot of end indexes, so the easier way was to use string[beginning:][:length], it saved me a lot of trouble.


回答 11

使用硬编码的索引本身可能是一团糟。

为了避免这种情况,Python提供了一个内置对象slice()

string = "my company has 1000$ on profit, but I lost 500$ gambling."

如果我们想知道我剩下多少钱。

正常解决方案:

final = int(string[15:19]) - int(string[43:46])
print(final)
>>>500

使用切片:

EARNINGS = slice(15, 19)
LOSSES = slice(43, 46)
final = int(string[EARNINGS]) - int(string[LOSSES])
print(final)
>>>500

使用切片可以获得可读性。

Using hardcoded indexes itself can be a mess.

In order to avoid that, Python offers a built-in object slice().

string = "my company has 1000$ on profit, but I lost 500$ gambling."

If we want to know how many money I got left.

Normal solution:

final = int(string[15:19]) - int(string[43:46])
print(final)
>>>500

Using slices:

EARNINGS = slice(15, 19)
LOSSES = slice(43, 46)
final = int(string[EARNINGS]) - int(string[LOSSES])
print(final)
>>>500

Using slice you gain readability.


如何在Python的终端中打印彩色文本?

问题:如何在Python的终端中打印彩色文本?

如何在Python中将彩色文本输出到终端?代表实体块的最佳Unicode符号是什么?

How can I output colored text to the terminal, in Python? What is the best Unicode symbol to represent a solid block?


回答 0

这在某种程度上取决于您所使用的平台。最常见的方法是打印ANSI转义序列。对于一个简单的示例,这是Blender构建脚本中的一些python代码:

class bcolors:
    HEADER = '\033[95m'
    OKBLUE = '\033[94m'
    OKGREEN = '\033[92m'
    WARNING = '\033[93m'
    FAIL = '\033[91m'
    ENDC = '\033[0m'
    BOLD = '\033[1m'
    UNDERLINE = '\033[4m'

要使用这样的代码,您可以执行以下操作

print(bcolors.WARNING + "Warning: No active frommets remain. Continue?" + bcolors.ENDC)

或者,使用Python3.6 +:

print(f"{bcolors.WARNING}Warning: No active frommets remain. Continue?{bcolors.ENDC}")

这将在包括OS X,Linux和Windows的Unix上运行(前提是您使用ANSICON,或者在Windows 10中,前提是您启用了VT100仿真)。有用于设置颜色,移动光标等的ansi代码。

如果您要对此进行复杂化处理(听起来就像是在编写游戏),则应查看“ curses”模块,该模块为您处理了许多复杂的部分。在Python的诅咒HOWTO是一个很好的介绍。

如果您没有使用扩展的ASCII码(即不在PC上),那么您将只能使用127以下的ASCII字符,而“#”或“ @”可能是您最好的选择。如果可以确保您的终端使用的是IBM 扩展的ascii字符集,那么您还有更多选择。字符176、177、178和219是“块字符”。

一些现代的基于文本的程序,例如“矮人要塞”,以图形模式模拟文本模式,并使用经典PC字体的图像。您可以在Dwarf Fortress Wiki see(用户制作的tileset)上找到一些可以使用的位图。

文本模式设计大赛已在文本模式下做图形更多的资源。

嗯..我认为这个答案有点过头了。不过,我正在计划一个史诗般的基于文本的冒险游戏。祝您彩色文字好运!

This somewhat depends on what platform you are on. The most common way to do this is by printing ANSI escape sequences. For a simple example, here’s some python code from the blender build scripts:

class bcolors:
    HEADER = '\033[95m'
    OKBLUE = '\033[94m'
    OKGREEN = '\033[92m'
    WARNING = '\033[93m'
    FAIL = '\033[91m'
    ENDC = '\033[0m'
    BOLD = '\033[1m'
    UNDERLINE = '\033[4m'

To use code like this, you can do something like

print(bcolors.WARNING + "Warning: No active frommets remain. Continue?" + bcolors.ENDC)

or, with Python3.6+:

print(f"{bcolors.WARNING}Warning: No active frommets remain. Continue?{bcolors.ENDC}")

This will work on unixes including OS X, linux and windows (provided you use ANSICON, or in Windows 10 provided you enable VT100 emulation). There are ansi codes for setting the color, moving the cursor, and more.

If you are going to get complicated with this (and it sounds like you are if you are writing a game), you should look into the “curses” module, which handles a lot of the complicated parts of this for you. The Python Curses HowTO is a good introduction.

If you are not using extended ASCII (i.e. not on a PC), you are stuck with the ascii characters below 127, and ‘#’ or ‘@’ is probably your best bet for a block. If you can ensure your terminal is using a IBM extended ascii character set, you have many more options. Characters 176, 177, 178 and 219 are the “block characters”.

Some modern text-based programs, such as “Dwarf Fortress”, emulate text mode in a graphical mode, and use images of the classic PC font. You can find some of these bitmaps that you can use on the Dwarf Fortress Wiki see (user-made tilesets).

The Text Mode Demo Contest has more resources for doing graphics in text mode.

Hmm.. I think got a little carried away on this answer. I am in the midst of planning an epic text-based adventure game, though. Good luck with your colored text!


回答 1

我很惊讶没有人提到Python termcolor模块。用法很简单:

from termcolor import colored

print colored('hello', 'red'), colored('world', 'green')

或在Python 3中:

print(colored('hello', 'red'), colored('world', 'green'))

但是,对于游戏编程和您要执行的“彩色块”来说,它可能不够复杂…

I’m surprised no one has mentioned the Python termcolor module. Usage is pretty simple:

from termcolor import colored

print colored('hello', 'red'), colored('world', 'green')

Or in Python 3:

print(colored('hello', 'red'), colored('world', 'green'))

It may not be sophisticated enough, however, for game programming and the “colored blocks” that you want to do…


回答 2

答案是Colorama,用于Python中的所有跨平台着色。

Python 3.6示例屏幕截图: 屏幕截图示例

The answer is Colorama for all cross-platform coloring in Python.

