标签归档：python-2.5

知识问答

如何装饰一堂课？

问题：如何装饰一堂课？

在Python 2.5中，有没有办法创建装饰类的装饰器？具体来说，我想使用装饰器将成员添加到类中，并更改构造函数以获取该成员的值。

寻找类似以下的内容（在“ Foo类：”上存在语法错误：

def getId(self): return self.__id

class addID(original_class):
    def __init__(self, id, *args, **kws):
        self.__id = id
        self.getId = getId
        original_class.__init__(self, *args, **kws)

@addID
class Foo:
    def __init__(self, value1):
        self.value1 = value1

if __name__ == '__main__':
    foo1 = Foo(5,1)
    print foo1.value1, foo1.getId()
    foo2 = Foo(15,2)
    print foo2.value1, foo2.getId()

我想我真正想要的是在Python中执行类似C＃接口的方法。我想我应该改变我的范式。

点击查看英文原文

In Python 2.5, is there a way to create a decorator that decorates a class? Specifically, I want to use a decorator to add a member to a class and change the constructor to take a value for that member.

Looking for something like the following (which has a syntax error on ‘class Foo:’:

def getId(self): return self.__id

class addID(original_class):
    def __init__(self, id, *args, **kws):
        self.__id = id
        self.getId = getId
        original_class.__init__(self, *args, **kws)

@addID
class Foo:
    def __init__(self, value1):
        self.value1 = value1

if __name__ == '__main__':
    foo1 = Foo(5,1)
    print foo1.value1, foo1.getId()
    foo2 = Foo(15,2)
    print foo2.value1, foo2.getId()

I guess what I’m really after is a way to do something like a C# interface in Python. I need to switch my paradigm I suppose.

回答 0

我的观点是您可能希望考虑一个子类，而不是您概述的方法。但是，不知道您的特定情况，YMMV :-)

您正在考虑的是元类。__new__元类中的函数将传递该类的完整建议定义，然后可以在创建类之前将其重写。那时，您可以将构造函数细分为一个新的。

例：

def substitute_init(self, id, *args, **kwargs):
    pass

class FooMeta(type):

    def __new__(cls, name, bases, attrs):
        attrs['__init__'] = substitute_init
        return super(FooMeta, cls).__new__(cls, name, bases, attrs)

class Foo(object):

    __metaclass__ = FooMeta

    def __init__(self, value1):
        pass

替换构造函数可能有点麻烦，但是语言确实为这种深入的内省和动态修改提供了支持。

点击查看英文原文

I would second the notion that you may wish to consider a subclass instead of the approach you’ve outlined. However, not knowing your specific scenario, YMMV :-)

What you’re thinking of is a metaclass. The __new__ function in a metaclass is passed the full proposed definition of the class, which it can then rewrite before the class is created. You can, at that time, sub out the constructor for a new one.

Example:

def substitute_init(self, id, *args, **kwargs):
    pass

class FooMeta(type):

    def __new__(cls, name, bases, attrs):
        attrs['__init__'] = substitute_init
        return super(FooMeta, cls).__new__(cls, name, bases, attrs)

class Foo(object):

    __metaclass__ = FooMeta

    def __init__(self, value1):
        pass

Replacing the constructor is perhaps a bit dramatic, but the language does provide support for this kind of deep introspection and dynamic modification.

回答 1

除了类装饰器是否是您的问题的正确解决方案的问题之外：

在Python 2.6和更高版本中，有带有@语法的类装饰器，因此您可以编写：

@addID
class Foo:
    pass

在旧版本中，您可以使用另一种方法：

class Foo:
    pass

Foo = addID(Foo)

但是请注意，这与函数装饰器的工作原理相同，并且装饰器应返回新（或修改后的原始）类，这不是您在示例中所做的。addID装饰器如下所示：

def addID(original_class):
    orig_init = original_class.__init__
    # Make copy of original __init__, so we can call it without recursion

    def __init__(self, id, *args, **kws):
        self.__id = id
        self.getId = getId
        orig_init(self, *args, **kws) # Call the original __init__

    original_class.__init__ = __init__ # Set the class' __init__ to the new one
    return original_class

然后，您可以按照上述方式为Python版本使用适当的语法。

但是我同意其他人的观点，如果你想重写继承，继承更适合__init__。

点击查看英文原文

Apart from the question whether class decorators are the right solution to your problem:

In Python 2.6 and higher, there are class decorators with the @-syntax, so you can write:

@addID
class Foo:
    pass

In older versions, you can do it another way:

class Foo:
    pass

Foo = addID(Foo)

Note however that this works the same as for function decorators, and that the decorator should return the new (or modified original) class, which is not what you’re doing in the example. The addID decorator would look like this:

def addID(original_class):
    orig_init = original_class.__init__
    # Make copy of original __init__, so we can call it without recursion

    def __init__(self, id, *args, **kws):
        self.__id = id
        self.getId = getId
        orig_init(self, *args, **kws) # Call the original __init__

    original_class.__init__ = __init__ # Set the class' __init__ to the new one
    return original_class

You could then use the appropriate syntax for your Python version as described above.

But I agree with others that inheritance is better suited if you want to override __init__.

回答 2

没有人解释过您可以动态定义类。因此，您可以使用一个装饰器来定义（并返回）一个子类：

def addId(cls):

    class AddId(cls):

        def __init__(self, id, *args, **kargs):
            super(AddId, self).__init__(*args, **kargs)
            self.__id = id

        def getId(self):
            return self.__id

    return AddId

可以在Python 2中使用它（来自Blckknght的评论，它解释了为什么您应该在2.6+中继续这样做），如下所示：

class Foo:
    pass

FooId = addId(Foo)

并且在Python 3中是这样的（但要小心使用 super()在类中）：

@addId
class Foo:
    pass

所以，你可以有你的蛋糕和吃它-继承和装饰！

点击查看英文原文

No one has explained that you can dynamically define classes. So you can have a decorator that defines (and returns) a subclass:

def addId(cls):

    class AddId(cls):

        def __init__(self, id, *args, **kargs):
            super(AddId, self).__init__(*args, **kargs)
            self.__id = id

        def getId(self):
            return self.__id

    return AddId

Which can be used in Python 2 (the comment from Blckknght which explains why you should continue to do this in 2.6+) like this:

class Foo:
    pass

FooId = addId(Foo)

And in Python 3 like this (but be careful to use super() in your classes):

@addId
class Foo:
    pass

So you can have your cake and eat it – inheritance and decorators!

回答 3

这不是一个好习惯，因此，没有机制可以做到这一点。完成所需内容的正确方法是继承。

查看类文档。

一个小例子：

class Employee(object):

    def __init__(self, age, sex, siblings=0):
        self.age = age
        self.sex = sex    
        self.siblings = siblings

    def born_on(self):    
        today = datetime.date.today()

        return today - datetime.timedelta(days=self.age*365)


class Boss(Employee):    
    def __init__(self, age, sex, siblings=0, bonus=0):
        self.bonus = bonus
        Employee.__init__(self, age, sex, siblings)

这样老板就拥有了一切Employee，还有他自己的__init__方法和自己的成员。

点击查看英文原文

That’s not a good practice and there is no mechanism to do that because of that. The right way to accomplish what you want is inheritance.

Take a look into the class documentation.

A little example:

class Employee(object):

    def __init__(self, age, sex, siblings=0):
        self.age = age
        self.sex = sex    
        self.siblings = siblings

    def born_on(self):    
        today = datetime.date.today()

        return today - datetime.timedelta(days=self.age*365)


class Boss(Employee):    
    def __init__(self, age, sex, siblings=0, bonus=0):
        self.bonus = bonus
        Employee.__init__(self, age, sex, siblings)

This way Boss has everything Employee has, with also his own __init__ method and own members.

回答 4

我同意继承更适合提出的问题。

我发现这个问题在装饰类上确实很方便，谢谢大家。

这是另外两个基于其他答案的示例，包括继承如何影响Python 2.7中的内容（以及@wraps，它维护原始函数的文档字符串等）：

def dec(klass):
    old_foo = klass.foo
    @wraps(klass.foo)
    def decorated_foo(self, *args ,**kwargs):
        print('@decorator pre %s' % msg)
        old_foo(self, *args, **kwargs)
        print('@decorator post %s' % msg)
    klass.foo = decorated_foo
    return klass

@dec  # No parentheses
class Foo...

