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什么时候del在python中有用?

问题:什么时候del在python中有用?

我真的想不出python为什么需要del关键字的任何原因(大多数语言似乎没有类似的关键字)。例如,可以删除变量而不是删除变量None。从字典中删除时,del可以添加一个方法。

是否有任何理由保留del在python中,或者它是Python的垃圾收集日的痕迹?

I can’t really think of any reason why python needs the del keyword (and most languages seem to not have a similar keyword). For instance, rather than deleting a variable, one could just assign None to it. And when deleting from a dictionary, a del method could be added.

Is there any reason to keep del in python, or is it a vestige of Python’s pre-garbage collection days?


回答 0

首先,除了局部变量,您还可以进行其他操作

del list_item[4]
del dictionary["alpha"]

两者都应该明显有用。其次,del在局部变量上使用可使意图更清晰。相比:

del foo

foo = None

我知道在这种情况下del foo,目的是从范围中删除变量。目前尚不清楚这样foo = None做。如果有人刚分配,foo = None我可能认为这是无效代码。但是我立即知道某个编码人员del foo正在尝试做什么。

Firstly, you can del other things besides local variables

del list_item[4]
del dictionary["alpha"]

Both of which should be clearly useful. Secondly, using del on a local variable makes the intent clearer. Compare:

del foo

to

foo = None

I know in the case of del foo that the intent is to remove the variable from scope. It’s not clear that foo = None is doing that. If somebody just assigned foo = None I might think it was dead code. But I instantly know what somebody who codes del foo was trying to do.


回答 1

这是做什么的一部分del(来自Python语言参考):

删除名称会从本地或全局命名空间中删除该名称的绑定

分配None名称不会从命名空间中删除名称的绑定。

(我想可能存在一些关于删除名称绑定是否真正有用的参数,但这是另一个问题。)

There’s this part of what del does (from the Python Language Reference):

Deletion of a name removes the binding of that name from the local or global namespace

Assigning None to a name does not remove the binding of the name from the namespace.

(I suppose there could be some debate about whether removing a name binding is actually useful, but that’s another question.)


回答 2

我发现一个del有用的地方是清理for循环中的无关变量:

for x in some_list:
  do(x)
del x

现在,您可以确定,如果在for循环之外使用x,它将是未定义的。

One place I’ve found del useful is cleaning up extraneous variables in for loops:

for x in some_list:
  do(x)
del x

Now you can be sure that x will be undefined if you use it outside the for loop.


回答 3

删除变量与将其设置为“无”不同

用删除变量名del可能很少使用,但是如果没有关键字就无法轻易实现。如果您可以通过编写来创建变量名a=1,那么理论上可以通过删除a来撤消该操作。

在某些情况下,它可以简化调试过程,因为尝试访问已删除的变量将引发NameError。

您可以删除类实例属性

Python使您可以编写如下内容:

class A(object):
    def set_a(self, a):
        self.a=a
a=A()
a.set_a(3)
if hasattr(a, "a"):
    print("Hallo")

如果您选择向类实例动态添加属性,那么您当然希望能够通过编写来撤消它

del a.a

Deleting a variable is different than setting it to None

Deleting variable names with del is probably something used rarely, but it is something that could not trivially be achieved without a keyword. If you can create a variable name by writing a=1, it is nice that you can theoretically undo this by deleting a.

It can make debugging easier in some cases as trying to access a deleted variable will raise an NameError.

You can delete class instance attributes

Python lets you write something like:

class A(object):
    def set_a(self, a):
        self.a=a
a=A()
a.set_a(3)
if hasattr(a, "a"):
    print("Hallo")

If you choose to dynamically add attributes to a class instance, you certainly want to be able to undo it by writing

del a.a

回答 4

有一个特定的示例说明您在检查异常时应使用的时间del(可能还有其他,但我知道这是sys.exc_info()一时的事)。此函数返回一个元组,引发的异常类型,消息和回溯。

前两个值通常足以诊断错误并采取措施,但第三个值包含引发异常与捕获异常之间的整个调用堆栈。特别是如果您做类似的事情

try:
    do_evil()
except:
    exc_type, exc_value, tb = sys.exc_info()
    if something(exc_value):
        raise

追溯,tb最终出现在调用堆栈的本地中,从而创建了无法进行垃圾回收的循环引用。因此,执行以下操作很重要:

try:
    do_evil()
except:
    exc_type, exc_value, tb = sys.exc_info()
    del tb
    if something(exc_value):
        raise

打破循环参考。在很多情况下,你想打电话sys.exc_info(),像元类魔术,回溯有用的,所以你必须确保你清理之前,你都不可能离开异常处理程序。如果不需要回溯,则应立即将其删除,或者直接执行以下操作:

exc_type, exc_value = sys.exc_info()[:2]

为了避免所有这一切。

There is a specific example of when you should use del (there may be others, but I know about this one off hand) when you are using sys.exc_info() to inspect an exception. This function returns a tuple, the type of exception that was raised, the message, and a traceback.

The first two values are usually sufficient to diagnose an error and act on it, but the third contains the entire call stack between where the exception was raised and where the the exception is caught. In particular, if you do something like

try:
    do_evil()
except:
    exc_type, exc_value, tb = sys.exc_info()
    if something(exc_value):
        raise

the traceback, tb ends up in the locals of the call stack, creating a circular reference that cannot be garbage collected. Thus, it is important to do:

try:
    do_evil()
except:
    exc_type, exc_value, tb = sys.exc_info()
    del tb
    if something(exc_value):
        raise

to break the circular reference. In many cases where you would want to call sys.exc_info(), like with metaclass magic, the traceback is useful, so you have to make sure that you clean it up before you can possibly leave the exception handler. If you don’t need the traceback, you should delete it immediately, or just do:

exc_type, exc_value = sys.exc_info()[:2]

To avoid it all together.


回答 5

只是另一种想法。

在像Django这样的框架中调试http应用程序时,调用堆栈充满了以前使用的无用且混乱的变量,尤其是当列表很长时,对于开发人员而言可能非常痛苦。因此,此时,命名空间控制可能会很有用。

Just another thinking.

When debugging http applications in framework like Django, the call stack full of useless and messed up variables previously used, especially when it’s a very long list, could be very painful for developers. so, at this point, namespace controlling could be useful.


回答 6

与将变量分配给None相比,显式使用“ del”也是一种更好的做法。如果尝试删除不存在的变量,则会遇到运行时错误,但如果尝试将不存在的变量设置为None,Python会静默将新变量设置为None,而将变量保留为想删除它在哪里。因此,del将帮助您尽早发现错误

Using “del” explicitly is also better practice than assigning a variable to None. If you attempt to del a variable that doesn’t exist, you’ll get a runtime error but if you attempt to set a variable that doesn’t exist to None, Python will silently set a new variable to None, leaving the variable you wanted deleted where it was. So del will help you catch your mistakes earlier


回答 7

要为以上答案添加几点: del x

x指示的定义r -> or指向对象的引用o),但del x更改r而不是o。这是对对象(而不是与关联的对象)的引用(指针)的操作x。区分ro是此处的关键。

  • 它将从中删除locals()
  • globals()如果x属于它,将其删除。
  • 将其从堆栈框架中删除(从物理上删除引用,但对象本身位于对象池中,而不位于堆栈框架中)。
  • 从当前作用域中删除它。限制局部变量定义的范围非常有用,否则会导致问题。
  • 它更多地是关于名称的声明而不是内容的定义。
  • 它影响x属于的地方,而不影响x指向的地方。内存中唯一的物理更改是这样。例如,如果x在字典或列表中,则将其(作为参考)从那里(不一定从对象池中)删除。在此示例中,它所属的字典是locals()与重叠的堆栈框架()globals()

To add a few points to above answers: del x

Definition of x indicates r -> o (a reference r pointing to an object o) but del x changes r rather than o. It is an operation on the reference (pointer) to object rather than the object associated with x. Distinguishing between r and o is key here.