A Python 3.6 example screenshot: example screenshot


回答 3

打印一个以颜色/样式开头的字符串,然后打印该字符串,然后通过以下命令结束颜色/样式更改'\x1b[0m'

print('\x1b[6;30;42m' + 'Success!' + '\x1b[0m')

成功的绿色背景示例

使用以下代码获取外壳程序文本的格式选项表:

def print_format_table():
    """
    prints table of formatted text format options
    """
    for style in range(8):
        for fg in range(30,38):
            s1 = ''
            for bg in range(40,48):
                format = ';'.join([str(style), str(fg), str(bg)])
                s1 += '\x1b[%sm %s \x1b[0m' % (format, format)
            print(s1)
        print('\n')

print_format_table()

浅色示例(完整)

在此处输入图片说明

暗光示例(部分)

输出的顶部

Print a string that starts a color/style, then the string, then end the color/style change with '\x1b[0m':

print('\x1b[6;30;42m' + 'Success!' + '\x1b[0m')

Success with green background example

Get a table of format options for shell text with following code:

def print_format_table():
    """
    prints table of formatted text format options
    """
    for style in range(8):
        for fg in range(30,38):
            s1 = ''
            for bg in range(40,48):
                format = ';'.join([str(style), str(fg), str(bg)])
                s1 += '\x1b[%sm %s \x1b[0m' % (format, format)
            print(s1)
        print('\n')

print_format_table()

Light-on-dark example (complete)

enter image description here

Dark-on-light example (partial)

top part of output


回答 4

定义一个以颜色开头的字符串和一个以颜色结尾的字符串,然后打印您的文本,其中起始字符串在前面,结尾字符串在结尾。

CRED = '\033[91m'
CEND = '\033[0m'
print(CRED + "Error, does not compute!" + CEND)

这会产生bash,并urxvt带有Zenburn风格的配色方案:

输出颜色

通过实验,我们可以获得更多的颜色:

颜色矩阵

注意:\33[5m\33[6m闪烁。

这样,我们可以创建完整的颜色集合:

CEND      = '\33[0m'
CBOLD     = '\33[1m'
CITALIC   = '\33[3m'
CURL      = '\33[4m'
CBLINK    = '\33[5m'
CBLINK2   = '\33[6m'
CSELECTED = '\33[7m'

CBLACK  = '\33[30m'
CRED    = '\33[31m'
CGREEN  = '\33[32m'
CYELLOW = '\33[33m'
CBLUE   = '\33[34m'
CVIOLET = '\33[35m'
CBEIGE  = '\33[36m'
CWHITE  = '\33[37m'

CBLACKBG  = '\33[40m'
CREDBG    = '\33[41m'
CGREENBG  = '\33[42m'
CYELLOWBG = '\33[43m'
CBLUEBG   = '\33[44m'
CVIOLETBG = '\33[45m'
CBEIGEBG  = '\33[46m'
CWHITEBG  = '\33[47m'

CGREY    = '\33[90m'
CRED2    = '\33[91m'
CGREEN2  = '\33[92m'
CYELLOW2 = '\33[93m'
CBLUE2   = '\33[94m'
CVIOLET2 = '\33[95m'
CBEIGE2  = '\33[96m'
CWHITE2  = '\33[97m'

CGREYBG    = '\33[100m'
CREDBG2    = '\33[101m'
CGREENBG2  = '\33[102m'
CYELLOWBG2 = '\33[103m'
CBLUEBG2   = '\33[104m'
CVIOLETBG2 = '\33[105m'
CBEIGEBG2  = '\33[106m'
CWHITEBG2  = '\33[107m'

这是生成测试的代码:

x = 0
for i in range(24):
  colors = ""
  for j in range(5):
    code = str(x+j)
    colors = colors + "\33[" + code + "m\\33[" + code + "m\033[0m "
  print(colors)
  x=x+5

Define a string that starts a color and a string that ends the color, then print your text with the start string at the front and the end string at the end.

CRED = '\033[91m'
CEND = '\033[0m'
print(CRED + "Error, does not compute!" + CEND)

This produces the following in bash, in urxvt with a Zenburn-style color scheme:

output colors

Through experimentation, we can get more colors:

color matrix

Note: \33[5m and \33[6m are blinking.

This way we can create a full color collection:

CEND      = '\33[0m'
CBOLD     = '\33[1m'
CITALIC   = '\33[3m'
CURL      = '\33[4m'
CBLINK    = '\33[5m'
CBLINK2   = '\33[6m'
CSELECTED = '\33[7m'

CBLACK  = '\33[30m'
CRED    = '\33[31m'
CGREEN  = '\33[32m'
CYELLOW = '\33[33m'
CBLUE   = '\33[34m'
CVIOLET = '\33[35m'
CBEIGE  = '\33[36m'
CWHITE  = '\33[37m'

CBLACKBG  = '\33[40m'
CREDBG    = '\33[41m'
CGREENBG  = '\33[42m'
CYELLOWBG = '\33[43m'
CBLUEBG   = '\33[44m'
CVIOLETBG = '\33[45m'
CBEIGEBG  = '\33[46m'
CWHITEBG  = '\33[47m'

CGREY    = '\33[90m'
CRED2    = '\33[91m'
CGREEN2  = '\33[92m'
CYELLOW2 = '\33[93m'
CBLUE2   = '\33[94m'
CVIOLET2 = '\33[95m'
CBEIGE2  = '\33[96m'
CWHITE2  = '\33[97m'

CGREYBG    = '\33[100m'
CREDBG2    = '\33[101m'
CGREENBG2  = '\33[102m'
CYELLOWBG2 = '\33[103m'
CBLUEBG2   = '\33[104m'
CVIOLETBG2 = '\33[105m'
CBEIGEBG2  = '\33[106m'
CWHITEBG2  = '\33[107m'

Here is the code to generate the test:

x = 0
for i in range(24):
  colors = ""
  for j in range(5):
    code = str(x+j)
    colors = colors + "\33[" + code + "m\\33[" + code + "m\033[0m "
  print(colors)
  x=x+5

回答 5

您想了解ANSI转义序列。这是一个简单的示例:

CSI="\x1B["
print(CSI+"31;40m" + "Colored Text" + CSI + "0m")

有关更多信息,请参见http://en.wikipedia.org/wiki/ANSI_escape_code

对于块字符,请尝试使用\ u2588这样的Unicode字符:

print(u"\u2588")

放在一起:

print(CSI+"31;40m" + u"\u2588" + CSI + "0m")

You want to learn about ANSI escape sequences. Here’s a brief example:

CSI="\x1B["
print(CSI+"31;40m" + "Colored Text" + CSI + "0m")

For more info see http://en.wikipedia.org/wiki/ANSI_escape_code

For a block character, try a unicode character like \u2588:

print(u"\u2588")

Putting it all together:

print(CSI+"31;40m" + u"\u2588" + CSI + "0m")

回答 6

我之所以做出回应,是因为我找到了一种在Windows 10上使用ANSI代码的方法,这样您就可以更改文本的颜色而无需任何内置模块:

进行此工作的行是os.system("")或任何其他系统调用,它使您可以在终端中打印ANSI代码:

import os

os.system("")

# Group of Different functions for different styles
class style():
    BLACK = '\033[30m'
    RED = '\033[31m'
    GREEN = '\033[32m'
    YELLOW = '\033[33m'
    BLUE = '\033[34m'
    MAGENTA = '\033[35m'
    CYAN = '\033[36m'
    WHITE = '\033[37m'
    UNDERLINE = '\033[4m'
    RESET = '\033[0m'

print(style.YELLOW + "Hello, World!")