通常，您想向装饰器添加参数：

from functools import wraps

def dec(msg='default'):
    def decorator(klass):
        old_foo = klass.foo
        @wraps(klass.foo)
        def decorated_foo(self, *args ,**kwargs):
            print('@decorator pre %s' % msg)
            old_foo(self, *args, **kwargs)
            print('@decorator post %s' % msg)
        klass.foo = decorated_foo
        return klass
    return decorator

@dec('foo decorator')  # You must add parentheses now, even if they're empty
class Foo(object):
    def foo(self, *args, **kwargs):
        print('foo.foo()')

@dec('subfoo decorator')
class SubFoo(Foo):
    def foo(self, *args, **kwargs):
        print('subfoo.foo() pre')
        super(SubFoo, self).foo(*args, **kwargs)
        print('subfoo.foo() post')

@dec('subsubfoo decorator')
class SubSubFoo(SubFoo):
    def foo(self, *args, **kwargs):
        print('subsubfoo.foo() pre')
        super(SubSubFoo, self).foo(*args, **kwargs)
        print('subsubfoo.foo() post')

SubSubFoo().foo()

输出：

@decorator pre subsubfoo decorator
subsubfoo.foo() pre
@decorator pre subfoo decorator
subfoo.foo() pre
@decorator pre foo decorator
foo.foo()
@decorator post foo decorator
subfoo.foo() post
@decorator post subfoo decorator
subsubfoo.foo() post
@decorator post subsubfoo decorator

我使用了一个函数装饰器，因为我发现它们更加简洁。这是一个装饰类的类：

class Dec(object):

    def __init__(self, msg):
        self.msg = msg

    def __call__(self, klass):
        old_foo = klass.foo
        msg = self.msg
        def decorated_foo(self, *args, **kwargs):
            print('@decorator pre %s' % msg)
            old_foo(self, *args, **kwargs)
            print('@decorator post %s' % msg)
        klass.foo = decorated_foo
        return klass

一个更强大的版本，用于检查这些括号，并在装饰的类上不存在该方法时起作用：

from inspect import isclass

def decorate_if(condition, decorator):
    return decorator if condition else lambda x: x

def dec(msg):
    # Only use if your decorator's first parameter is never a class
    assert not isclass(msg)

    def decorator(klass):
        old_foo = getattr(klass, 'foo', None)

        @decorate_if(old_foo, wraps(klass.foo))
        def decorated_foo(self, *args ,**kwargs):
            print('@decorator pre %s' % msg)
            if callable(old_foo):
                old_foo(self, *args, **kwargs)
            print('@decorator post %s' % msg)

        klass.foo = decorated_foo
        return klass

    return decorator

在assert该装饰尚未使用支票没有括号。如果包含，则将要装饰的类传递给msg装饰器的参数，该参数将引发AssertionError。

@decorate_if仅将decoratorif condition求值应用于True。

使用getattr，callable测试和@decorate_if可以使装饰器foo()在被装饰的类上不存在该方法时也不会中断。

点击查看英文原文

I’d agree inheritance is a better fit for the problem posed.

I found this question really handy though on decorating classes, thanks all.

Here’s another couple of examples, based on other answers, including how inheritance affects things in Python 2.7, (and @wraps, which maintains the original function’s docstring, etc.):

def dec(klass):
    old_foo = klass.foo
    @wraps(klass.foo)
    def decorated_foo(self, *args ,**kwargs):
        print('@decorator pre %s' % msg)
        old_foo(self, *args, **kwargs)
        print('@decorator post %s' % msg)
    klass.foo = decorated_foo
    return klass

@dec  # No parentheses
class Foo...

Often you want to add parameters to your decorator:

from functools import wraps

def dec(msg='default'):
    def decorator(klass):
        old_foo = klass.foo
        @wraps(klass.foo)
        def decorated_foo(self, *args ,**kwargs):
            print('@decorator pre %s' % msg)
            old_foo(self, *args, **kwargs)
            print('@decorator post %s' % msg)
        klass.foo = decorated_foo
        return klass
    return decorator

@dec('foo decorator')  # You must add parentheses now, even if they're empty
class Foo(object):
    def foo(self, *args, **kwargs):
        print('foo.foo()')

@dec('subfoo decorator')
class SubFoo(Foo):
    def foo(self, *args, **kwargs):
        print('subfoo.foo() pre')
        super(SubFoo, self).foo(*args, **kwargs)
        print('subfoo.foo() post')

@dec('subsubfoo decorator')
class SubSubFoo(SubFoo):
    def foo(self, *args, **kwargs):
        print('subsubfoo.foo() pre')
        super(SubSubFoo, self).foo(*args, **kwargs)
        print('subsubfoo.foo() post')

SubSubFoo().foo()

Outputs:

@decorator pre subsubfoo decorator
subsubfoo.foo() pre
@decorator pre subfoo decorator
subfoo.foo() pre
@decorator pre foo decorator
foo.foo()
@decorator post foo decorator
subfoo.foo() post
@decorator post subfoo decorator
subsubfoo.foo() post
@decorator post subsubfoo decorator

I’ve used a function decorator, as I find them more concise. Here’s a class to decorate a class:

class Dec(object):

    def __init__(self, msg):
        self.msg = msg

    def __call__(self, klass):
        old_foo = klass.foo
        msg = self.msg
        def decorated_foo(self, *args, **kwargs):
            print('@decorator pre %s' % msg)
            old_foo(self, *args, **kwargs)
            print('@decorator post %s' % msg)
        klass.foo = decorated_foo
        return klass

A more robust version that checks for those parentheses, and works if the methods don’t exist on the decorated class:

from inspect import isclass

def decorate_if(condition, decorator):
    return decorator if condition else lambda x: x

def dec(msg):
    # Only use if your decorator's first parameter is never a class
    assert not isclass(msg)

    def decorator(klass):
        old_foo = getattr(klass, 'foo', None)

        @decorate_if(old_foo, wraps(klass.foo))
        def decorated_foo(self, *args ,**kwargs):
            print('@decorator pre %s' % msg)
            if callable(old_foo):
                old_foo(self, *args, **kwargs)
            print('@decorator post %s' % msg)

        klass.foo = decorated_foo
        return klass

    return decorator

The assert checks that the decorator has not been used without parentheses. If it has, then the class being decorated is passed to the msg parameter of the decorator, which raises an AssertionError.

@decorate_if only applies the decorator if condition evaluates to True.

The getattr, callable test, and @decorate_if are used so that the decorator doesn’t break if the foo() method doesn’t exist on the class being decorated.

回答 5

实际上，这里有一个很好的类装饰器实现：

https://github.com/agiliq/Django-parsley/blob/master/parsley/decorators.py

我实际上认为这是一个非常有趣的实现。因为它继承了它装饰的类的子类，所以在诸如isinstance检查。

它还有另外一个好处：它并不罕见__init__在自定义声明Django表单进行修改或补充，self.fields因此，最好的改变self.fields发生后，所有的__init__已经跑了有问题的类。

非常聪明。

但是，在您的类中，您实际上希望装饰更改构造函数，但我认为这不是类装饰器的好用例。

点击查看英文原文

There’s actually a pretty good implementation of a class decorator here:

https://github.com/agiliq/Django-parsley/blob/master/parsley/decorators.py

I actually think this is a pretty interesting implementation. Because it subclasses the class it decorates, it will behave exactly like this class in things like isinstance checks.

It has an added benefit: it’s not uncommon for the __init__ statement in a custom django Form to make modifications or additions to self.fields so it’s better for changes to self.fields to happen after all of __init__ has run for the class in question.

Very clever.

However, in your class you actually want the decoration to alter the constructor, which I don’t think is a good use case for a class decorator.