  • It removes it from locals().
  • Removes it from globals() if x belongs there.
  • Removes it from the stack frame (removes the reference physically from it, but the object itself resides in object pool and not in the stack frame).
  • Removes it from the current scope. It is very useful to limit the span of definition of a local variable, which otherwise can cause problems.
  • It is more about declaration of the name rather than definition of content.
  • It affects where x belongs to, not where x points to. The only physical change in memory is this. For example if x is in a dictionary or list, it (as a reference) is removed from there(and not necessarily from the object pool). In this example, the dictionary it belongs is the stack frame (locals()), which overlaps with globals().

回答 8

使用numpy.load后强制关闭文件:

利基用法也许,但我发现它在numpy.load用于读取文件时很有用。我会不时地更新文件,并且需要将具有相同名称的文件复制到目录中。

我曾经del发布过文件,并允许我复制到新文件中。

请注意,我想避免使用with上下文管理器,因为我在命令行上玩弄图并且不想太多地按下Tab键!

看到这个问题。

Force closing a file after using numpy.load:

A niche usage perhaps but I found it useful when using numpy.load to read a file. Every once in a while I would update the file and need to copy a file with the same name to the directory.

I used del to release the file and allow me to copy in the new file.

Note I want to avoid the with context manager as I was playing around with plots on the command line and didn’t want to be pressing tab a lot!

See this question.


回答 9

del通常在__init__.py文件中看到。__init__.py文件中定义的所有全局变量都将自动“导出”(将包含在中from module import *)。避免这种情况的一种方法是定义__all__,但这会变得混乱,而且并非所有人都使用它。

例如,如果您的代码__init__.py

import sys
if sys.version_info < (3,):
    print("Python 2 not supported")

然后,您的模块将导出sys名称。你应该写

import sys
if sys.version_info < (3,):
    print("Python 2 not supported")

del sys

del is often seen in __init__.py files. Any global variable that is defined in an __init__.py file is automatically “exported” (it will be included in a from module import *). One way to avoid this is to define __all__, but this can get messy and not everyone uses it.

For example, if you had code in __init__.py like

import sys
if sys.version_info < (3,):
    print("Python 2 not supported")

Then your module would export the sys name. You should instead write

import sys
if sys.version_info < (3,):
    print("Python 2 not supported")

del sys

回答 10

作为一个例子什么del可用于,我觉得有用我的情况是这样的:

def f(a, b, c=3):
    return '{} {} {}'.format(a, b, c)

def g(**kwargs):
    if 'c' in kwargs and kwargs['c'] is None:
        del kwargs['c']

    return f(**kwargs)

# g(a=1, b=2, c=None) === '1 2 3'
# g(a=1, b=2) === '1 2 3'
# g(a=1, b=2, c=4) === '1 2 4'

这两个功能都可以在不同的封装/模块和程序员不需要知道默认值参数cf实际拥有。因此,通过将kwargs与del结合使用,可以通过将其设置为None来说“我想要c的默认值”(或者在这种情况下也可以保留它)。

您可以使用类似的方法做同样的事情:

def g(a, b, c=None):
    kwargs = {'a': a,
              'b': b}
    if c is not None:
        kwargs['c'] = c

    return f(**kwargs)

但是,我发现前面的示例更加干燥和优雅。

As an example of what del can be used for, I find it useful i situations like this:

def f(a, b, c=3):
    return '{} {} {}'.format(a, b, c)

def g(**kwargs):
    if 'c' in kwargs and kwargs['c'] is None:
        del kwargs['c']

    return f(**kwargs)

# g(a=1, b=2, c=None) === '1 2 3'
# g(a=1, b=2) === '1 2 3'
# g(a=1, b=2, c=4) === '1 2 4'

These two functions can be in different packages/modules and the programmer doesn’t need to know what default value argument c in f actually have. So by using kwargs in combination with del you can say “I want the default value on c” by setting it to None (or in this case also leave it).

You could do the same thing with something like:

def g(a, b, c=None):
    kwargs = {'a': a,
              'b': b}
    if c is not None:
        kwargs['c'] = c

    return f(**kwargs)

However I find the previous example more DRY and elegant.


回答 11

什么时候del在python中有用?

您可以使用它删除数组的单个元素,而不是切片语法x[i:i+1]=[]。例如,如果您在其中os.walk并希望删除目录中的元素,这可能会很有用。不过,我认为关键字对此无济于事,因为一个人只能制作一个[].remove(index)方法(该.remove方法实际上是搜索并删除第一个实例值)。

When is del useful in python?

You can use it to remove a single element of an array instead of the slice syntax x[i:i+1]=[]. This may be useful if for example you are in os.walk and wish to delete an element in the directory. I would not consider a keyword useful for this though, since one could just make a [].remove(index) method (the .remove method is actually search-and-remove-first-instance-of-value).


回答 12

del当使用Numpy处理大数据时,我发现对于伪手动内存管理很有用。例如:

for image_name in large_image_set:
    large_image = io.imread(image_name)
    height, width, depth = large_image.shape
    large_mask = np.all(large_image == <some_condition>)
    # Clear memory, make space
    del large_image; gc.collect()

    large_processed_image = np.zeros((height, width, depth))
    large_processed_image[large_mask] = (new_value)
    io.imsave("processed_image.png", large_processed_image)

    # Clear memory, make space
    del large_mask, large_processed_image; gc.collect()

这可能是由于Python GC无法跟上系统疯狂地交换脚本而使脚本陷入停顿状态,并且在松散的内存阈值下它可以完美平滑地运行,从而留出了很大的可用空间来浏览机器和代码,而它的工作。

I’ve found del to be useful for pseudo-manual memory management when handling large data with Numpy. For example:

for image_name in large_image_set:
    large_image = io.imread(image_name)
    height, width, depth = large_image.shape
    large_mask = np.all(large_image == <some_condition>)
    # Clear memory, make space
    del large_image; gc.collect()

    large_processed_image = np.zeros((height, width, depth))
    large_processed_image[large_mask] = (new_value)
    io.imsave("processed_image.png", large_processed_image)

    # Clear memory, make space
    del large_mask, large_processed_image; gc.collect()

This can be the difference between bringing a script to a grinding halt as the system swaps like mad when the Python GC can’t keep up, and it running perfectly smooth below a loose memory threshold that leaves plenty of headroom to use the machine to browse and code while it’s working.