注意:尽管此选项与其他Windows选项具有相同的选项,但是Windows即使使用此技巧也无法完全支持ANSI代码。并非所有的文本装饰颜色都起作用,并且所有“明亮”颜色(代码90-97和100-107)显示的颜色与常规颜色相同(代码30-37和40-47)

编辑:感谢@jl查找更短的方法。

tl; dros.system("")在文件顶部附近添加。

Python版本: 3.6.7

I’m responding because I have found out a way to use ANSI codes on Windows 10, so that you can change the colour of text without any modules that aren’t built in:

The line that makes this work is os.system(""), or any other system call, which allows you to print ANSI codes in the Terminal:

import os

os.system("")

# Group of Different functions for different styles
class style():
    BLACK = '\033[30m'
    RED = '\033[31m'
    GREEN = '\033[32m'
    YELLOW = '\033[33m'
    BLUE = '\033[34m'
    MAGENTA = '\033[35m'
    CYAN = '\033[36m'
    WHITE = '\033[37m'
    UNDERLINE = '\033[4m'
    RESET = '\033[0m'

print(style.YELLOW + "Hello, World!")

Note: Although this gives the same options as other Windows options, Windows does not full support ANSI codes, even with this trick. Not all the text decoration colours work and all the ‘bright’ colours (Codes 90-97 and 100-107) display the same as the regular colours (Codes 30-37 and 40-47)

Edit: Thanks to @j-l for finding an even shorter method.

tl;dr: Add os.system("") near the top of your file.

Python Version: 3.6.7


回答 7

我最喜欢的方法是使用Blessings库(完整披露:我写了它)。例如:

from blessings import Terminal

t = Terminal()
print t.red('This is red.')
print t.bold_bright_red_on_black('Bright red on black')

要打印彩色砖,最可靠的方法是使用背景色打印空间。我用这种技术绘制了鼻子渐进式的进度条

print t.on_green(' ')

您也可以在特定位置打印:

with t.location(0, 5):
    print t.on_yellow(' ')

如果您在游戏过程中不得不考虑其他终端功能,也可以这样做。您可以使用Python的标准字符串格式来保持可读性:

print '{t.clear_eol}You just cleared a {t.bold}whole{t.normal} line!'.format(t=t)

Blessings的好处在于,它尽其所能在各种终端上工作,而不仅仅是(绝大多数)ANSI颜色终端。它还在使代码简洁明了的同时,将无法读取的转义序列保留在代码之外。玩得开心!

My favorite way is with the Blessings library (full disclosure: I wrote it). For example:

from blessings import Terminal

t = Terminal()
print t.red('This is red.')
print t.bold_bright_red_on_black('Bright red on black')

To print colored bricks, the most reliable way is to print spaces with background colors. I use this technique to draw the progress bar in nose-progressive:

print t.on_green(' ')

You can print in specific locations as well:

with t.location(0, 5):
    print t.on_yellow(' ')

If you have to muck with other terminal capabilities in the course of your game, you can do that as well. You can use Python’s standard string formatting to keep it readable:

print '{t.clear_eol}You just cleared a {t.bold}whole{t.normal} line!'.format(t=t)

The nice thing about Blessings is that it does its best to work on all sorts of terminals, not just the (overwhelmingly common) ANSI-color ones. It also keeps unreadable escape sequences out of your code while remaining concise to use. Have fun!


回答 8

sty与colorama相似,但较为冗长,支持8位24位(rgb)颜色,允许您注册自己的样式,支持静音,非常灵活,文档丰富,并且更多。

例子:

from sty import fg, bg, ef, rs

foo = fg.red + 'This is red text!' + fg.rs
bar = bg.blue + 'This has a blue background!' + bg.rs
baz = ef.italic + 'This is italic text' + rs.italic
qux = fg(201) + 'This is pink text using 8bit colors' + fg.rs
qui = fg(255, 10, 10) + 'This is red text using 24bit colors.' + fg.rs

# Add custom colors:

from sty import Style, RgbFg

fg.orange = Style(RgbFg(255, 150, 50))

buf = fg.orange + 'Yay, Im orange.' + fg.rs

print(foo, bar, baz, qux, qui, buf, sep='\n')

印刷品:

在此处输入图片说明

演示: 在此处输入图片说明

sty is similar to colorama, but it’s less verbose, supports 8bit and 24bit (rgb) colors, allows you to register your own styles, supports muting, is really flexible, well documented and more.

Examples:

from sty import fg, bg, ef, rs

foo = fg.red + 'This is red text!' + fg.rs
bar = bg.blue + 'This has a blue background!' + bg.rs
baz = ef.italic + 'This is italic text' + rs.italic
qux = fg(201) + 'This is pink text using 8bit colors' + fg.rs
qui = fg(255, 10, 10) + 'This is red text using 24bit colors.' + fg.rs