回答 6

这是一个示例，它回答了返回类参数的问题。而且，它仍然尊重继承链，即仅返回类本身的参数。get_params作为一个简单的示例，添加了该功能，但是借助inspect模块，可以添加其他功能。

import inspect 

class Parent:
    @classmethod
    def get_params(my_class):
        return list(inspect.signature(my_class).parameters.keys())

class OtherParent:
    def __init__(self, a, b, c):
        pass

class Child(Parent, OtherParent):
    def __init__(self, x, y, z):
        pass

print(Child.get_params())
>>['x', 'y', 'z']
点击查看英文原文

Here is an example which answers the question of returning the parameters of a class. Moreover, it still respects the chain of inheritance, i.e. only the parameters of the class itself are returned. The function get_params is added as a simple example, but other functionalities can be added thanks to the inspect module. 

import inspect 

class Parent:
    @classmethod
    def get_params(my_class):
        return list(inspect.signature(my_class).parameters.keys())

class OtherParent:
    def __init__(self, a, b, c):
        pass

class Child(Parent, OtherParent):
    def __init__(self, x, y, z):
        pass

print(Child.get_params())
>>['x', 'y', 'z']

回答 7

Django具有method_decorator一个装饰器，可以将任何装饰器转换为方法装饰器，您可以看到它是如何实现的django.utils.decorators：

https://github.com/django/django/blob/50cf183d219face91822c75fa0a15fe2fe3cb32d/django/utils/decorators.py#L53

https://docs.djangoproject.com/zh-CN/3.0/topics/class-based-views/intro/#decorating-the-class

点击查看英文原文

Django has method_decorator which is a decorator that turns any decorator into a method decorator, you can see how it’s implemented in django.utils.decorators:

https://github.com/django/django/blob/50cf183d219face91822c75fa0a15fe2fe3cb32d/django/utils/decorators.py#L53

https://docs.djangoproject.com/en/3.0/topics/class-based-views/intro/#decorating-the-class

知识问答

从__future__导入absolute_import实际起什么作用？

问题：从__future__导入absolute_import实际起什么作用？

我已经回答了有关Python中绝对导入的问题，我认为通过阅读Python 2.5 changelog和随附的PEP可以理解。但是，在安装Python 2.5并尝试制作一个正确使用的示例时from __future__ import absolute_import，我意识到事情还不清楚。

直接从上面链接的更改日志，此语句准确总结了我对绝对导入更改的理解：

假设您有一个像这样的包目录：

pkg/
pkg/__init__.py
pkg/main.py
pkg/string.py

这定义了一个名为的包，pkg其中包含pkg.main和pkg.string子模块。

考虑main.py模块中的代码。如果执行该语句会import string怎样？在Python 2.4和更早的版本，它会先看看在包的目录进行相对进口，发现包装/ string.py，进口该文件的内容pkg.string模块，并且该模块被绑定到名字"string"的pkg.main模块的命名空间。

所以我创建了这个确切的目录结构：

$ ls -R
.:
pkg/

./pkg:
__init__.py  main.py  string.py

__init__.py并且string.py是空的。main.py包含以下代码：

import string
print string.ascii_uppercase

不出所料，使用Python 2.5运行此命令失败，并显示AttributeError：

$ python2.5 pkg/main.py
Traceback (most recent call last):
  File "pkg/main.py", line 2, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

但是，在2.5更新日志中，我们发现了这一点（添加了重点）：

在Python 2.5中，您可以import使用from __future__ import absolute_import指令将的行为切换为绝对导入。在将来的版本（可能是Python 2.7）中，此绝对导入行为将成为默认设置。一旦将绝对导入设置为默认设置，import string就将始终找到标准库的版本。

因此pkg/main2.py，我创建了，main.py但与将来的import指令相同。现在看起来像这样：

from __future__ import absolute_import
import string
print string.ascii_uppercase

但是，使用Python 2.5运行它会失败AttributeError：

$ python2.5 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 3, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

这个漂亮的断然反驳该语句import string将始终找到STD-库版本绝对进口启用。而且，尽管警告了绝对导入已计划成为“新的默认”行为，但无论使用还是不使用__future__指令，我都使用Python 2.7遇到了相同的问题：

$ python2.7 pkg/main.py
Traceback (most recent call last):
  File "pkg/main.py", line 2, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

$ python2.7 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 3, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

以及Python 3.5（有无）（假设print两个文件中的语句均已更改）：

$ python3.5 pkg/main.py
Traceback (most recent call last):
  File "pkg/main.py", line 2, in <module>
    print(string.ascii_uppercase)
AttributeError: module 'string' has no attribute 'ascii_uppercase'

$ python3.5 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 3, in <module>
    print(string.ascii_uppercase)
AttributeError: module 'string' has no attribute 'ascii_uppercase'

我已经测试了其他变化。相反的string.py，我已经创建了一个空的模块-一个新的目录string仅包含一个空的__init__.py-而不是从发放的进口main.py，我有cd“d到pkg并直接从REPL运行的进口。这些变体（或它们的组合）都没有改变上面的结果。我无法将其与我已阅读的有关__future__指令和绝对导入的内容相协调。

在我看来，这很容易通过以下方式进行解释（这是来自Python 2文档，但该语句在适用于Python 3的同一文档中保持不变）：

系统路径

（…）

在程序启动时进行初始化，该列表的第一项path[0]是包含用于调用Python解释器的脚本的目录。如果脚本目录不可用（例如，如果解释器是交互式调用的，或者从标准输入中读取了脚本），path[0]则为空字符串，该字符串将Python首先引导到当前目录中的搜索模块。

那我想念什么呢？为什么该__future__声明似乎不按其要求行事？在文档的这两部分之间以及所描述的行为与实际行为之间，这种矛盾的解决方案是什么？

点击查看英文原文

I have answered a question regarding absolute imports in Python, which I thought I understood based on reading the Python 2.5 changelog and accompanying PEP. However, upon installing Python 2.5 and attempting to craft an example of properly using from __future__ import absolute_import, I realize things are not so clear.

Straight from the changelog linked above, this statement accurately summarized my understanding of the absolute import change:

Let’s say you have a package directory like this:

pkg/
pkg/__init__.py
pkg/main.py
pkg/string.py

This defines a package named pkg containing the pkg.main and pkg.string submodules.

Consider the code in the main.py module. What happens if it executes the statement import string? In Python 2.4 and earlier, it will first look in the package’s directory to perform a relative import, finds pkg/string.py, imports the contents of that file as the pkg.string module, and that module is bound to the name "string" in the pkg.main module’s namespace.

So I created this exact directory structure:

$ ls -R
.:
pkg/

./pkg:
__init__.py  main.py  string.py

__init__.py and string.py are empty. main.py contains the following code:

import string
print string.ascii_uppercase

As expected, running this with Python 2.5 fails with an AttributeError:

$ python2.5 pkg/main.py
Traceback (most recent call last):
  File "pkg/main.py", line 2, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

However, further along in the 2.5 changelog, we find this (emphasis added):

In Python 2.5, you can switch import‘s behaviour to absolute imports using a from __future__ import absolute_import directive. This absolute-import behaviour will become the default in a future version (probably Python 2.7). Once absolute imports are the default, import string will always find the standard library’s version.

I thus created pkg/main2.py, identical to main.py but with the additional future import directive. It now looks like this:

from __future__ import absolute_import
import string
print string.ascii_uppercase

Running this with Python 2.5, however… fails with an AttributeError:

$ python2.5 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 3, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

This pretty flatly contradicts the statement that import string will always find the std-lib version with absolute imports enabled. What’s more, despite the warning that absolute imports are scheduled to become the “new default” behavior, I hit this same problem using both Python 2.7, with or without the __future__ directive:

$ python2.7 pkg/main.py
Traceback (most recent call last):
  File "pkg/main.py", line 2, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

$ python2.7 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 3, in <module>
    print string.ascii_uppercase
AttributeError: 'module' object has no attribute 'ascii_uppercase'

as well as Python 3.5, with or without (assuming the print statement is changed in both files):

$ python3.5 pkg/main.py
Traceback (most recent call last):
  File "pkg/main.py", line 2, in <module>
    print(string.ascii_uppercase)
AttributeError: module 'string' has no attribute 'ascii_uppercase'

$ python3.5 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 3, in <module>
    print(string.ascii_uppercase)
AttributeError: module 'string' has no attribute 'ascii_uppercase'

I have tested other variations of this. Instead of string.py, I have created an empty module — a directory named string containing only an empty __init__.py — and instead of issuing imports from main.py, I have cd‘d to pkg and run imports directly from the REPL. Neither of these variations (nor a combination of them) changed the results above. I cannot reconcile this with what I have read about the __future__ directive and absolute imports.

It seems to me that this is easily explicable by the following (this is from the Python 2 docs but this statement remains unchanged in the same docs for Python 3):

sys.path

(…)

As initialized upon program startup, the first item of this list, path[0], is the directory containing the script that was used to invoke the Python interpreter. If the script directory is not available (e.g. if the interpreter is invoked interactively or if the script is read from standard input), path[0] is the empty string, which directs Python to search modules in the current directory first.

So what am I missing? Why does the __future__ statement seemingly not do what it says, and what is the resolution of this contradiction between these two sections of documentation, as well as between described and actual behavior?