回答 13

我认为del具有自己的语法的原因之一是,在某些情况下,鉴于del操作绑定或变量而不是它引用的值,用函数替换它可能很难。因此,如果要创建del的函数版本,则需要传入上下文。del foo将需要变为globals()。remove(’foo’)或locals()。remove(’foo’),从而变得混乱而且可读性较差。我仍然说,鉴于del似乎很少使用,摆脱它会很好。但是删除语言功能/缺陷可能很痛苦。也许python 4会删除它:)

I think one of the reasons that del has its own syntax is that replacing it with a function might be hard in certain cases given it operates on the binding or variable and not the value it references. Thus if a function version of del were to be created a context would need to be passed in. del foo would need to become globals().remove(‘foo’) or locals().remove(‘foo’) which gets messy and less readable. Still I say getting rid of del would be good given its seemingly rare use. But removing language features/flaws can be painful. Maybe python 4 will remove it :)


回答 14

我想详细说明可接受的答案,以突出显示将变量设置为None与使用删除变量之间的细微差别del

给定变量foo = 'bar',并提供以下函数定义:

def test_var(var):
    if var:
        print('variable tested true')
    else:
        print('variable tested false')

最初声明后,test_var(foo)Yield将variable tested true达到预期。

现在尝试:

foo = None
test_var(foo)

产生variable tested false

将此行为与:

del foo
test_var(foo)

现在提高了NameError: name 'foo' is not defined

I would like to elaborate on the accepted answer to highlight the nuance between setting a variable to None versus removing it with del:

Given the variable foo = 'bar', and the following function definition:

def test_var(var):
    if var:
        print('variable tested true')
    else:
        print('variable tested false')

Once initially declared, test_var(foo) yields variable tested true as expected.

Now try:

foo = None
test_var(foo)

which yields variable tested false.

Contrast this behavior with:

del foo
test_var(foo)

which now raises NameError: name 'foo' is not defined.


回答 15

又一小生用法:在pyroot与ROOT5或ROOT6,“删除”可以是,以去除被称为无再现有C ++对象Python对象是有用的。这允许pyroot的动态查找来找到一个同名的C ++对象,并将其绑定到python名称。因此,您可能会遇到以下情况:

import ROOT as R
input_file = R.TFile('inputs/___my_file_name___.root')
tree = input_file.Get('r')
tree.Draw('hy>>hh(10,0,5)')
R.gPad.Close()
R.hy # shows that hy is still available. It can even be redrawn at this stage.
tree.Draw('hy>>hh(3,0,3)') # overwrites the C++ object in ROOT's namespace
R.hy # shows that R.hy is None, since the C++ object it pointed to is gone
del R.hy
R.hy # now finds the new C++ object

希望此利基市场将通过ROOT7的精明对象管理而关闭。

Yet another niche usage: In pyroot with ROOT5 or ROOT6, “del” may be useful to remove a python object that referred to a no-longer existing C++ object. This allows the dynamic lookup of pyroot to find an identically-named C++ object and bind it to the python name. So you can have a scenario such as:

import ROOT as R
input_file = R.TFile('inputs/___my_file_name___.root')
tree = input_file.Get('r')
tree.Draw('hy>>hh(10,0,5)')
R.gPad.Close()
R.hy # shows that hy is still available. It can even be redrawn at this stage.
tree.Draw('hy>>hh(3,0,3)') # overwrites the C++ object in ROOT's namespace
R.hy # shows that R.hy is None, since the C++ object it pointed to is gone
del R.hy
R.hy # now finds the new C++ object

Hopefully, this niche will be closed with ROOT7’s saner object management.


回答 16

“ del”命令对于控制数组中的数据非常有用,例如:

elements = ["A", "B", "C", "D"]
# Remove first element.
del elements[:1]
print(elements)

输出:

[‘B’,’C’,’D’]

The “del” command is very useful for controlling data in an array, for example:

elements = ["A", "B", "C", "D"]
# Remove first element.
del elements[:1]
print(elements)

Output:

[‘B’, ‘C’, ‘D’]


回答 17

一旦我不得不使用:

del serial
serial = None

因为仅使用:

serial = None

释放串口的速度不够快,无法立即再次打开它。从那堂课中,我学到了del真正的意思:“现在就GC!等到完成为止”,这在很多情况下非常有用。当然,您可能会有一个system.gc.del_this_and_wait_balbalbalba(obj)

Once I had to use:

del serial
serial = None

because using only:

serial = None

didn’t release the serial port fast enough to immediately open it again. From that lesson I learned that del really meant: “GC this NOW! and wait until it’s done” and that is really useful in a lot of situations. Of course, you may have a system.gc.del_this_and_wait_balbalbalba(obj).


回答 18

del在许多语言中都等同于“未设置”,并且作为从另一种语言到python的交叉参考点。人们倾向于寻找命令,它们执行的操作与以前使用第一语言时所做的相同……将var更改为“”或没有任何操作并不会真正从范围中删除该var。它只是清空其值,该var本身的名称仍将存储在内存中…为什么?!在占用大量内存的脚本中..保持垃圾桶的“否”和“反之……”每种语言都具有某种形式的“未设置/删除” var函数..为什么不使用python?

del is the equivalent of “unset” in many languages and as a cross reference point moving from another language to python.. people tend to look for commands that do the same thing that they used to do in their first language… also setting a var to “” or none doesn’t really remove the var from scope..it just empties its value the name of the var itself would still be stored in memory…why?!? in a memory intensive script..keeping trash behind its just a no no and anyways…every language out there has some form of an “unset/delete” var function..why not python?


回答 19

python中的每个对象都有一个标识符,类型,与之关联的引用计数,当我们使用del时,引用计数会减少,当引用计数变为零时,它很可能成为垃圾回收的候选对象。与将标识符设置为“无”相比,这可以区分del。在后面的情况下,它只是意味着该对象被遗忘了(直到我们超出范围,在这种情况下,计数减少了),现在标识符仅指向其他某个对象(内存位置)。

Every object in python has an identifier, Type, reference count associated with it, when we use del the reference count is reduced, when the reference count becomes zero it is a potential candidate for getting garbage collected. This differentiates the del when compared to setting an identifier to None. In later case it simply means the object is just left out wild( until we are out of scope in which case the count is reduced) and simply now the identifier point to some other object(memory location).


如何使用泡菜保存字典?

问题:如何使用泡菜保存字典?

我已经仔细阅读了Python文档提供的信息,但仍然有些困惑。有人可以张贴示例代码来编写新文件,然后使用pickle将字典转储到其中吗?

I have looked through the information that the Python docs give, but I’m still a little confused. Could somebody post sample code that would write a new file then use pickle to dump a dictionary into it?


回答 0

尝试这个:

import pickle

a = {'hello': 'world'}

with open('filename.pickle', 'wb') as handle:
    pickle.dump(a, handle, protocol=pickle.HIGHEST_PROTOCOL)

with open('filename.pickle', 'rb') as handle:
    b = pickle.load(handle)

print a == b

Try this:

import pickle

a = {'hello': 'world'}

with open('filename.pickle', 'wb') as handle:
    pickle.dump(a, handle, protocol=pickle.HIGHEST_PROTOCOL)

with open('filename.pickle', 'rb') as handle:
    b = pickle.load(handle)

print a == b

回答 1

import pickle

your_data = {'foo': 'bar'}

# Store data (serialize)
with open('filename.pickle', 'wb') as handle:
    pickle.dump(your_data, handle, protocol=pickle.HIGHEST_PROTOCOL)

# Load data (deserialize)
with open('filename.pickle', 'rb') as handle:
    unserialized_data = pickle.load(handle)

print(your_data == unserialized_data)

的优点HIGHEST_PROTOCOL是文件变小。这使得脱皮有时更快。

重要提示:泡菜的最大文件大小约为2GB。

替代方式

import mpu
your_data = {'foo': 'bar'}
mpu.io.write('filename.pickle', data)
unserialized_data = mpu.io.read('filename.pickle')

替代格式

对于您的应用程序,以下内容可能很重要:

  • 其他编程语言的支持
  • 阅读/写作表现
  • 紧凑度(文件大小)

另请参阅:数据序列化格式的比较

如果您想寻找一种制作配置文件的方法,则可能需要阅读我的短文《Python中的配置文件》。

import pickle

your_data = {'foo': 'bar'}

# Store data (serialize)
with open('filename.pickle', 'wb') as handle:
    pickle.dump(your_data, handle, protocol=pickle.HIGHEST_PROTOCOL)

# Load data (deserialize)
with open('filename.pickle', 'rb') as handle:
    unserialized_data = pickle.load(handle)

print(your_data == unserialized_data)

The advantage of HIGHEST_PROTOCOL is that files get smaller. This makes unpickling sometimes much faster.