# Add custom colors:

from sty import Style, RgbFg

fg.orange = Style(RgbFg(255, 150, 50))

buf = fg.orange + 'Yay, Im orange.' + fg.rs

print(foo, bar, baz, qux, qui, buf, sep='\n')

prints:

enter image description here

Demo: enter image description here


回答 9

使用for循环生成一个具有所有颜色的类,以将每种颜色的组合最多迭代到100,然后编写一个带有python颜色的类。请按我的意愿复制并粘贴GPLv2:

class colors:
    '''Colors class:
    reset all colors with colors.reset
    two subclasses fg for foreground and bg for background.
    use as colors.subclass.colorname.
    i.e. colors.fg.red or colors.bg.green
    also, the generic bold, disable, underline, reverse, strikethrough,
    and invisible work with the main class
    i.e. colors.bold
    '''
    reset='\033[0m'
    bold='\033[01m'
    disable='\033[02m'
    underline='\033[04m'
    reverse='\033[07m'
    strikethrough='\033[09m'
    invisible='\033[08m'
    class fg:
        black='\033[30m'
        red='\033[31m'
        green='\033[32m'
        orange='\033[33m'
        blue='\033[34m'
        purple='\033[35m'
        cyan='\033[36m'
        lightgrey='\033[37m'
        darkgrey='\033[90m'
        lightred='\033[91m'
        lightgreen='\033[92m'
        yellow='\033[93m'
        lightblue='\033[94m'
        pink='\033[95m'
        lightcyan='\033[96m'
    class bg:
        black='\033[40m'
        red='\033[41m'
        green='\033[42m'
        orange='\033[43m'
        blue='\033[44m'
        purple='\033[45m'
        cyan='\033[46m'
        lightgrey='\033[47m'

generated a class with all the colors using a for loop to iterate every combination of color up to 100, then wrote a class with python colors. Copy and paste as you will, GPLv2 by me:

class colors:
    '''Colors class:
    reset all colors with colors.reset
    two subclasses fg for foreground and bg for background.
    use as colors.subclass.colorname.
    i.e. colors.fg.red or colors.bg.green
    also, the generic bold, disable, underline, reverse, strikethrough,
    and invisible work with the main class
    i.e. colors.bold
    '''
    reset='\033[0m'
    bold='\033[01m'
    disable='\033[02m'
    underline='\033[04m'
    reverse='\033[07m'
    strikethrough='\033[09m'
    invisible='\033[08m'
    class fg:
        black='\033[30m'
        red='\033[31m'
        green='\033[32m'
        orange='\033[33m'
        blue='\033[34m'
        purple='\033[35m'
        cyan='\033[36m'
        lightgrey='\033[37m'
        darkgrey='\033[90m'
        lightred='\033[91m'
        lightgreen='\033[92m'
        yellow='\033[93m'
        lightblue='\033[94m'
        pink='\033[95m'
        lightcyan='\033[96m'
    class bg:
        black='\033[40m'
        red='\033[41m'
        green='\033[42m'
        orange='\033[43m'
        blue='\033[44m'
        purple='\033[45m'
        cyan='\033[46m'
        lightgrey='\033[47m'

回答 10

试试这个简单的代码

def prRed(prt): print("\033[91m {}\033[00m" .format(prt))
def prGreen(prt): print("\033[92m {}\033[00m" .format(prt))
def prYellow(prt): print("\033[93m {}\033[00m" .format(prt))
def prLightPurple(prt): print("\033[94m {}\033[00m" .format(prt))
def prPurple(prt): print("\033[95m {}\033[00m" .format(prt))
def prCyan(prt): print("\033[96m {}\033[00m" .format(prt))
def prLightGray(prt): print("\033[97m {}\033[00m" .format(prt))
def prBlack(prt): print("\033[98m {}\033[00m" .format(prt))

prGreen("Hello world")

Try this simple code

def prRed(prt): print("\033[91m {}\033[00m" .format(prt))
def prGreen(prt): print("\033[92m {}\033[00m" .format(prt))
def prYellow(prt): print("\033[93m {}\033[00m" .format(prt))
def prLightPurple(prt): print("\033[94m {}\033[00m" .format(prt))
def prPurple(prt): print("\033[95m {}\033[00m" .format(prt))
def prCyan(prt): print("\033[96m {}\033[00m" .format(prt))
def prLightGray(prt): print("\033[97m {}\033[00m" .format(prt))
def prBlack(prt): print("\033[98m {}\033[00m" .format(prt))

prGreen("Hello world")

回答 11

在Windows上,您可以使用模块“ win32console”(在某些Python发行版中可用)或模块“ ctypes”(Python 2.5及更高版本)来访问Win32 API。

要查看完整的代码,同时支持方式,见色控制台报告代码Testoob

ctypes示例:

import ctypes

# Constants from the Windows API
STD_OUTPUT_HANDLE = -11
FOREGROUND_RED    = 0x0004 # text color contains red.

def get_csbi_attributes(handle):
    # Based on IPython's winconsole.py, written by Alexander Belchenko
    import struct
    csbi = ctypes.create_string_buffer(22)
    res = ctypes.windll.kernel32.GetConsoleScreenBufferInfo(handle, csbi)
    assert res

    (bufx, bufy, curx, cury, wattr,
    left, top, right, bottom, maxx, maxy) = struct.unpack("hhhhHhhhhhh", csbi.raw)
    return wattr


handle = ctypes.windll.kernel32.GetStdHandle(STD_OUTPUT_HANDLE)
reset = get_csbi_attributes(handle)

ctypes.windll.kernel32.SetConsoleTextAttribute(handle, FOREGROUND_RED)
print "Cherry on top"
ctypes.windll.kernel32.SetConsoleTextAttribute(handle, reset)

On Windows you can use module ‘win32console’ (available in some Python distributions) or module ‘ctypes’ (Python 2.5 and up) to access the Win32 API.

To see complete code that supports both ways, see the color console reporting code from Testoob.

ctypes example:

import ctypes

# Constants from the Windows API
STD_OUTPUT_HANDLE = -11
FOREGROUND_RED    = 0x0004 # text color contains red.

def get_csbi_attributes(handle):
    # Based on IPython's winconsole.py, written by Alexander Belchenko
    import struct
    csbi = ctypes.create_string_buffer(22)
    res = ctypes.windll.kernel32.GetConsoleScreenBufferInfo(handle, csbi)
    assert res

    (bufx, bufy, curx, cury, wattr,
    left, top, right, bottom, maxx, maxy) = struct.unpack("hhhhHhhhhhh", csbi.raw)
    return wattr


handle = ctypes.windll.kernel32.GetStdHandle(STD_OUTPUT_HANDLE)
reset = get_csbi_attributes(handle)

ctypes.windll.kernel32.SetConsoleTextAttribute(handle, FOREGROUND_RED)
print "Cherry on top"
ctypes.windll.kernel32.SetConsoleTextAttribute(handle, reset)

回答 12

基于@joeld的答案非常简单

class PrintInColor:
    RED = '\033[91m'
    GREEN = '\033[92m'
    YELLOW = '\033[93m'
    LIGHT_PURPLE = '\033[94m'
    PURPLE = '\033[95m'
    END = '\033[0m'