回答 0

变更日志的字眼很草率。from __future__ import absolute_import并不关心某些东西是否是标准库的一部分，import string也不会总是为您提供启用绝对导入的标准库模块。

from __future__ import absolute_import表示如果您愿意import string，Python将始终寻找顶级string模块，而不是current_package.string。但是，它不会影响Python用于确定string模块文件的逻辑。当你做

python pkg/script.py

pkg/script.py看起来不像是Python软件包的一部分。按照正常步骤，将pkg目录添加到路径，.py目录中的所有文件pkg看起来都像顶级模块。import string发现pkg/string.py不是因为它正在执行相对导入，而是因为它pkg/string.py似乎是顶级模块string。这不是标准库string模块的事实并没有出现。

要将文件作为pkg软件包的一部分运行，您可以

python -m pkg.script

在这种情况下，该pkg目录将不会添加到路径中。但是，当前目录将被添加到路径中。

您还可以添加一些样板，pkg/script.py以使Python pkg即使在作为文件运行时也将其视为包的一部分：

if __name__ == '__main__' and __package__ is None:
    __package__ = 'pkg'

但是，这不会影响sys.path。您需要进行一些其他处理才能pkg从路径中删除目录，并且如果pkg的父目录不在路径中，则也需要将其粘贴在路径中。

点击查看英文原文

The changelog is sloppily worded. from __future__ import absolute_import does not care about whether something is part of the standard library, and import string will not always give you the standard-library module with absolute imports on.

from __future__ import absolute_import means that if you import string, Python will always look for a top-level string module, rather than current_package.string. However, it does not affect the logic Python uses to decide what file is the string module. When you do

python pkg/script.py

pkg/script.py doesn’t look like part of a package to Python. Following the normal procedures, the pkg directory is added to the path, and all .py files in the pkg directory look like top-level modules. import string finds pkg/string.py not because it’s doing a relative import, but because pkg/string.py appears to be the top-level module string. The fact that this isn’t the standard-library string module doesn’t come up.

To run the file as part of the pkg package, you could do

python -m pkg.script

In this case, the pkg directory will not be added to the path. However, the current directory will be added to the path.

You can also add some boilerplate to pkg/script.py to make Python treat it as part of the pkg package even when run as a file:

if __name__ == '__main__' and __package__ is None:
    __package__ = 'pkg'

However, this won’t affect sys.path. You’ll need some additional handling to remove the pkg directory from the path, and if pkg‘s parent directory isn’t on the path, you’ll need to stick that on the path too.

回答 1

绝对导入和相对导入之间的差异仅在您从包中导入一个模块并且该模块从该包中导入另一个子模块时才起作用。看到不同：

$ mkdir pkg
$ touch pkg/__init__.py
$ touch pkg/string.py
$ echo 'import string;print(string.ascii_uppercase)' > pkg/main1.py
$ python2
Python 2.7.9 (default, Dec 13 2014, 18:02:08) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pkg.main1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "pkg/main1.py", line 1, in <module>
    import string;print(string.ascii_uppercase)
AttributeError: 'module' object has no attribute 'ascii_uppercase'
>>> 
$ echo 'from __future__ import absolute_import;import string;print(string.ascii_uppercase)' > pkg/main2.py
$ python2
Python 2.7.9 (default, Dec 13 2014, 18:02:08) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pkg.main2
ABCDEFGHIJKLMNOPQRSTUVWXYZ
>>>

特别是：

$ python2 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 1, in <module>
    from __future__ import absolute_import;import string;print(string.ascii_uppercase)
AttributeError: 'module' object has no attribute 'ascii_uppercase'
$ python2
Python 2.7.9 (default, Dec 13 2014, 18:02:08) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pkg.main2
ABCDEFGHIJKLMNOPQRSTUVWXYZ
>>> 
$ python2 -m pkg.main2
ABCDEFGHIJKLMNOPQRSTUVWXYZ

请注意，python2 pkg/main2.py启动python2和导入后的行为不同pkg.main2（等同于使用-m开关）。

如果要运行程序包的子模块，请始终使用该-m开关，该开关可防止解释器链接sys.path列表并正确处理子模块的语义。

另外，我更喜欢对包子模块使用显式相对导入，因为它们在失败的情况下提供了更多的语义和更好的错误消息。

点击查看英文原文

The difference between absolute and relative imports come into play only when you import a module from a package and that module imports an other submodule from that package. See the difference:

$ mkdir pkg
$ touch pkg/__init__.py
$ touch pkg/string.py
$ echo 'import string;print(string.ascii_uppercase)' > pkg/main1.py
$ python2
Python 2.7.9 (default, Dec 13 2014, 18:02:08) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pkg.main1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "pkg/main1.py", line 1, in <module>
    import string;print(string.ascii_uppercase)
AttributeError: 'module' object has no attribute 'ascii_uppercase'
>>> 
$ echo 'from __future__ import absolute_import;import string;print(string.ascii_uppercase)' > pkg/main2.py
$ python2
Python 2.7.9 (default, Dec 13 2014, 18:02:08) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pkg.main2
ABCDEFGHIJKLMNOPQRSTUVWXYZ
>>>

In particular:

$ python2 pkg/main2.py
Traceback (most recent call last):
  File "pkg/main2.py", line 1, in <module>
    from __future__ import absolute_import;import string;print(string.ascii_uppercase)
AttributeError: 'module' object has no attribute 'ascii_uppercase'
$ python2
Python 2.7.9 (default, Dec 13 2014, 18:02:08) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pkg.main2
ABCDEFGHIJKLMNOPQRSTUVWXYZ
>>> 
$ python2 -m pkg.main2
ABCDEFGHIJKLMNOPQRSTUVWXYZ

Note that python2 pkg/main2.py has a different behaviour then launching python2 and then importing pkg.main2 (which is equivalent to using the -m switch).

If you ever want to run a submodule of a package always use the -m switch which prevents the interpreter for chaining the sys.path list and correctly handles the semantics of the submodule.

Also, I much prefer using explicit relative imports for package submodules since they provide more semantics and better error messages in case of failure.

知识问答

如何生成列表的所有排列？

问题：如何生成列表的所有排列？

如何在Python中生成列表的所有排列，而与列表中元素的类型无关？

例如：

permutations([])
[]

permutations([1])
[1]

permutations([1, 2])
[1, 2]
[2, 1]

permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
点击查看英文原文

How do you generate all the permutations of a list in Python, independently of the type of elements in that list?

For example:

permutations([])
[]

permutations([1])
[1]

permutations([1, 2])
[1, 2]
[2, 1]

permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

回答 0

从Python 2.6开始（如果您使用的是Python 3），您可以使用标准库工具：itertools.permutations。

import itertools
list(itertools.permutations([1, 2, 3]))

如果您出于某种原因使用旧版Python（<2.6），或者只是想知道它的工作原理，那么这是一种不错的方法，摘自 http://code.activestate.com/recipes/252178/：

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                # nb elements[0:1] works in both string and list contexts
                yield perm[:i] + elements[0:1] + perm[i:]

的文档中列出了几种其他方法itertools.permutations。这是一个：

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

另一个基于itertools.product：

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)
点击查看英文原文

Starting with Python 2.6 (and if you’re on Python 3) you have a standard-library tool for this: itertools.permutations.

import itertools
list(itertools.permutations([1, 2, 3]))

If you’re using an older Python (<2.6) for some reason or are just curious to know how it works, here’s one nice approach, taken from http://code.activestate.com/recipes/252178/:

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                # nb elements[0:1] works in both string and list contexts
                yield perm[:i] + elements[0:1] + perm[i:]

A couple of alternative approaches are listed in the documentation of itertools.permutations. Here’s one:

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

And another, based on itertools.product:

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

回答 1

在Python 2.6及更高版本中：

import itertools
itertools.permutations([1,2,3])

（作为生成器返回。用于list(permutations(l))作为列表返回。）

点击查看英文原文

And in Python 2.6 onwards:

import itertools
itertools.permutations([1,2,3])

(returned as a generator. Use list(permutations(l)) to return as a list.)