Important notice: The maximum file size of pickle is about 2GB.

Alternative way

import mpu
your_data = {'foo': 'bar'}
mpu.io.write('filename.pickle', data)
unserialized_data = mpu.io.read('filename.pickle')

Alternative Formats

For your application, the following might be important:

  • Support by other programming languages
  • Reading / writing performance
  • Compactness (file size)

See also: Comparison of data serialization formats

In case you are rather looking for a way to make configuration files, you might want to read my short article Configuration files in Python


回答 2

# Save a dictionary into a pickle file.
import pickle

favorite_color = {"lion": "yellow", "kitty": "red"}  # create a dictionary
pickle.dump(favorite_color, open("save.p", "wb"))  # save it into a file named save.p

# -------------------------------------------------------------
# Load the dictionary back from the pickle file.
import pickle

favorite_color = pickle.load(open("save.p", "rb"))
# favorite_color is now {"lion": "yellow", "kitty": "red"}
# Save a dictionary into a pickle file.
import pickle

favorite_color = {"lion": "yellow", "kitty": "red"}  # create a dictionary
pickle.dump(favorite_color, open("save.p", "wb"))  # save it into a file named save.p

# -------------------------------------------------------------
# Load the dictionary back from the pickle file.
import pickle

favorite_color = pickle.load(open("save.p", "rb"))
# favorite_color is now {"lion": "yellow", "kitty": "red"}

回答 3

通常,dict除非仅包含简单的对象(例如字符串和整数),否则酸洗a 将失败。

Python 2.7.9 (default, Dec 11 2014, 01:21:43) 
[GCC 4.2.1 Compatible Apple Clang 4.1 ((tags/Apple/clang-421.11.66))] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from numpy import *
>>> type(globals())     
<type 'dict'>
>>> import pickle
>>> pik = pickle.dumps(globals())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 649, in save_dict
    self._batch_setitems(obj.iteritems())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 663, in _batch_setitems
    save(v)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 306, in save
    rv = reduce(self.proto)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/copy_reg.py", line 70, in _reduce_ex
    raise TypeError, "can't pickle %s objects" % base.__name__
TypeError: can't pickle module objects
>>> 

即使是非常简单的方法dict也会经常失败。它仅取决于内容。

>>> d = {'x': lambda x:x}
>>> pik = pickle.dumps(d)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 649, in save_dict
    self._batch_setitems(obj.iteritems())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 663, in _batch_setitems
    save(v)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 748, in save_global
    (obj, module, name))
pickle.PicklingError: Can't pickle <function <lambda> at 0x102178668>: it's not found as __main__.<lambda>

但是,如果使用更好的序列化器(例如dill或)cloudpickle,则可以对大多数词典进行腌制:

>>> import dill
>>> pik = dill.dumps(d)

或者,如果您想将dict文件保存到文件中…

>>> with open('save.pik', 'w') as f:
...   dill.dump(globals(), f)
... 

后一个示例与此处发布的任何其他好的答案相同(除了忽略商品内容的可腌性之外dict)。

In general, pickling a dict will fail unless you have only simple objects in it, like strings and integers.

Python 2.7.9 (default, Dec 11 2014, 01:21:43) 
[GCC 4.2.1 Compatible Apple Clang 4.1 ((tags/Apple/clang-421.11.66))] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from numpy import *
>>> type(globals())     
<type 'dict'>
>>> import pickle
>>> pik = pickle.dumps(globals())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 649, in save_dict
    self._batch_setitems(obj.iteritems())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 663, in _batch_setitems
    save(v)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 306, in save
    rv = reduce(self.proto)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/copy_reg.py", line 70, in _reduce_ex
    raise TypeError, "can't pickle %s objects" % base.__name__
TypeError: can't pickle module objects
>>> 

Even a really simple dict will often fail. It just depends on the contents.

>>> d = {'x': lambda x:x}
>>> pik = pickle.dumps(d)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 649, in save_dict
    self._batch_setitems(obj.iteritems())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 663, in _batch_setitems
    save(v)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 748, in save_global
    (obj, module, name))
pickle.PicklingError: Can't pickle <function <lambda> at 0x102178668>: it's not found as __main__.<lambda>

However, if you use a better serializer like dill or cloudpickle, then most dictionaries can be pickled:

>>> import dill
>>> pik = dill.dumps(d)

Or if you want to save your dict to a file…

>>> with open('save.pik', 'w') as f:
...   dill.dump(globals(), f)
... 

The latter example is identical to any of the other good answers posted here (which aside from neglecting the picklability of the contents of the dict are good).


回答 4

>>> import pickle
>>> with open("/tmp/picklefile", "wb") as f:
...     pickle.dump({}, f)
... 

通常,最好使用cPickle实现

>>> import cPickle as pickle
>>> help(pickle.dump)
Help on built-in function dump in module cPickle:

dump(...)
    dump(obj, file, protocol=0) -- Write an object in pickle format to the given file.

    See the Pickler docstring for the meaning of optional argument proto.
>>> import pickle
>>> with open("/tmp/picklefile", "wb") as f:
...     pickle.dump({}, f)
... 

normally it’s preferable to use the cPickle implementation

>>> import cPickle as pickle
>>> help(pickle.dump)
Help on built-in function dump in module cPickle:

dump(...)
    dump(obj, file, protocol=0) -- Write an object in pickle format to the given file.

    See the Pickler docstring for the meaning of optional argument proto.

回答 5

如果您只想将字典存储在单个文件中,请pickle像这样使用

import pickle

a = {'hello': 'world'}

with open('filename.pickle', 'wb') as handle:
    pickle.dump(a, handle)

with open('filename.pickle', 'rb') as handle:
    b = pickle.load(handle)

如果要在多个文件中保存和还原多个词典以进行缓存和存储更复杂的数据,请使用anycache。它可以完成您需要的所有其他工作pickle

from anycache import anycache

@anycache(cachedir='path/to/files')
def myfunc(hello):
    return {'hello', hello}

Anycache myfunc根据不同文件的参数存储不同的结果,cachedir然后重新加载它们。

有关更多详细信息,请参见文档

If you just want to store the dict in a single file, use pickle like that

import pickle

a = {'hello': 'world'}

with open('filename.pickle', 'wb') as handle:
    pickle.dump(a, handle)

with open('filename.pickle', 'rb') as handle:
    b = pickle.load(handle)

If you want to save and restore multiple dictionaries in multiple files for caching and store more complex data, use anycache. It does all the other stuff you need around pickle

from anycache import anycache

@anycache(cachedir='path/to/files')
def myfunc(hello):
    return {'hello', hello}

Anycache stores the different myfunc results depending on the arguments to different files in cachedir and reloads them.

See the documentation for any further details.