    @classmethod
    def red(cls, s, **kwargs):
        print(cls.RED + s + cls.END, **kwargs)

    @classmethod
    def green(cls, s, **kwargs):
        print(cls.GREEN + s + cls.END, **kwargs)

    @classmethod
    def yellow(cls, s, **kwargs):
        print(cls.YELLOW + s + cls.END, **kwargs)

    @classmethod
    def lightPurple(cls, s, **kwargs):
        print(cls.LIGHT_PURPLE + s + cls.END, **kwargs)

    @classmethod
    def purple(cls, s, **kwargs):
        print(cls.PURPLE + s + cls.END, **kwargs)

然后就

PrintInColor.red('hello', end=' ')
PrintInColor.green('world')

Stupidly simple based on @joeld’s answer

class PrintInColor:
    RED = '\033[91m'
    GREEN = '\033[92m'
    YELLOW = '\033[93m'
    LIGHT_PURPLE = '\033[94m'
    PURPLE = '\033[95m'
    END = '\033[0m'

    @classmethod
    def red(cls, s, **kwargs):
        print(cls.RED + s + cls.END, **kwargs)

    @classmethod
    def green(cls, s, **kwargs):
        print(cls.GREEN + s + cls.END, **kwargs)

    @classmethod
    def yellow(cls, s, **kwargs):
        print(cls.YELLOW + s + cls.END, **kwargs)

    @classmethod
    def lightPurple(cls, s, **kwargs):
        print(cls.LIGHT_PURPLE + s + cls.END, **kwargs)

    @classmethod
    def purple(cls, s, **kwargs):
        print(cls.PURPLE + s + cls.END, **kwargs)

Then just

PrintInColor.red('hello', end=' ')
PrintInColor.green('world')

回答 13

我已经将@joeld答案包装到具有全局函数的模块中,可以在代码的任何地方使用它。

文件:log.py

HEADER = '\033[95m'
OKBLUE = '\033[94m'
OKGREEN = '\033[92m'
WARNING = '\033[93m'
FAIL = '\033[91m'
ENDC = '\033[0m'
BOLD = "\033[1m"

def disable():
    HEADER = ''
    OKBLUE = ''
    OKGREEN = ''
    WARNING = ''
    FAIL = ''
    ENDC = ''

def infog( msg):
    print OKGREEN + msg + ENDC

def info( msg):
    print OKBLUE + msg + ENDC

def warn( msg):
    print WARNING + msg + ENDC

def err( msg):
    print FAIL + msg + ENDC

用途如下:

 import log
    log.info("Hello World")
    log.err("System Error")

I have wrapped @joeld answer into a module with global functions that I can use anywhere in my code.

file: log.py

HEADER = '\033[95m'
OKBLUE = '\033[94m'
OKGREEN = '\033[92m'
WARNING = '\033[93m'
FAIL = '\033[91m'
ENDC = '\033[0m'
BOLD = "\033[1m"

def disable():
    HEADER = ''
    OKBLUE = ''
    OKGREEN = ''
    WARNING = ''
    FAIL = ''
    ENDC = ''

def infog( msg):
    print OKGREEN + msg + ENDC

def info( msg):
    print OKBLUE + msg + ENDC

def warn( msg):
    print WARNING + msg + ENDC

def err( msg):
    print FAIL + msg + ENDC

use as follows:

 import log
    log.info("Hello World")
    log.err("System Error")

回答 14

对于Windows,除非使用win32api,否则无法使用颜色打印到控制台。

对于Linux,这就像使用print一样简单,这里概述了转义序列:

色彩

对于要像盒子一样打印的字符,实际上取决于您在控制台窗口中使用的字体。井字符号效果很好,但是取决于字体:

#

For Windows you cannot print to console with colors unless you’re using the win32api.

For Linux it’s as simple as using print, with the escape sequences outlined here:

Colors

For the character to print like a box, it really depends on what font you are using for the console window. The pound symbol works well, but it depends on the font:

#

回答 15

# Pure Python 3.x demo, 256 colors
# Works with bash under Linux and MacOS

fg = lambda text, color: "\33[38;5;" + str(color) + "m" + text + "\33[0m"
bg = lambda text, color: "\33[48;5;" + str(color) + "m" + text + "\33[0m"

def print_six(row, format, end="\n"):
    for col in range(6):
        color = row*6 + col - 2
        if color>=0:
            text = "{:3d}".format(color)
            print (format(text,color), end=" ")
        else:
            print(end="    ")   # four spaces
    print(end=end)

for row in range(0, 43):
    print_six(row, fg, " ")
    print_six(row, bg)

# Simple usage: print(fg("text", 160))

具有改变前景和背景的文本,颜色0..141 具有更改的前景和背景的文本,颜色142..255

# Pure Python 3.x demo, 256 colors
# Works with bash under Linux and MacOS

fg = lambda text, color: "\33[38;5;" + str(color) + "m" + text + "\33[0m"
bg = lambda text, color: "\33[48;5;" + str(color) + "m" + text + "\33[0m"

def print_six(row, format, end="\n"):
    for col in range(6):
        color = row*6 + col - 2
        if color>=0:
            text = "{:3d}".format(color)
            print (format(text,color), end=" ")
        else:
            print(end="    ")   # four spaces
    print(end=end)

for row in range(0, 43):
    print_six(row, fg, " ")
    print_six(row, bg)

# Simple usage: print(fg("text", 160))

Text with altering foreground and background, colors 0..141 Text with altering foreground and background, colors 142..255


回答 16

我最终这样做了,我觉得那是最干净的:

formatters = {             
    'RED': '\033[91m',     
    'GREEN': '\033[92m',   
    'END': '\033[0m',      
}

print 'Master is currently {RED}red{END}!'.format(**formatters)
print 'Help make master {GREEN}green{END} again!'.format(**formatters)

I ended up doing this, I felt it was cleanest:

formatters = {             
    'RED': '\033[91m',     
    'GREEN': '\033[92m',   
    'END': '\033[0m',      
}

print 'Master is currently {RED}red{END}!'.format(**formatters)
print 'Help make master {GREEN}green{END} again!'.format(**formatters)