回答 2

以下代码仅适用于Python 2.6及更高版本

首先，导入itertools：

import itertools

排列（顺序很重要）：

print list(itertools.permutations([1,2,3,4], 2))
[(1, 2), (1, 3), (1, 4),
(2, 1), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 4),
(4, 1), (4, 2), (4, 3)]

组合（顺序无关紧要）：

print list(itertools.combinations('123', 2))
[('1', '2'), ('1', '3'), ('2', '3')]

笛卡尔积（具有多个可迭代项）：

print list(itertools.product([1,2,3], [4,5,6]))
[(1, 4), (1, 5), (1, 6),
(2, 4), (2, 5), (2, 6),
(3, 4), (3, 5), (3, 6)]

笛卡尔积（本身具有一个可迭代的）：

print list(itertools.product([1,2], repeat=3))
[(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2),
(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)]
点击查看英文原文

The following code with Python 2.6 and above ONLY

First, import itertools:

import itertools

Permutation (order matters):

print list(itertools.permutations([1,2,3,4], 2))
[(1, 2), (1, 3), (1, 4),
(2, 1), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 4),
(4, 1), (4, 2), (4, 3)]

Combination (order does NOT matter):

print list(itertools.combinations('123', 2))
[('1', '2'), ('1', '3'), ('2', '3')]

Cartesian product (with several iterables):

print list(itertools.product([1,2,3], [4,5,6]))
[(1, 4), (1, 5), (1, 6),
(2, 4), (2, 5), (2, 6),
(3, 4), (3, 5), (3, 6)]

Cartesian product (with one iterable and itself):

print list(itertools.product([1,2], repeat=3))
[(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2),
(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)]

回答 3

def permutations(head, tail=''):
    if len(head) == 0: print tail
    else:
        for i in range(len(head)):
            permutations(head[0:i] + head[i+1:], tail+head[i])

称为：

permutations('abc')
点击查看英文原文 
def permutations(head, tail=''):
    if len(head) == 0: print tail
    else:
        for i in range(len(head)):
            permutations(head[0:i] + head[i+1:], tail+head[i])

called as:

permutations('abc')

回答 4

#!/usr/bin/env python

def perm(a, k=0):
   if k == len(a):
      print a
   else:
      for i in xrange(k, len(a)):
         a[k], a[i] = a[i] ,a[k]
         perm(a, k+1)
         a[k], a[i] = a[i], a[k]

perm([1,2,3])

输出：

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 2, 1]
[3, 1, 2]

当我交换列表的内容时，需要一个可变的序列类型作为输入。例如，perm(list("ball"))将工作，perm("ball")不会因为您不能更改字符串。

此Python实现受Horowitz，Sahni和Rajasekeran的《计算机算法》一书中介绍的算法的启发。

点击查看英文原文 
#!/usr/bin/env python

def perm(a, k=0):
   if k == len(a):
      print a
   else:
      for i in xrange(k, len(a)):
         a[k], a[i] = a[i] ,a[k]
         perm(a, k+1)
         a[k], a[i] = a[i], a[k]

perm([1,2,3])

Output:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 2, 1]
[3, 1, 2]

As I’m swapping the content of the list it’s required a mutable sequence type as input. E.g. perm(list("ball")) will work and perm("ball") won’t because you can’t change a string.

This Python implementation is inspired by the algorithm presented in the book Computer Algorithms by Horowitz, Sahni and Rajasekeran.

回答 5

此解决方案实现了一个生成器，以避免将所有排列保留在内存中：

def permutations (orig_list):
    if not isinstance(orig_list, list):
        orig_list = list(orig_list)

    yield orig_list

    if len(orig_list) == 1:
        return

    for n in sorted(orig_list):
        new_list = orig_list[:]
        pos = new_list.index(n)
        del(new_list[pos])
        new_list.insert(0, n)
        for resto in permutations(new_list[1:]):
            if new_list[:1] + resto <> orig_list:
                yield new_list[:1] + resto
点击查看英文原文

This solution implements a generator, to avoid holding all the permutations on memory:

def permutations (orig_list):
    if not isinstance(orig_list, list):
        orig_list = list(orig_list)

    yield orig_list

    if len(orig_list) == 1:
        return

    for n in sorted(orig_list):
        new_list = orig_list[:]
        pos = new_list.index(n)
        del(new_list[pos])
        new_list.insert(0, n)
        for resto in permutations(new_list[1:]):
            if new_list[:1] + resto <> orig_list:
                yield new_list[:1] + resto

回答 6

以实用的风格

def addperm(x,l):
    return [ l[0:i] + [x] + l[i:]  for i in range(len(l)+1) ]

def perm(l):
    if len(l) == 0:
        return [[]]
    return [x for y in perm(l[1:]) for x in addperm(l[0],y) ]

print perm([ i for i in range(3)])

结果：

[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]]
点击查看英文原文

In a functional style

def addperm(x,l):
    return [ l[0:i] + [x] + l[i:]  for i in range(len(l)+1) ]

def perm(l):
    if len(l) == 0:
        return [[]]
    return [x for y in perm(l[1:]) for x in addperm(l[0],y) ]

print perm([ i for i in range(3)])

The result:

[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]]

回答 7

以下代码是给定列表的就地排列，实现为生成器。由于仅返回对列表的引用，因此不应在生成器外部修改列表。该解决方案是非递归的，因此使用低内存。输入列表中元素的多个副本也可以很好地工作。

def permute_in_place(a):
    a.sort()
    yield list(a)

    if len(a) <= 1:
        return

    first = 0
    last = len(a)
    while 1:
        i = last - 1

        while 1:
            i = i - 1
            if a[i] < a[i+1]:
                j = last - 1
                while not (a[i] < a[j]):
                    j = j - 1
                a[i], a[j] = a[j], a[i] # swap the values
                r = a[i+1:last]
                r.reverse()
                a[i+1:last] = r
                yield list(a)
                break
            if i == first:
                a.reverse()
                return

if __name__ == '__main__':
    for n in range(5):
        for a in permute_in_place(range(1, n+1)):
            print a
        print

    for a in permute_in_place([0, 0, 1, 1, 1]):
        print a
    print
点击查看英文原文

The following code is an in-place permutation of a given list, implemented as a generator. Since it only returns references to the list, the list should not be modified outside the generator. The solution is non-recursive, so uses low memory. Work well also with multiple copies of elements in the input list.

def permute_in_place(a):
    a.sort()
    yield list(a)

    if len(a) <= 1:
        return

    first = 0
    last = len(a)
    while 1:
        i = last - 1

        while 1:
            i = i - 1
            if a[i] < a[i+1]:
                j = last - 1
                while not (a[i] < a[j]):
                    j = j - 1
                a[i], a[j] = a[j], a[i] # swap the values
                r = a[i+1:last]
                r.reverse()
                a[i+1:last] = r
                yield list(a)
                break
            if i == first:
                a.reverse()
                return

if __name__ == '__main__':
    for n in range(5):
        for a in permute_in_place(range(1, n+1)):
            print a
        print

    for a in permute_in_place([0, 0, 1, 1, 1]):
        print a
    print

回答 8

我认为一种很明显的方式可能是：

def permutList(l):
    if not l:
            return [[]]
    res = []
    for e in l:
            temp = l[:]
            temp.remove(e)
            res.extend([[e] + r for r in permutList(temp)])

    return res
点击查看英文原文

A quite obvious way in my opinion might be also:

def permutList(l):
    if not l:
            return [[]]
    res = []
    for e in l:
            temp = l[:]
            temp.remove(e)
            res.extend([[e] + r for r in permutList(temp)])

    return res

回答 9

list2Perm = [1, 2.0, 'three']
listPerm = [[a, b, c]
            for a in list2Perm
            for b in list2Perm
            for c in list2Perm
            if ( a != b and b != c and a != c )
            ]
print listPerm

输出：

[
    [1, 2.0, 'three'], 
    [1, 'three', 2.0], 
    [2.0, 1, 'three'], 
    [2.0, 'three', 1], 
    ['three', 1, 2.0], 
    ['three', 2.0, 1]
]
点击查看英文原文 
list2Perm = [1, 2.0, 'three']
listPerm = [[a, b, c]
            for a in list2Perm
            for b in list2Perm
            for c in list2Perm
            if ( a != b and b != c and a != c )
            ]
print listPerm

Output:

[
    [1, 2.0, 'three'], 
    [1, 'three', 2.0], 
    [2.0, 1, 'three'], 
    [2.0, 'three', 1], 
    ['three', 1, 2.0], 
    ['three', 2.0, 1]
]

回答 10

我使用了基于阶乘数系统的算法-对于长度为n的列表，您可以逐项组合每个排列项，并从每个阶段剩下的项中进行选择。第一项有n个选择，第二项有n-1个，最后一项只有n个，因此可以将阶乘数字系统中数字的数字用作索引。这样，数字0到n！-1对应于字典顺序中所有可能的排列。

from math import factorial
def permutations(l):
    permutations=[]
    length=len(l)
    for x in xrange(factorial(length)):
        available=list(l)
        newPermutation=[]
        for radix in xrange(length, 0, -1):
            placeValue=factorial(radix-1)
            index=x/placeValue
            newPermutation.append(available.pop(index))
            x-=index*placeValue
        permutations.append(newPermutation)
    return permutations

permutations(range(3))