回答 6

将Python数据(例如字典)转储到pickle文件的简单方法。

import pickle

your_dictionary = {}

pickle.dump(your_dictionary, open('pickle_file_name.p', 'wb'))

Simple way to dump a Python data (e.g. dictionary) to a pickle file.

import pickle

your_dictionary = {}

pickle.dump(your_dictionary, open('pickle_file_name.p', 'wb'))

回答 7

import pickle

dictobj = {'Jack' : 123, 'John' : 456}

filename = "/foldername/filestore"

fileobj = open(filename, 'wb')

pickle.dump(dictobj, fileobj)

fileobj.close()
import pickle

dictobj = {'Jack' : 123, 'John' : 456}

filename = "/foldername/filestore"

fileobj = open(filename, 'wb')

pickle.dump(dictobj, fileobj)

fileobj.close()

回答 8

我发现酸洗令人困惑(可能是因为我很胖)。我发现这可行,但是:

myDictionaryString=str(myDictionary)

然后可以将其写入文本文件。我遇到错误并告诉我将整数写入.dat文件时,我放弃尝试使用pickle。很抱歉没有使用泡菜。

I’ve found pickling confusing (possibly because I’m thick). I found that this works, though:

myDictionaryString=str(myDictionary)

Which you can then write to a text file. I gave up trying to use pickle as I was getting errors telling me to write integers to a .dat file. I apologise for not using pickle.


更改字典中键的名称

问题:更改字典中键的名称

我想更改Python字典中条目的键。

有没有简单的方法可以做到这一点?

I want to change the key of an entry in a Python dictionary.

Is there a straightforward way to do this?


回答 0

只需2个步骤即可轻松完成:

dictionary[new_key] = dictionary[old_key]
del dictionary[old_key]

或第一步

dictionary[new_key] = dictionary.pop(old_key)

KeyError如果dictionary[old_key]未定义,它将引发。请注意,这删除dictionary[old_key]

>>> dictionary = { 1: 'one', 2:'two', 3:'three' }
>>> dictionary['ONE'] = dictionary.pop(1)
>>> dictionary
{2: 'two', 3: 'three', 'ONE': 'one'}
>>> dictionary['ONE'] = dictionary.pop(1)
Traceback (most recent call last):
  File "<input>", line 1, in <module>
KeyError: 1

Easily done in 2 steps:

dictionary[new_key] = dictionary[old_key]
del dictionary[old_key]

Or in 1 step:

dictionary[new_key] = dictionary.pop(old_key)

which will raise KeyError if dictionary[old_key] is undefined. Note that this will delete dictionary[old_key].

>>> dictionary = { 1: 'one', 2:'two', 3:'three' }
>>> dictionary['ONE'] = dictionary.pop(1)
>>> dictionary
{2: 'two', 3: 'three', 'ONE': 'one'}
>>> dictionary['ONE'] = dictionary.pop(1)
Traceback (most recent call last):
  File "<input>", line 1, in <module>
KeyError: 1

回答 1

如果要更改所有键:

d = {'x':1, 'y':2, 'z':3}
d1 = {'x':'a', 'y':'b', 'z':'c'}

In [10]: dict((d1[key], value) for (key, value) in d.items())
Out[10]: {'a': 1, 'b': 2, 'c': 3}

如果要更改单个键:可以采用上述任何建议。

if you want to change all the keys:

d = {'x':1, 'y':2, 'z':3}
d1 = {'x':'a', 'y':'b', 'z':'c'}

In [10]: dict((d1[key], value) for (key, value) in d.items())
Out[10]: {'a': 1, 'b': 2, 'c': 3}

if you want to change single key: You can go with any of the above suggestion.


回答 2

流行的

>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>> 

pop’n’fresh

>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>> 

回答 3

在python 2.7及更高版本中,您可以使用字典理解:这是我在使用DictReader读取CSV时遇到的示例。用户已在所有列名后添加“:”

ori_dict = {'key1:' : 1, 'key2:' : 2, 'key3:' : 3}

摆脱键后面的“:”:

corrected_dict = { k.replace(':', ''): v for k, v in ori_dict.items() }

In python 2.7 and higher, you can use dictionary comprehension: This is an example I encountered while reading a CSV using a DictReader. The user had suffixed all the column names with ‘:’

ori_dict = {'key1:' : 1, 'key2:' : 2, 'key3:' : 3}

to get rid of the trailing ‘:’ in the keys:

corrected_dict = { k.replace(':', ''): v for k, v in ori_dict.items() }


回答 4

由于键是字典用于查找值的对象,因此您无法真正更改它们。您可以做的最接近的操作是保存与旧密钥关联的值,将其删除,然后添加带有替换密钥和已保存值的新条目。其他几个答案说明了可以完成此操作的不同方式。

Since keys are what dictionaries use to lookup values, you can’t really change them. The closest thing you can do is to save the value associated with the old key, delete it, then add a new entry with the replacement key and the saved value. Several of the other answers illustrate different ways this can be accomplished.


回答 5

如果您有复杂的字典,则表示该字典中有一个字典或列表:

myDict = {1:"one",2:{3:"three",4:"four"}}
myDict[2][5] = myDict[2].pop(4)
print myDict

Output
{1: 'one', 2: {3: 'three', 5: 'four'}}

If you have a complex dict, it means there is a dict or list within the dict:

myDict = {1:"one",2:{3:"three",4:"four"}}
myDict[2][5] = myDict[2].pop(4)
print myDict

Output
{1: 'one', 2: {3: 'three', 5: 'four'}}

回答 6

没有直接的方法可以执行此操作,但是您可以删除然后分配

d = {1:2,3:4}

d[newKey] = d[1]
del d[1]

或进行批量密钥更改:

d = dict((changeKey(k), v) for k, v in d.items())

No direct way to do this, but you can delete-then-assign

d = {1:2,3:4}

d[newKey] = d[1]
del d[1]

or do mass key changes:

d = dict((changeKey(k), v) for k, v in d.items())

回答 7

转换字典中的所有键

假设这是您的字典:

>>> sample = {'person-id': '3', 'person-name': 'Bob'}

要将所有破折号转换为示例字典键中的下划线:

>>> sample = {key.replace('-', '_'): sample.pop(key) for key in sample.keys()}
>>> sample
>>> {'person_id': '3', 'person_name': 'Bob'}

To convert all the keys in the dictionary

Suppose this is your dictionary:

>>> sample = {'person-id': '3', 'person-name': 'Bob'}

To convert all the dashes to underscores in the sample dictionary key:

>>> sample = {key.replace('-', '_'): sample.pop(key) for key in sample.keys()}
>>> sample
>>> {'person_id': '3', 'person_name': 'Bob'}

回答 8

d = {1:2,3:4}

假设我们想将键更改为列表元素p = [‘a’,’b’]。以下代码将执行以下操作:

d=dict(zip(p,list(d.values()))) 

我们得到

{'a': 2, 'b': 4}
d = {1:2,3:4}

suppose that we want to change the keys to the list elements p=[‘a’ , ‘b’]. the following code will do:

d=dict(zip(p,list(d.values()))) 

and we get

{'a': 2, 'b': 4}

回答 9

您可以将相同的值与许多键相关联,或者只删除一个键并重新添加具有相同值的新键。

例如,如果您有键->值:

red->1
blue->2
green->4

没有理由您无法添加purple->2或删除red->1并添加orange->1

You can associate the same value with many keys, or just remove a key and re-add a new key with the same value.