回答 17

使用https://pypi.python.org/pypi/lazyme 在@joeld答案上构建pip install -U lazyme

from lazyme.string import color_print
>>> color_print('abc')
abc
>>> color_print('abc', color='pink')
abc
>>> color_print('abc', color='red')
abc
>>> color_print('abc', color='yellow')
abc
>>> color_print('abc', color='green')
abc
>>> color_print('abc', color='blue', underline=True)
abc
>>> color_print('abc', color='blue', underline=True, bold=True)
abc
>>> color_print('abc', color='pink', underline=True, bold=True)
abc

屏幕截图:

在此处输入图片说明


color_print使用新的格式化程序对进行了一些更新,例如:

>>> from lazyme.string import palette, highlighter, formatter
>>> from lazyme.string import color_print
>>> palette.keys() # Available colors.
['pink', 'yellow', 'cyan', 'magenta', 'blue', 'gray', 'default', 'black', 'green', 'white', 'red']
>>> highlighter.keys() # Available highlights.
['blue', 'pink', 'gray', 'black', 'yellow', 'cyan', 'green', 'magenta', 'white', 'red']
>>> formatter.keys() # Available formatter, 
['hide', 'bold', 'italic', 'default', 'fast_blinking', 'faint', 'strikethrough', 'underline', 'blinking', 'reverse']

注:italicfast blinkingstrikethrough可能无法在所有终端上使用,在Mac / Ubuntu上也无法使用。

例如

>>> color_print('foo bar', color='pink', highlight='white')
foo bar
>>> color_print('foo bar', color='pink', highlight='white', reverse=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', bold=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', faint=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', faint=True, reverse=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', underline=True, reverse=True)
foo bar

屏幕截图:

在此处输入图片说明

Building on @joeld answer, using https://pypi.python.org/pypi/lazyme pip install -U lazyme :

from lazyme.string import color_print
>>> color_print('abc')
abc
>>> color_print('abc', color='pink')
abc
>>> color_print('abc', color='red')
abc
>>> color_print('abc', color='yellow')
abc
>>> color_print('abc', color='green')
abc
>>> color_print('abc', color='blue', underline=True)
abc
>>> color_print('abc', color='blue', underline=True, bold=True)
abc
>>> color_print('abc', color='pink', underline=True, bold=True)
abc

Screenshot:

enter image description here


Some updates to the color_print with new formatters, e.g.:

>>> from lazyme.string import palette, highlighter, formatter
>>> from lazyme.string import color_print
>>> palette.keys() # Available colors.
['pink', 'yellow', 'cyan', 'magenta', 'blue', 'gray', 'default', 'black', 'green', 'white', 'red']
>>> highlighter.keys() # Available highlights.
['blue', 'pink', 'gray', 'black', 'yellow', 'cyan', 'green', 'magenta', 'white', 'red']
>>> formatter.keys() # Available formatter, 
['hide', 'bold', 'italic', 'default', 'fast_blinking', 'faint', 'strikethrough', 'underline', 'blinking', 'reverse']

Note: italic, fast blinking and strikethrough may not work on all terminals, doesn’t work on Mac / Ubuntu.

E.g.

>>> color_print('foo bar', color='pink', highlight='white')
foo bar
>>> color_print('foo bar', color='pink', highlight='white', reverse=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', bold=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', faint=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', faint=True, reverse=True)
foo bar
>>> color_print('foo bar', color='pink', highlight='white', underline=True, reverse=True)
foo bar

Screenshot:

enter image description here


回答 18

def black(text):
    print('\033[30m', text, '\033[0m', sep='')

def red(text):
    print('\033[31m', text, '\033[0m', sep='')

def green(text):
    print('\033[32m', text, '\033[0m', sep='')

def yellow(text):
    print('\033[33m', text, '\033[0m', sep='')

def blue(text):
    print('\033[34m', text, '\033[0m', sep='')

def magenta(text):
    print('\033[35m', text, '\033[0m', sep='')

def cyan(text):
    print('\033[36m', text, '\033[0m', sep='')

def gray(text):
    print('\033[90m', text, '\033[0m', sep='')


black("BLACK")
red("RED")
green("GREEN")
yellow("YELLOW")
blue("BLACK")
magenta("MAGENTA")
cyan("CYAN")
gray("GRAY")

在线尝试

def black(text):
    print('\033[30m', text, '\033[0m', sep='')

def red(text):
    print('\033[31m', text, '\033[0m', sep='')

def green(text):
    print('\033[32m', text, '\033[0m', sep='')

def yellow(text):
    print('\033[33m', text, '\033[0m', sep='')

def blue(text):
    print('\033[34m', text, '\033[0m', sep='')

def magenta(text):
    print('\033[35m', text, '\033[0m', sep='')

def cyan(text):
    print('\033[36m', text, '\033[0m', sep='')

def gray(text):
    print('\033[90m', text, '\033[0m', sep='')


black("BLACK")
red("RED")
green("GREEN")
yellow("YELLOW")
blue("BLACK")
magenta("MAGENTA")
cyan("CYAN")
gray("GRAY")

Try online


回答 19

请注意,with关键字与需要重置的修饰符(使用Python 3和Colorama)混合的程度如何:

from colorama import Fore, Style
import sys

class Highlight:
  def __init__(self, clazz, color):
    self.color = color
    self.clazz = clazz
  def __enter__(self):
    print(self.color, end="")
  def __exit__(self, type, value, traceback):
    if self.clazz == Fore:
      print(Fore.RESET, end="")
    else:
      assert self.clazz == Style
      print(Style.RESET_ALL, end="")
    sys.stdout.flush()

with Highlight(Fore, Fore.GREEN):
  print("this is highlighted")
print("this is not")

note how well the with keyword mixes with modifiers like these that need to be reset (using Python 3 and Colorama):

from colorama import Fore, Style
import sys

class Highlight:
  def __init__(self, clazz, color):
    self.color = color
    self.clazz = clazz
  def __enter__(self):
    print(self.color, end="")
  def __exit__(self, type, value, traceback):
    if self.clazz == Fore:
      print(Fore.RESET, end="")
    else:
      assert self.clazz == Style
      print(Style.RESET_ALL, end="")
    sys.stdout.flush()

with Highlight(Fore, Fore.GREEN):
  print("this is highlighted")
print("this is not")

回答 20

您可以使用curses库的Python实现:http : //docs.python.org/library/curses.html

另外,运行此命令,您将找到您的盒子:

for i in range(255):
    print i, chr(i)

You can use the Python implementation of the curses library: http://docs.python.org/library/curses.html

Also, run this and you’ll find your box:

for i in range(255):
    print i, chr(i)

回答 21

您可以使用CLINT:

from clint.textui import colored
print colored.red('some warning message')
print colored.green('nicely done!')