输出：

[[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]

该方法是非递归的，但是在我的计算机上它会稍微慢一些，并且当n！时xrange会引发错误！太大而无法转换为C长整数（对我来说n = 13）。当我需要它时就足够了，但是从长远来看这不是itertools.permutations。

点击查看英文原文

I used an algorithm based on the factorial number system– For a list of length n, you can assemble each permutation item by item, selecting from the items left at each stage. You have n choices for the first item, n-1 for the second, and only one for the last, so you can use the digits of a number in the factorial number system as the indices. This way the numbers 0 through n!-1 correspond to all possible permutations in lexicographic order.

from math import factorial
def permutations(l):
    permutations=[]
    length=len(l)
    for x in xrange(factorial(length)):
        available=list(l)
        newPermutation=[]
        for radix in xrange(length, 0, -1):
            placeValue=factorial(radix-1)
            index=x/placeValue
            newPermutation.append(available.pop(index))
            x-=index*placeValue
        permutations.append(newPermutation)
    return permutations

permutations(range(3))

output:

[[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]

This method is non-recursive, but it is slightly slower on my computer and xrange raises an error when n! is too large to be converted to a C long integer (n=13 for me). It was enough when I needed it, but it’s no itertools.permutations by a long shot.

回答 11

请注意，此算法具有n factorial时间复杂度，其中n是输入列表的长度

打印运行结果：

global result
result = [] 

def permutation(li):
if li == [] or li == None:
    return

if len(li) == 1:
    result.append(li[0])
    print result
    result.pop()
    return

for i in range(0,len(li)):
    result.append(li[i])
    permutation(li[:i] + li[i+1:])
    result.pop()

例： 

permutation([1,2,3])

输出：

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
点击查看英文原文

Note that this algorithm has an n factorial time complexity, where n is the length of the input list

Print the results on the run:

global result
result = [] 

def permutation(li):
if li == [] or li == None:
    return

if len(li) == 1:
    result.append(li[0])
    print result
    result.pop()
    return

for i in range(0,len(li)):
    result.append(li[i])
    permutation(li[:i] + li[i+1:])
    result.pop()

Example: 

permutation([1,2,3])

Output:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

回答 12

正如tzwenn的回答，确实可以迭代每个排列的第一个元素。但是，以这种方式编写此解决方案效率更高：

def all_perms(elements):
    if len(elements) <= 1:
        yield elements  # Only permutation possible = no permutation
    else:
        # Iteration over the first element in the result permutation:
        for (index, first_elmt) in enumerate(elements):
            other_elmts = elements[:index]+elements[index+1:]
            for permutation in all_perms(other_elmts): 
                yield [first_elmt] + permutation

该解决方案的速度提高了约30％，这显然归功于递归以len(elements) <= 1代替0。yield就像Riccardo Reyes的解决方案一样，它使用生成器函数（通过），因此内存效率也更高。

点击查看英文原文

One can indeed iterate over the first element of each permutation, as in tzwenn’s answer. It is however more efficient to write this solution this way:

def all_perms(elements):
    if len(elements) <= 1:
        yield elements  # Only permutation possible = no permutation
    else:
        # Iteration over the first element in the result permutation:
        for (index, first_elmt) in enumerate(elements):
            other_elmts = elements[:index]+elements[index+1:]
            for permutation in all_perms(other_elmts): 
                yield [first_elmt] + permutation

This solution is about 30 % faster, apparently thanks to the recursion ending at len(elements) <= 1 instead of 0. It is also much more memory-efficient, as it uses a generator function (through yield), like in Riccardo Reyes’s solution.

回答 13

这是受Haskell实现的启发，该实现使用列表理解： 

def permutation(list):
    if len(list) == 0:
        return [[]]
    else:
        return [[x] + ys for x in list for ys in permutation(delete(list, x))]

def delete(list, item):
    lc = list[:]
    lc.remove(item)
    return lc
点击查看英文原文

This is inspired by the Haskell implementation using list comprehension: 

def permutation(list):
    if len(list) == 0:
        return [[]]
    else:
        return [[x] + ys for x in list for ys in permutation(delete(list, x))]

def delete(list, item):
    lc = list[:]
    lc.remove(item)
    return lc

回答 14

常规执行（无收益-将在内存中做所有事情）：

def getPermutations(array):
    if len(array) == 1:
        return [array]
    permutations = []
    for i in range(len(array)): 
        # get all perm's of subarray w/o current item
        perms = getPermutations(array[:i] + array[i+1:])  
        for p in perms:
            permutations.append([array[i], *p])
    return permutations

收益实施：

def getPermutations(array):
    if len(array) == 1:
        yield array
    else:
        for i in range(len(array)):
            perms = getPermutations(array[:i] + array[i+1:])
            for p in perms:
                yield [array[i], *p]

基本思想是遍历数组中所有元素的第1个位置，然后在第2个位置中遍历所有其余元素，而第1个位置没有选择元素，依此类推。您可以使用recursion进行操作，其中停止条件是到达由1个元素组成的数组-在这种情况下，您将返回该数组。

点击查看英文原文

Regular implementation (no yield – will do everything in memory):

def getPermutations(array):
    if len(array) == 1:
        return [array]
    permutations = []
    for i in range(len(array)): 
        # get all perm's of subarray w/o current item
        perms = getPermutations(array[:i] + array[i+1:])  
        for p in perms:
            permutations.append([array[i], *p])
    return permutations

Yield implementation:

def getPermutations(array):
    if len(array) == 1:
        yield array
    else:
        for i in range(len(array)):
            perms = getPermutations(array[:i] + array[i+1:])
            for p in perms:
                yield [array[i], *p]

The basic idea is to go over all the elements in the array for the 1st position, and then in 2nd position go over all the rest of the elements without the chosen element for the 1st, etc. You can do this with recursion, where the stop criteria is getting to an array of 1 element – in which case you return that array.

回答 15

为了提高性能，从Knuth（p22）那里得到了一个麻木的解决方案：

from numpy import empty, uint8
from math import factorial

def perms(n):
    f = 1
    p = empty((2*n-1, factorial(n)), uint8)
    for i in range(n):
        p[i, :f] = i
        p[i+1:2*i+1, :f] = p[:i, :f]  # constitution de blocs
        for j in range(i):
            p[:i+1, f*(j+1):f*(j+2)] = p[j+1:j+i+2, :f]  # copie de blocs
        f = f*(i+1)
    return p[:n, :]

复制大块内存可以节省时间-比list(itertools.permutations(range(n))以下方法快20倍：

In [1]: %timeit -n10 list(permutations(range(10)))
10 loops, best of 3: 815 ms per loop

In [2]: %timeit -n100 perms(10) 
100 loops, best of 3: 40 ms per loop
点击查看英文原文

For performance, a numpy solution inspired by Knuth, (p22) :

from numpy import empty, uint8
from math import factorial

def perms(n):
    f = 1
    p = empty((2*n-1, factorial(n)), uint8)
    for i in range(n):
        p[i, :f] = i
        p[i+1:2*i+1, :f] = p[:i, :f]  # constitution de blocs
        for j in range(i):
            p[:i+1, f*(j+1):f*(j+2)] = p[j+1:j+i+2, :f]  # copie de blocs
        f = f*(i+1)
    return p[:n, :]

Copying large blocs of memory saves time – it’s 20x faster than list(itertools.permutations(range(n)) :

In [1]: %timeit -n10 list(permutations(range(10)))
10 loops, best of 3: 815 ms per loop

In [2]: %timeit -n100 perms(10) 
100 loops, best of 3: 40 ms per loop

回答 16

from __future__ import print_function

def perm(n):
    p = []
    for i in range(0,n+1):
        p.append(i)
    while True:
        for i in range(1,n+1):
            print(p[i], end=' ')
        print("")
        i = n - 1
        found = 0
        while (not found and i>0):
            if p[i]<p[i+1]:
                found = 1
            else:
                i = i - 1
        k = n
        while p[i]>p[k]:
            k = k - 1
        aux = p[i]
        p[i] = p[k]
        p[k] = aux
        for j in range(1,(n-i)/2+1):
            aux = p[i+j]
            p[i+j] = p[n-j+1]
            p[n-j+1] = aux
        if not found:
            break

perm(5)
点击查看英文原文 
from __future__ import print_function

def perm(n):
    p = []
    for i in range(0,n+1):
        p.append(i)
    while True:
        for i in range(1,n+1):
            print(p[i], end=' ')
        print("")
        i = n - 1
        found = 0
        while (not found and i>0):
            if p[i]<p[i+1]:
                found = 1
            else:
                i = i - 1
        k = n
        while p[i]>p[k]:
            k = k - 1
        aux = p[i]
        p[i] = p[k]
        p[k] = aux
        for j in range(1,(n-i)/2+1):
            aux = p[i+j]
            p[i+j] = p[n-j+1]
            p[n-j+1] = aux
        if not found:
            break

perm(5)