For example, if you have keys->values:

red->1
blue->2
green->4

there’s no reason you can’t add purple->2 or remove red->1 and add orange->1


回答 10

如果一次更改所有按键。在这里,我将阻止所有键。

a = {'making' : 1, 'jumping' : 2, 'climbing' : 1, 'running' : 2}
b = {ps.stem(w) : a[w] for w in a.keys()}
print(b)
>>> {'climb': 1, 'jump': 2, 'make': 1, 'run': 2} #output

In case of changing all the keys at once. Here I am stemming the keys.

a = {'making' : 1, 'jumping' : 2, 'climbing' : 1, 'running' : 2}
b = {ps.stem(w) : a[w] for w in a.keys()}
print(b)
>>> {'climb': 1, 'jump': 2, 'make': 1, 'run': 2} #output

回答 11

我还没有看到这个确切的答案:

dict['key'] = value

您甚至可以对对象属性执行此操作。通过执行以下操作使它们成为字典:

dict = vars(obj)

然后,您可以像字典一样操作对象属性:

dict['attribute'] = value

I haven’t seen this exact answer:

dict['key'] = value

You can even do this to object attributes. Make them into a dictionary by doing this:

dict = vars(obj)

Then you can manipulate the object attributes like you would a dictionary:

dict['attribute'] = value

将字典转换为JSON

问题:将字典转换为JSON

r = {'is_claimed': 'True', 'rating': 3.5}
r = json.dumps(r)
file.write(str(r['rating']))

我无法访问JSON中的数据。我究竟做错了什么?

TypeError: string indices must be integers, not str
r = {'is_claimed': 'True', 'rating': 3.5}
r = json.dumps(r)
file.write(str(r['rating']))

I am not able to access my data in the JSON. What am I doing wrong?

TypeError: string indices must be integers, not str

回答 0

json.dumps()将字典转换为str对象,而不是json(dict)对象!因此,您必须使用方法将其加载strdictjson.loads()

请参阅json.dumps()作为保存方法和json.loads()检索方法。

这是代码示例,可以帮助您进一步了解它:

import json

r = {'is_claimed': 'True', 'rating': 3.5}
r = json.dumps(r)
loaded_r = json.loads(r)
loaded_r['rating'] #Output 3.5
type(r) #Output str
type(loaded_r) #Output dict

json.dumps() converts a dictionary to str object, not a json(dict) object! So you have to load your str into a dict to use it by using json.loads() method

See json.dumps() as a save method and json.loads() as a retrieve method.

This is the code sample which might help you understand it more:

import json

r = {'is_claimed': 'True', 'rating': 3.5}
r = json.dumps(r)
loaded_r = json.loads(r)
loaded_r['rating'] #Output 3.5
type(r) #Output str
type(loaded_r) #Output dict

回答 1

json.dumps()返回python字典的JSON字符串表示形式。查看文件

您不能这样做,r['rating']因为r是字符串,不再是dict

也许你的意思是

r = {'is_claimed': 'True', 'rating': 3.5}
json = json.dumps(r) # note i gave it a different name
file.write(str(r['rating']))

json.dumps() returns the JSON string representation of the python dict. See the docs

You can’t do r['rating'] because r is a string, not a dict anymore

Perhaps you meant something like

r = {'is_claimed': 'True', 'rating': 3.5}
json = json.dumps(r) # note i gave it a different name
file.write(str(r['rating']))

回答 2

无需通过使用将其转换为字符串 json.dumps()

r = {'is_claimed': 'True', 'rating': 3.5}
file.write(r['is_claimed'])
file.write(str(r['rating']))

您可以直接从dict对象获取值。

No need to convert it in a string by using json.dumps()

r = {'is_claimed': 'True', 'rating': 3.5}
file.write(r['is_claimed'])
file.write(str(r['rating']))

You can get the values directly from the dict object.


回答 3

将r定义为字典应该可以解决这个问题:

>>> r: dict = {'is_claimed': 'True', 'rating': 3.5}
>>> print(r['rating'])
3.5
>>> type(r)
<class 'dict'>

Defining r as a dictionary should do the trick:

>>> r: dict = {'is_claimed': 'True', 'rating': 3.5}
>>> print(r['rating'])
3.5
>>> type(r)
<class 'dict'>

将字典作为关键字参数传递给函数

问题:将字典作为关键字参数传递给函数

我想使用字典在python中调用一个函数。

这是一些代码:

d = dict(param='test')

def f(param):
    print(param)

f(d)

这可以打印,{'param': 'test'}但我希望只打印test

我希望它可以类似地工作以获取更多参数:

d = dict(p1=1, p2=2)
def f2(p1, p2):
    print(p1, p2)
f2(d)

这可能吗?

I’d like to call a function in python using a dictionary.

Here is some code:

d = dict(param='test')

def f(param):
    print(param)

f(d)

This prints {'param': 'test'} but I’d like it to just print test.

I’d like it to work similarly for more parameters:

d = dict(p1=1, p2=2)
def f2(p1, p2):
    print(p1, p2)
f2(d)

Is this possible?


回答 0

最后自己解决了。很简单,我只是缺少**运算符来解开字典

因此,我的示例变为:

d = dict(p1=1, p2=2)
def f2(p1,p2):
    print p1, p2
f2(**d)

Figured it out for myself in the end. It is simple, I was just missing the ** operator to unpack the dictionary

So my example becomes:

d = dict(p1=1, p2=2)
def f2(p1,p2):
    print p1, p2
f2(**d)

回答 1

In[1]: def myfunc(a=1, b=2):
In[2]:    print(a, b)

In[3]: mydict = {'a': 100, 'b': 200}

In[4]: myfunc(**mydict)
100 200

一些可能有助于了解的其他详细信息(阅读本文并经过测试后遇到的问题):

  1. 该函数可以具有字典中包含的参数
  2. 不能覆盖字典中已经存在的参数
  3. 字典不能包含函数中没有的参数。

例子:

数字1:该函数可以具有字典中未包含的参数

In[5]: mydict = {'a': 100}
In[6]: myfunc(**mydict)
100 2

数字2:您不能覆盖字典中已经存在的参数

In[7]: mydict = {'a': 100, 'b': 200}
In[8]: myfunc(a=3, **mydict)

TypeError: myfunc() got multiple values for keyword argument 'a'

数字3:字典中不能包含函数中没有的参数。

In[9]:  mydict = {'a': 100, 'b': 200, 'c': 300}
In[10]: myfunc(**mydict)

TypeError: myfunc() got an unexpected keyword argument 'c'

根据注释中的要求,第3号的解决方案是根据函数中可用的关键字参数来过滤字典:

In[11]: import inspect
In[12]: mydict = {'a': 100, 'b': 200, 'c': 300}
In[13]: filtered_mydict = {k: v for k, v in mydict.items() if k in [p.name for p in inspect.signature(myfunc).parameters.values()]}
In[14]: myfunc(**filtered_mydict)
100 200

另一种选择是在函数中接受(并忽略)其他kwarg:

In[15]: def myfunc2(a=None, **kwargs):
In[16]:    print(a)

In[17]: mydict = {'a': 100, 'b': 200, 'c': 300}

In[18]: myfunc2(**mydict)
100

请注意,除了可以有效地使用位置参数和列表或元组之外,还可以使用与kwargs相同的方式,这是一个更高级的示例,其中同时包含位置和关键字args:

In[19]: def myfunc3(a, *posargs, b=2, **kwargs):
In[20]:    print(a, b)
In[21]:    print(posargs)
In[22]:    print(kwargs)

In[23]: mylist = [10, 20, 30]
In[24]: mydict = {'b': 200, 'c': 300}

In[25]: myfunc3(*mylist, **mydict)
10 200
(20, 30)
{'c': 300}
In[1]: def myfunc(a=1, b=2):
In[2]:    print(a, b)

In[3]: mydict = {'a': 100, 'b': 200}

In[4]: myfunc(**mydict)
100 200

A few extra details that might be helpful to know (questions I had after reading this and went and tested):

  1. The function can have parameters that are not included in the dictionary
  2. You can not override a parameter that is already in the dictionary
  3. The dictionary can not have parameters that aren’t in the function.