从GitHub获取它

You could use CLINT:

from clint.textui import colored
print colored.red('some warning message')
print colored.green('nicely done!')

Get it from GitHub.


回答 22

如果您正在编写游戏,也许您想更改背景颜色并仅使用空格?例如:

print " "+ "\033[01;41m" + " " +"\033[01;46m"  + "  " + "\033[01;42m"

If you are programming a game perhaps you would like to change the background color and use only spaces? For example:

print " "+ "\033[01;41m" + " " +"\033[01;46m"  + "  " + "\033[01;42m"

回答 23

我知道我迟到了。但是,我有一个名为ColorIt。非常简单。

这里有些例子:

from ColorIt import *

# Use this to ensure that ColorIt will be usable by certain command line interfaces
initColorIt()

# Foreground
print (color ('This text is red', colors.RED))
print (color ('This text is orange', colors.ORANGE))
print (color ('This text is yellow', colors.YELLOW))
print (color ('This text is green', colors.GREEN))
print (color ('This text is blue', colors.BLUE))
print (color ('This text is purple', colors.PURPLE))
print (color ('This text is white', colors.WHITE))

# Background
print (background ('This text has a background that is red', colors.RED))
print (background ('This text has a background that is orange', colors.ORANGE))
print (background ('This text has a background that is yellow', colors.YELLOW))
print (background ('This text has a background that is green', colors.GREEN))
print (background ('This text has a background that is blue', colors.BLUE))
print (background ('This text has a background that is purple', colors.PURPLE))
print (background ('This text has a background that is white', colors.WHITE))

# Custom
print (color ("This color has a custom grey text color", (150, 150, 150))
print (background ("This color has a custom grey background", (150, 150, 150))

# Combination
print (background (color ("This text is blue with a white background", colors.BLUE), colors.WHITE))

这给您:

ColorIt的图片

还值得注意的是,这是跨平台的,并且已经在Mac,Linux和Windows上进行了测试。

您可能想尝试一下:https : //github.com/CodeForeverAndEver/ColorIt

注意:几天后将添加闪烁,斜体,粗体等。

I know that I am late. But, I have a library called ColorIt. It is super simple.

Here are some examples:

from ColorIt import *

# Use this to ensure that ColorIt will be usable by certain command line interfaces
initColorIt()

# Foreground
print (color ('This text is red', colors.RED))
print (color ('This text is orange', colors.ORANGE))
print (color ('This text is yellow', colors.YELLOW))
print (color ('This text is green', colors.GREEN))
print (color ('This text is blue', colors.BLUE))
print (color ('This text is purple', colors.PURPLE))
print (color ('This text is white', colors.WHITE))

# Background
print (background ('This text has a background that is red', colors.RED))
print (background ('This text has a background that is orange', colors.ORANGE))
print (background ('This text has a background that is yellow', colors.YELLOW))
print (background ('This text has a background that is green', colors.GREEN))
print (background ('This text has a background that is blue', colors.BLUE))
print (background ('This text has a background that is purple', colors.PURPLE))
print (background ('This text has a background that is white', colors.WHITE))

# Custom
print (color ("This color has a custom grey text color", (150, 150, 150))
print (background ("This color has a custom grey background", (150, 150, 150))

# Combination
print (background (color ("This text is blue with a white background", colors.BLUE), colors.WHITE))

This gives you:

Picture of ColorIt

It’s also worth noting that this is cross platform and has been tested on mac, linux, and windows.

You might want to try it out: https://github.com/CodeForeverAndEver/ColorIt

Note: Blinking, italics, bold, etc. will be added in a few days.


回答 24

如果您使用的是Windows,那么就到这里!

# display text on a Windows console
# Windows XP with Python27 or Python32
from ctypes import windll
# needed for Python2/Python3 diff
try:
    input = raw_input
except:
    pass
STD_OUTPUT_HANDLE = -11
stdout_handle = windll.kernel32.GetStdHandle(STD_OUTPUT_HANDLE)
# look at the output and select the color you want
# for instance hex E is yellow on black
# hex 1E is yellow on blue
# hex 2E is yellow on green and so on
for color in range(0, 75):
     windll.kernel32.SetConsoleTextAttribute(stdout_handle, color)
     print("%X --> %s" % (color, "Have a fine day!"))
     input("Press Enter to go on ... ")

If you are using Windows, then here you go!

# display text on a Windows console
# Windows XP with Python27 or Python32
from ctypes import windll
# needed for Python2/Python3 diff
try:
    input = raw_input
except:
    pass
STD_OUTPUT_HANDLE = -11
stdout_handle = windll.kernel32.GetStdHandle(STD_OUTPUT_HANDLE)
# look at the output and select the color you want
# for instance hex E is yellow on black
# hex 1E is yellow on blue
# hex 2E is yellow on green and so on
for color in range(0, 75):
     windll.kernel32.SetConsoleTextAttribute(stdout_handle, color)
     print("%X --> %s" % (color, "Have a fine day!"))
     input("Press Enter to go on ... ")

回答 25

如果您使用的是Django

>>> from django.utils.termcolors import colorize
>>> print colorize("Hello World!", fg="blue", bg='red',
...                 opts=('bold', 'blink', 'underscore',))
Hello World!
>>> help(colorize)

快照:

图片

(我通常在运行服务器终端上使用彩色输出进行调试,所以我添加了它。)

您可以测试它是否已安装在您的计算机中:
$ python -c "import django; print django.VERSION"
要安装它,请检查:如何安装Django

试试看!!

If you are using Django

>>> from django.utils.termcolors import colorize
>>> print colorize("Hello World!", fg="blue", bg='red',
...                 opts=('bold', 'blink', 'underscore',))
Hello World!
>>> help(colorize)

snapshot:

image

(I generally use colored output for debugging on runserver terminal so I added it.)

You can test if it is installed in your machine:
$ python -c "import django; print django.VERSION"
To install it check: How to install Django

Give it a Try!!