回答 17

这是一种适用于列表的算法，无需创建新的中间列表，类似于https://stackoverflow.com/a/108651/184528上Ber的解决方案。

def permute(xs, low=0):
    if low + 1 >= len(xs):
        yield xs
    else:
        for p in permute(xs, low + 1):
            yield p        
        for i in range(low + 1, len(xs)):        
            xs[low], xs[i] = xs[i], xs[low]
            for p in permute(xs, low + 1):
                yield p        
            xs[low], xs[i] = xs[i], xs[low]

for p in permute([1, 2, 3, 4]):
    print p

您可以在此处亲自尝试该代码：http : //repl.it/J9v

点击查看英文原文

Here is an algorithm that works on a list without creating new intermediate lists similar to Ber’s solution at https://stackoverflow.com/a/108651/184528. 

def permute(xs, low=0):
    if low + 1 >= len(xs):
        yield xs
    else:
        for p in permute(xs, low + 1):
            yield p        
        for i in range(low + 1, len(xs)):        
            xs[low], xs[i] = xs[i], xs[low]
            for p in permute(xs, low + 1):
                yield p        
            xs[low], xs[i] = xs[i], xs[low]

for p in permute([1, 2, 3, 4]):
    print p

You can try the code out for yourself here: http://repl.it/J9v

回答 18

递归的美丽：

>>> import copy
>>> def perm(prefix,rest):
...      for e in rest:
...              new_rest=copy.copy(rest)
...              new_prefix=copy.copy(prefix)
...              new_prefix.append(e)
...              new_rest.remove(e)
...              if len(new_rest) == 0:
...                      print new_prefix + new_rest
...                      continue
...              perm(new_prefix,new_rest)
... 
>>> perm([],['a','b','c','d'])
['a', 'b', 'c', 'd']
['a', 'b', 'd', 'c']
['a', 'c', 'b', 'd']
['a', 'c', 'd', 'b']
['a', 'd', 'b', 'c']
['a', 'd', 'c', 'b']
['b', 'a', 'c', 'd']
['b', 'a', 'd', 'c']
['b', 'c', 'a', 'd']
['b', 'c', 'd', 'a']
['b', 'd', 'a', 'c']
['b', 'd', 'c', 'a']
['c', 'a', 'b', 'd']
['c', 'a', 'd', 'b']
['c', 'b', 'a', 'd']
['c', 'b', 'd', 'a']
['c', 'd', 'a', 'b']
['c', 'd', 'b', 'a']
['d', 'a', 'b', 'c']
['d', 'a', 'c', 'b']
['d', 'b', 'a', 'c']
['d', 'b', 'c', 'a']
['d', 'c', 'a', 'b']
['d', 'c', 'b', 'a']
点击查看英文原文

The beauty of recursion:

>>> import copy
>>> def perm(prefix,rest):
...      for e in rest:
...              new_rest=copy.copy(rest)
...              new_prefix=copy.copy(prefix)
...              new_prefix.append(e)
...              new_rest.remove(e)
...              if len(new_rest) == 0:
...                      print new_prefix + new_rest
...                      continue
...              perm(new_prefix,new_rest)
... 
>>> perm([],['a','b','c','d'])
['a', 'b', 'c', 'd']
['a', 'b', 'd', 'c']
['a', 'c', 'b', 'd']
['a', 'c', 'd', 'b']
['a', 'd', 'b', 'c']
['a', 'd', 'c', 'b']
['b', 'a', 'c', 'd']
['b', 'a', 'd', 'c']
['b', 'c', 'a', 'd']
['b', 'c', 'd', 'a']
['b', 'd', 'a', 'c']
['b', 'd', 'c', 'a']
['c', 'a', 'b', 'd']
['c', 'a', 'd', 'b']
['c', 'b', 'a', 'd']
['c', 'b', 'd', 'a']
['c', 'd', 'a', 'b']
['c', 'd', 'b', 'a']
['d', 'a', 'b', 'c']
['d', 'a', 'c', 'b']
['d', 'b', 'a', 'c']
['d', 'b', 'c', 'a']
['d', 'c', 'a', 'b']
['d', 'c', 'b', 'a']

回答 19

此算法是最有效的算法，它避免了在递归调用中进行数组传递和操作，在Python 2、3中有效：

def permute(items):
    length = len(items)
    def inner(ix=[]):
        do_yield = len(ix) == length - 1
        for i in range(0, length):
            if i in ix: #avoid duplicates
                continue
            if do_yield:
                yield tuple([items[y] for y in ix + [i]])
            else:
                for p in inner(ix + [i]):
                    yield p
    return inner()

用法：

for p in permute((1,2,3)):
    print(p)

(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
点击查看英文原文

This algorithm is the most effective one, it avoids of array passing and manipulation in recursive calls, works in Python 2, 3:

def permute(items):
    length = len(items)
    def inner(ix=[]):
        do_yield = len(ix) == length - 1
        for i in range(0, length):
            if i in ix: #avoid duplicates
                continue
            if do_yield:
                yield tuple([items[y] for y in ix + [i]])
            else:
                for p in inner(ix + [i]):
                    yield p
    return inner()

Usage:

for p in permute((1,2,3)):
    print(p)

(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

回答 20

def pzip(c, seq):
    result = []
    for item in seq:
        for i in range(len(item)+1):
            result.append(item[i:]+c+item[:i])
    return result


def perm(line):
    seq = [c for c in line]
    if len(seq) <=1 :
        return seq
    else:
        return pzip(seq[0], perm(seq[1:]))
点击查看英文原文 
def pzip(c, seq):
    result = []
    for item in seq:
        for i in range(len(item)+1):
            result.append(item[i:]+c+item[:i])
    return result


def perm(line):
    seq = [c for c in line]
    if len(seq) <=1 :
        return seq
    else:
        return pzip(seq[0], perm(seq[1:]))

回答 21

另一种方法（无库）

def permutation(input):
    if len(input) == 1:
        return input if isinstance(input, list) else [input]

    result = []
    for i in range(len(input)):
        first = input[i]
        rest = input[:i] + input[i + 1:]
        rest_permutation = permutation(rest)
        for p in rest_permutation:
            result.append(first + p)
    return result

输入可以是字符串或列表

print(permutation('abcd'))
print(permutation(['a', 'b', 'c', 'd']))
点击查看英文原文

ANOTHER APPROACH (without libs)

def permutation(input):
    if len(input) == 1:
        return input if isinstance(input, list) else [input]

    result = []
    for i in range(len(input)):
        first = input[i]
        rest = input[:i] + input[i + 1:]
        rest_permutation = permutation(rest)
        for p in rest_permutation:
            result.append(first + p)
    return result

Input can be a string or a list

print(permutation('abcd'))
print(permutation(['a', 'b', 'c', 'd']))

回答 22

免责声明：软件包作者提供的无形插件。:)

该猪手包是从大多数实现不同，它产生不实际包含的排列，而是描述的排列和各位置之间的映射关系的顺序，使其能够工作，排列非常大名单“，如图所示伪名单在这个演示中，执行了一个非常瞬时的操作并在“包含”字母中所有字母排列的伪列表中进行查找，而没有使用比典型的Web页面更多的内存或处理。

无论如何，要生成排列列表，我们可以执行以下操作。

import trotter

my_permutations = trotter.Permutations(3, [1, 2, 3])

print(my_permutations)

for p in my_permutations:
    print(p)

输出：

包含[1、2、3]的6个3排列的伪列表。
[1,2,3]
[1、3、2]
[3，1，2]
[3，2，1]
[2，3，1]
[2，1，3]
点击查看英文原文

Disclaimer: shapeless plug by package author. :)

The trotter package is different from most implementations in that it generates pseudo lists that don’t actually contain permutations but rather describe mappings between permutations and respective positions in an ordering, making it possible to work with very large ‘lists’ of permutations, as shown in this demo which performs pretty instantaneous operations and look-ups in a pseudo-list ‘containing’ all the permutations of the letters in the alphabet, without using more memory or processing than a typical web page.