Examples:

Number 1: The function can have parameters that are not included in the dictionary

In[5]: mydict = {'a': 100}
In[6]: myfunc(**mydict)
100 2

Number 2: You can not override a parameter that is already in the dictionary

In[7]: mydict = {'a': 100, 'b': 200}
In[8]: myfunc(a=3, **mydict)

TypeError: myfunc() got multiple values for keyword argument 'a'

Number 3: The dictionary can not have parameters that aren’t in the function.

In[9]:  mydict = {'a': 100, 'b': 200, 'c': 300}
In[10]: myfunc(**mydict)

TypeError: myfunc() got an unexpected keyword argument 'c'

As requested in comments, a solution to Number 3 is to filter the dictionary based on the keyword arguments available in the function:

In[11]: import inspect
In[12]: mydict = {'a': 100, 'b': 200, 'c': 300}
In[13]: filtered_mydict = {k: v for k, v in mydict.items() if k in [p.name for p in inspect.signature(myfunc).parameters.values()]}
In[14]: myfunc(**filtered_mydict)
100 200

Another option is to accept (and ignore) additional kwargs in your function:

In[15]: def myfunc2(a=None, **kwargs):
In[16]:    print(a)

In[17]: mydict = {'a': 100, 'b': 200, 'c': 300}

In[18]: myfunc2(**mydict)
100

Notice further than you can use positional arguments and lists or tuples in effectively the same way as kwargs, here’s a more advanced example incorporating both positional and keyword args:

In[19]: def myfunc3(a, *posargs, b=2, **kwargs):
In[20]:    print(a, b)
In[21]:    print(posargs)
In[22]:    print(kwargs)

In[23]: mylist = [10, 20, 30]
In[24]: mydict = {'b': 200, 'c': 300}

In[25]: myfunc3(*mylist, **mydict)
10 200
(20, 30)
{'c': 300}

回答 2

在python中,这称为“拆包”,您​​可以在本教程中找到有关它的信息。我同意,它的文档很烂,尤其是因为它是如此的有用。

In python, this is called “unpacking”, and you can find a bit about it in the tutorial. The documentation of it sucks, I agree, especially because of how fantasically useful it is.


回答 3

在这里-可以进行其他任何迭代:

d = {'param' : 'test'}

def f(dictionary):
    for key in dictionary:
        print key

f(d)

Here ya go – works just any other iterable:

d = {'param' : 'test'}

def f(dictionary):
    for key in dictionary:
        print key

f(d)

来自对象字段的Python字典

问题:来自对象字段的Python字典

您是否知道是否有内置函数可以从任意对象构建字典?我想做这样的事情:

>>> class Foo:
...     bar = 'hello'
...     baz = 'world'
...
>>> f = Foo()
>>> props(f)
{ 'bar' : 'hello', 'baz' : 'world' }

注意:它不应包含方法。仅字段。

Do you know if there is a built-in function to build a dictionary from an arbitrary object? I’d like to do something like this:

>>> class Foo:
...     bar = 'hello'
...     baz = 'world'
...
>>> f = Foo()
>>> props(f)
{ 'bar' : 'hello', 'baz' : 'world' }

NOTE: It should not include methods. Only fields.


回答 0

请注意,Python 2.7中的最佳实践是使用新型类(Python 3不需要),即

class Foo(object):
   ...

同样,“对象”和“类”之间也存在差异。要从任意对象构建字典,只需使用即可__dict__。通常,您将在类级别声明您的方法,并在实例级别声明您的属性,因此__dict__应该没问题。例如:

>>> class A(object):
...   def __init__(self):
...     self.b = 1
...     self.c = 2
...   def do_nothing(self):
...     pass
...
>>> a = A()
>>> a.__dict__
{'c': 2, 'b': 1}

更好的方法(由robert建议在注释中使用)是内置vars函数:

>>> vars(a)
{'c': 2, 'b': 1}

另外,根据您要执行的操作,最好继承自dict。然后,您的Class已经是字典,并且如果您愿意,可以覆盖getattr和/或setattr调用并设置字典。例如:

class Foo(dict):
    def __init__(self):
        pass
    def __getattr__(self, attr):
        return self[attr]

    # etc...

Note that best practice in Python 2.7 is to use new-style classes (not needed with Python 3), i.e.

class Foo(object):
   ...

Also, there’s a difference between an ‘object’ and a ‘class’. To build a dictionary from an arbitrary object, it’s sufficient to use __dict__. Usually, you’ll declare your methods at class level and your attributes at instance level, so __dict__ should be fine. For example:

>>> class A(object):
...   def __init__(self):
...     self.b = 1
...     self.c = 2
...   def do_nothing(self):
...     pass
...
>>> a = A()
>>> a.__dict__
{'c': 2, 'b': 1}

A better approach (suggested by robert in comments) is the builtin vars function:

>>> vars(a)
{'c': 2, 'b': 1}

Alternatively, depending on what you want to do, it might be nice to inherit from dict. Then your class is already a dictionary, and if you want you can override getattr and/or setattr to call through and set the dict. For example:

class Foo(dict):
    def __init__(self):
        pass
    def __getattr__(self, attr):
        return self[attr]

    # etc...

回答 1

取而代之的是x.__dict__,它实际上更具有Pythonic的用法vars(x)

Instead of x.__dict__, it’s actually more pythonic to use vars(x).


回答 2

dir内置会给你对象的所有属性,包括特殊的方法,如__str____dict__和一大堆人,你可能不希望的。但是您可以执行以下操作:

>>> class Foo(object):
...     bar = 'hello'
...     baz = 'world'
...
>>> f = Foo()
>>> [name for name in dir(f) if not name.startswith('__')]
[ 'bar', 'baz' ]
>>> dict((name, getattr(f, name)) for name in dir(f) if not name.startswith('__')) 
{ 'bar': 'hello', 'baz': 'world' }

因此可以通过定义如下props函数将其扩展为仅返回数据属性而不是方法:

import inspect

def props(obj):
    pr = {}
    for name in dir(obj):
        value = getattr(obj, name)
        if not name.startswith('__') and not inspect.ismethod(value):
            pr[name] = value
    return pr

The dir builtin will give you all the object’s attributes, including special methods like __str__, __dict__ and a whole bunch of others which you probably don’t want. But you can do something like:

>>> class Foo(object):
...     bar = 'hello'
...     baz = 'world'
...
>>> f = Foo()
>>> [name for name in dir(f) if not name.startswith('__')]
[ 'bar', 'baz' ]
>>> dict((name, getattr(f, name)) for name in dir(f) if not name.startswith('__')) 
{ 'bar': 'hello', 'baz': 'world' }

So can extend this to only return data attributes and not methods, by defining your props function like this:

import inspect

def props(obj):
    pr = {}
    for name in dir(obj):
        value = getattr(obj, name)
        if not name.startswith('__') and not inspect.ismethod(value):
            pr[name] = value
    return pr

回答 3

我已经结合了两个答案:

dict((key, value) for key, value in f.__dict__.iteritems() 
    if not callable(value) and not key.startswith('__'))