回答 26

这是一个诅咒的例子:

import curses

def main(stdscr):
    stdscr.clear()
    if curses.has_colors():
        for i in xrange(1, curses.COLORS):
            curses.init_pair(i, i, curses.COLOR_BLACK)
            stdscr.addstr("COLOR %d! " % i, curses.color_pair(i))
            stdscr.addstr("BOLD! ", curses.color_pair(i) | curses.A_BOLD)
            stdscr.addstr("STANDOUT! ", curses.color_pair(i) | curses.A_STANDOUT)
            stdscr.addstr("UNDERLINE! ", curses.color_pair(i) | curses.A_UNDERLINE)
            stdscr.addstr("BLINK! ", curses.color_pair(i) | curses.A_BLINK)
            stdscr.addstr("DIM! ", curses.color_pair(i) | curses.A_DIM)
            stdscr.addstr("REVERSE! ", curses.color_pair(i) | curses.A_REVERSE)
    stdscr.refresh()
    stdscr.getch()

if __name__ == '__main__':
    print "init..."
    curses.wrapper(main)

Here’s a curses example:

import curses

def main(stdscr):
    stdscr.clear()
    if curses.has_colors():
        for i in xrange(1, curses.COLORS):
            curses.init_pair(i, i, curses.COLOR_BLACK)
            stdscr.addstr("COLOR %d! " % i, curses.color_pair(i))
            stdscr.addstr("BOLD! ", curses.color_pair(i) | curses.A_BOLD)
            stdscr.addstr("STANDOUT! ", curses.color_pair(i) | curses.A_STANDOUT)
            stdscr.addstr("UNDERLINE! ", curses.color_pair(i) | curses.A_UNDERLINE)
            stdscr.addstr("BLINK! ", curses.color_pair(i) | curses.A_BLINK)
            stdscr.addstr("DIM! ", curses.color_pair(i) | curses.A_DIM)
            stdscr.addstr("REVERSE! ", curses.color_pair(i) | curses.A_REVERSE)
    stdscr.refresh()
    stdscr.getch()

if __name__ == '__main__':
    print "init..."
    curses.wrapper(main)

回答 27

https://raw.github.com/fabric/fabric/master/fabric/colors.py

"""
.. versionadded:: 0.9.2

Functions for wrapping strings in ANSI color codes.

Each function within this module returns the input string ``text``, wrapped
with ANSI color codes for the appropriate color.

For example, to print some text as green on supporting terminals::

    from fabric.colors import green

    print(green("This text is green!"))

Because these functions simply return modified strings, you can nest them::

    from fabric.colors import red, green

    print(red("This sentence is red, except for " + \
          green("these words, which are green") + "."))

If ``bold`` is set to ``True``, the ANSI flag for bolding will be flipped on
for that particular invocation, which usually shows up as a bold or brighter
version of the original color on most terminals.
"""


def _wrap_with(code):

    def inner(text, bold=False):
        c = code
        if bold:
            c = "1;%s" % c
        return "\033[%sm%s\033[0m" % (c, text)
    return inner

red = _wrap_with('31')
green = _wrap_with('32')
yellow = _wrap_with('33')
blue = _wrap_with('34')
magenta = _wrap_with('35')
cyan = _wrap_with('36')
white = _wrap_with('37')

https://raw.github.com/fabric/fabric/master/fabric/colors.py

"""
.. versionadded:: 0.9.2

Functions for wrapping strings in ANSI color codes.

Each function within this module returns the input string ``text``, wrapped
with ANSI color codes for the appropriate color.

For example, to print some text as green on supporting terminals::

    from fabric.colors import green

    print(green("This text is green!"))

Because these functions simply return modified strings, you can nest them::

    from fabric.colors import red, green

    print(red("This sentence is red, except for " + \
          green("these words, which are green") + "."))

If ``bold`` is set to ``True``, the ANSI flag for bolding will be flipped on
for that particular invocation, which usually shows up as a bold or brighter
version of the original color on most terminals.
"""


def _wrap_with(code):

    def inner(text, bold=False):
        c = code
        if bold:
            c = "1;%s" % c
        return "\033[%sm%s\033[0m" % (c, text)
    return inner

red = _wrap_with('31')
green = _wrap_with('32')
yellow = _wrap_with('33')
blue = _wrap_with('34')
magenta = _wrap_with('35')
cyan = _wrap_with('36')
white = _wrap_with('37')

回答 28

asciimatics为构建文本UI和动画提供了可移植的支持:

#!/usr/bin/env python
from asciimatics.effects import RandomNoise  # $ pip install asciimatics
from asciimatics.renderers import SpeechBubble, Rainbow
from asciimatics.scene import Scene
from asciimatics.screen import Screen
from asciimatics.exceptions import ResizeScreenError


def demo(screen):
    render = Rainbow(screen, SpeechBubble('Rainbow'))
    effects = [RandomNoise(screen, signal=render)]
    screen.play([Scene(effects, -1)], stop_on_resize=True)

while True:
    try:
        Screen.wrapper(demo)
        break
    except ResizeScreenError:
        pass

Asciicast:

ASCII噪声中的彩虹色文本

asciimatics provides a portable support for building text UI and animations:

#!/usr/bin/env python
from asciimatics.effects import RandomNoise  # $ pip install asciimatics
from asciimatics.renderers import SpeechBubble, Rainbow
from asciimatics.scene import Scene
from asciimatics.screen import Screen
from asciimatics.exceptions import ResizeScreenError


def demo(screen):
    render = Rainbow(screen, SpeechBubble('Rainbow'))
    effects = [RandomNoise(screen, signal=render)]
    screen.play([Scene(effects, -1)], stop_on_resize=True)

while True:
    try:
        Screen.wrapper(demo)
        break
    except ResizeScreenError:
        pass

Asciicast:

rainbow-colored text among ascii noise


回答 29

另一个包装python 3打印功能的pypi模块:

https://pypi.python.org/pypi/colorprint

如果您也可以在python 2.x中使用它from __future__ import print。这是来自模块pypi页面的python 2示例:

from __future__ import print_function
from colorprint import *

print('Hello', 'world', color='blue', end='', sep=', ')
print('!', color='red', format=['bold', 'blink'])

输出“你好,世界!” 用蓝色和感叹号标记为红色和闪烁。

Yet another pypi module that wraps the python 3 print function:

https://pypi.python.org/pypi/colorprint

It’s usable in python 2.x if you also from __future__ import print. Here is a python 2 example from the modules pypi page:

from __future__ import print_function
from colorprint import *

print('Hello', 'world', color='blue', end='', sep=', ')
print('!', color='red', format=['bold', 'blink'])

Outputs “Hello, world!” with the words in blue and the exclamation mark bold red and blinking.