In any case, to generate a list of permutations, we can do the following.

import trotter

my_permutations = trotter.Permutations(3, [1, 2, 3])

print(my_permutations)

for p in my_permutations:
    print(p)

Output:

A pseudo-list containing 6 3-permutations of [1, 2, 3].
[1, 2, 3]
[1, 3, 2]
[3, 1, 2]
[3, 2, 1]
[2, 3, 1]
[2, 1, 3]

回答 23

生成所有可能的排列

我正在使用python3.4：

def calcperm(arr, size):
    result = set([()])
    for dummy_idx in range(size):
        temp = set()
        for dummy_lst in result:
            for dummy_outcome in arr:
                if dummy_outcome not in dummy_lst:
                    new_seq = list(dummy_lst)
                    new_seq.append(dummy_outcome)
                    temp.add(tuple(new_seq))
        result = temp
    return result

测试用例：

lst = [1, 2, 3, 4]
#lst = ["yellow", "magenta", "white", "blue"]
seq = 2
final = calcperm(lst, seq)
print(len(final))
print(final)
点击查看英文原文

Generate all possible permutations

I’m using python3.4:

def calcperm(arr, size):
    result = set([()])
    for dummy_idx in range(size):
        temp = set()
        for dummy_lst in result:
            for dummy_outcome in arr:
                if dummy_outcome not in dummy_lst:
                    new_seq = list(dummy_lst)
                    new_seq.append(dummy_outcome)
                    temp.add(tuple(new_seq))
        result = temp
    return result

Test Cases:

lst = [1, 2, 3, 4]
#lst = ["yellow", "magenta", "white", "blue"]
seq = 2
final = calcperm(lst, seq)
print(len(final))
print(final)

回答 24

为了节省大家的搜索和实验时间，以下是Python中的非递归置换解决方案，该解决方案也适用于Numba（自0.41版起）：

@numba.njit()
def permutations(A, k):
    r = [[i for i in range(0)]]
    for i in range(k):
        r = [[a] + b for a in A for b in r if (a in b)==False]
    return r
permutations([1,2,3],3)
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

给人印象的表现：

%timeit permutations(np.arange(5),5)

243 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
time: 406 ms

%timeit list(itertools.permutations(np.arange(5),5))
15.9 µs ± 8.61 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
time: 12.9 s

因此，仅当必须从njitted函数调用它时才使用此版本，否则，请选择itertools实现。

点击查看英文原文

To save you folks possible hours of searching and experimenting, here’s the non-recursive permutaions solution in Python which also works with Numba (as of v. 0.41):

@numba.njit()
def permutations(A, k):
    r = [[i for i in range(0)]]
    for i in range(k):
        r = [[a] + b for a in A for b in r if (a in b)==False]
    return r
permutations([1,2,3],3)
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

To give an impression about performance:

%timeit permutations(np.arange(5),5)

243 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
time: 406 ms

%timeit list(itertools.permutations(np.arange(5),5))
15.9 µs ± 8.61 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
time: 12.9 s

So use this version only if you have to call it from njitted function, otherwise prefer itertools implementation.

回答 25

我看到这些递归函数内部进行了很多迭代，而并非完全是纯函数递归…

因此对于那些甚至无法遵守一个循环的人来说，这是一个总的，完全不必要的完全递归解决方案

def all_insert(x, e, i=0):
    return [x[0:i]+[e]+x[i:]] + all_insert(x,e,i+1) if i<len(x)+1 else []

def for_each(X, e):
    return all_insert(X[0], e) + for_each(X[1:],e) if X else []

def permute(x):
    return [x] if len(x) < 2 else for_each( permute(x[1:]) , x[0])


perms = permute([1,2,3])
点击查看英文原文

I see a lot of iteration going on inside these recursive functions, not exactly pure recursion…

so for those of you who cannot abide by even a single loop, here’s a gross, totally unnecessary fully recursive solution

def all_insert(x, e, i=0):
    return [x[0:i]+[e]+x[i:]] + all_insert(x,e,i+1) if i<len(x)+1 else []

def for_each(X, e):
    return all_insert(X[0], e) + for_each(X[1:],e) if X else []

def permute(x):
    return [x] if len(x) < 2 else for_each( permute(x[1:]) , x[0])


perms = permute([1,2,3])

回答 26

另一个解决方案：

def permutation(flag, k =1 ):
    N = len(flag)
    for i in xrange(0, N):
        if flag[i] != 0:
            continue
        flag[i] = k 
        if k == N:
            print flag
        permutation(flag, k+1)
        flag[i] = 0

permutation([0, 0, 0])
点击查看英文原文

Another solution:

def permutation(flag, k =1 ):
    N = len(flag)
    for i in xrange(0, N):
        if flag[i] != 0:
            continue
        flag[i] = k 
        if k == N:
            print flag
        permutation(flag, k+1)
        flag[i] = 0

permutation([0, 0, 0])

回答 27

我的Python解决方案：

def permutes(input,offset):
    if( len(input) == offset ):
        return [''.join(input)]

    result=[]        
    for i in range( offset, len(input) ):
         input[offset], input[i] = input[i], input[offset]
         result = result + permutes(input,offset+1)
         input[offset], input[i] = input[i], input[offset]
    return result

# input is a "string"
# return value is a list of strings
def permutations(input):
    return permutes( list(input), 0 )

# Main Program
print( permutations("wxyz") )
点击查看英文原文

My Python Solution:

def permutes(input,offset):
    if( len(input) == offset ):
        return [''.join(input)]

    result=[]        
    for i in range( offset, len(input) ):
         input[offset], input[i] = input[i], input[offset]
         result = result + permutes(input,offset+1)
         input[offset], input[i] = input[i], input[offset]
    return result

# input is a "string"
# return value is a list of strings
def permutations(input):
    return permutes( list(input), 0 )

# Main Program
print( permutations("wxyz") )

回答 28

def permutation(word, first_char=None):
    if word == None or len(word) == 0: return []
    if len(word) == 1: return [word]

    result = []
    first_char = word[0]
    for sub_word in permutation(word[1:], first_char):
        result += insert(first_char, sub_word)
    return sorted(result)

def insert(ch, sub_word):
    arr = [ch + sub_word]
    for i in range(len(sub_word)):
        arr.append(sub_word[i:] + ch + sub_word[:i])
    return arr


assert permutation(None) == []
assert permutation('') == []
assert permutation('1')  == ['1']
assert permutation('12') == ['12', '21']

print permutation('abc')

输出：[‘abc’，’acb’，’bac’，’bca’，’cab’，’cba’]

点击查看英文原文 
def permutation(word, first_char=None):
    if word == None or len(word) == 0: return []
    if len(word) == 1: return [word]

    result = []
    first_char = word[0]
    for sub_word in permutation(word[1:], first_char):
        result += insert(first_char, sub_word)
    return sorted(result)

def insert(ch, sub_word):
    arr = [ch + sub_word]
    for i in range(len(sub_word)):
        arr.append(sub_word[i:] + ch + sub_word[:i])
    return arr


assert permutation(None) == []
assert permutation('') == []
assert permutation('1')  == ['1']
assert permutation('12') == ['12', '21']

print permutation('abc')

Output: [‘abc’, ‘acb’, ‘bac’, ‘bca’, ‘cab’, ‘cba’]

回答 29

使用 Counter

from collections import Counter

def permutations(nums):
    ans = [[]]
    cache = Counter(nums)

    for idx, x in enumerate(nums):
        result = []
        for items in ans:
            cache1 = Counter(items)
            for id, n in enumerate(nums):
                if cache[n] != cache1[n] and items + [n] not in result:
                    result.append(items + [n])

        ans = result
    return ans
permutations([1, 2, 2])
> [[1, 2, 2], [2, 1, 2], [2, 2, 1]]
点击查看英文原文

Using Counter

from collections import Counter

def permutations(nums):
    ans = [[]]
    cache = Counter(nums)

    for idx, x in enumerate(nums):
        result = []
        for items in ans:
            cache1 = Counter(items)
            for id, n in enumerate(nums):
                if cache[n] != cache1[n] and items + [n] not in result:
                    result.append(items + [n])

        ans = result
    return ans
permutations([1, 2, 2])
> [[1, 2, 2], [2, 1, 2], [2, 2, 1]]