I’ve settled with a combination of both answers:

dict((key, value) for key, value in f.__dict__.iteritems() 
    if not callable(value) and not key.startswith('__'))

回答 4

我以为我会花些时间向您展示如何通过转换对象来决定字典dict(obj)

class A(object):
    d = '4'
    e = '5'
    f = '6'

    def __init__(self):
        self.a = '1'
        self.b = '2'
        self.c = '3'

    def __iter__(self):
        # first start by grabbing the Class items
        iters = dict((x,y) for x,y in A.__dict__.items() if x[:2] != '__')

        # then update the class items with the instance items
        iters.update(self.__dict__)

        # now 'yield' through the items
        for x,y in iters.items():
            yield x,y

a = A()
print(dict(a)) 
# prints "{'a': '1', 'c': '3', 'b': '2', 'e': '5', 'd': '4', 'f': '6'}"

此代码的关键部分是 __iter__功能。

正如评论所解释的,我们要做的第一件事是获取Class项,并防止以’__’开头的任何东西。

一旦创建了它dict,就可以使用updatedict函数并传入实例__dict__

这些将为您提供完整的成员类+实例字典。现在剩下的就是迭代它们并产生回报。

另外,如果您打算大量使用它,则可以创建一个@iterable类装饰器。

def iterable(cls):
    def iterfn(self):
        iters = dict((x,y) for x,y in cls.__dict__.items() if x[:2] != '__')
        iters.update(self.__dict__)

        for x,y in iters.items():
            yield x,y

    cls.__iter__ = iterfn
    return cls

@iterable
class B(object):
    d = 'd'
    e = 'e'
    f = 'f'

    def __init__(self):
        self.a = 'a'
        self.b = 'b'
        self.c = 'c'

b = B()
print(dict(b))

I thought I’d take some time to show you how you can translate an object to dict via dict(obj).

class A(object):
    d = '4'
    e = '5'
    f = '6'

    def __init__(self):
        self.a = '1'
        self.b = '2'
        self.c = '3'

    def __iter__(self):
        # first start by grabbing the Class items
        iters = dict((x,y) for x,y in A.__dict__.items() if x[:2] != '__')

        # then update the class items with the instance items
        iters.update(self.__dict__)

        # now 'yield' through the items
        for x,y in iters.items():
            yield x,y

a = A()
print(dict(a)) 
# prints "{'a': '1', 'c': '3', 'b': '2', 'e': '5', 'd': '4', 'f': '6'}"

The key section of this code is the __iter__ function.

As the comments explain, the first thing we do is grab the Class items and prevent anything that starts with ‘__’.

Once you’ve created that dict, then you can use the update dict function and pass in the instance __dict__.

These will give you a complete class+instance dictionary of members. Now all that’s left is to iterate over them and yield the returns.

Also, if you plan on using this a lot, you can create an @iterable class decorator.

def iterable(cls):
    def iterfn(self):
        iters = dict((x,y) for x,y in cls.__dict__.items() if x[:2] != '__')
        iters.update(self.__dict__)

        for x,y in iters.items():
            yield x,y

    cls.__iter__ = iterfn
    return cls

@iterable
class B(object):
    d = 'd'
    e = 'e'
    f = 'f'

    def __init__(self):
        self.a = 'a'
        self.b = 'b'
        self.c = 'c'

b = B()
print(dict(b))

回答 5

要从任意对象构建字典,只需使用即可__dict__

这会错过对象从其类继承的属性。例如,

class c(object):
    x = 3
a = c()

hasattr(a,’x’)是true,但是’x’不会出现在a .__ dict__

To build a dictionary from an arbitrary object, it’s sufficient to use __dict__.

This misses attributes that the object inherits from its class. For example,

class c(object):
    x = 3
a = c()

hasattr(a, ‘x’) is true, but ‘x’ does not appear in a.__dict__


回答 6

答案较晚,但提供了完整性和对Google员工的好处:

def props(x):
    return dict((key, getattr(x, key)) for key in dir(x) if key not in dir(x.__class__))

这不会显示在类中定义的方法,但仍会显示字段,包括分配给lambda的字段或以双下划线开头的字段。

Late answer but provided for completeness and the benefit of googlers:

def props(x):
    return dict((key, getattr(x, key)) for key in dir(x) if key not in dir(x.__class__))

This will not show methods defined in the class, but it will still show fields including those assigned to lambdas or those which start with a double underscore.


回答 7

我认为最简单的方法是为该类创建一个getitem属性。如果需要写入对象,则可以创建一个自定义setattr。这是getitem的示例:

class A(object):
    def __init__(self):
        self.b = 1
        self.c = 2
    def __getitem__(self, item):
        return self.__dict__[item]

# Usage: 
a = A()
a.__getitem__('b')  # Outputs 1
a.__dict__  # Outputs {'c': 2, 'b': 1}
vars(a)  # Outputs {'c': 2, 'b': 1}

dict将对象属性生成到字典中,并且字典对象可用于获取所需的项目。

I think the easiest way is to create a getitem attribute for the class. If you need to write to the object, you can create a custom setattr . Here is an example for getitem:

class A(object):
    def __init__(self):
        self.b = 1
        self.c = 2
    def __getitem__(self, item):
        return self.__dict__[item]

# Usage: 
a = A()
a.__getitem__('b')  # Outputs 1
a.__dict__  # Outputs {'c': 2, 'b': 1}
vars(a)  # Outputs {'c': 2, 'b': 1}

dict generates the objects attributes into a dictionary and the dictionary object can be used to get the item you need.


回答 8

使用的缺点 __dict__是它很浅。它不会将任何子类转换为字典。

如果您使用的是Python3.5或更高版本,则可以使用jsons

>>> import jsons
>>> jsons.dump(f)
{'bar': 'hello', 'baz': 'world'}

A downside of using __dict__ is that it is shallow; it won’t convert any subclasses to dictionaries.

If you’re using Python3.5 or higher, you can use jsons:

>>> import jsons
>>> jsons.dump(f)
{'bar': 'hello', 'baz': 'world'}

回答 9

如果要列出部分属性,请覆盖__dict__

def __dict__(self):
    d = {
    'attr_1' : self.attr_1,
    ...
    }
    return d

# Call __dict__
d = instance.__dict__()

如果您instance获得了一些大块数据,并且想要d像消息队列一样推送到Redis ,这将很有帮助。

If you want to list part of your attributes, override __dict__:

def __dict__(self):
    d = {
    'attr_1' : self.attr_1,
    ...
    }
    return d

# Call __dict__
d = instance.__dict__()

This helps a lot if your instance get some large block data and you want to push d to Redis like message queue.


回答 10

PYTHON 3:

class DateTimeDecoder(json.JSONDecoder):

   def __init__(self, *args, **kargs):
        JSONDecoder.__init__(self, object_hook=self.dict_to_object,
                         *args, **kargs)

   def dict_to_object(self, d):
       if '__type__' not in d:
          return d

       type = d.pop('__type__')
       try:
          dateobj = datetime(**d)
          return dateobj
       except:
          d['__type__'] = type
          return d

def json_default_format(value):
    try:
        if isinstance(value, datetime):
            return {
                '__type__': 'datetime',
                'year': value.year,
                'month': value.month,
                'day': value.day,
                'hour': value.hour,
                'minute': value.minute,
                'second': value.second,
                'microsecond': value.microsecond,
            }
        if isinstance(value, decimal.Decimal):
            return float(value)
        if isinstance(value, Enum):
            return value.name
        else:
            return vars(value)
    except Exception as e:
        raise ValueError

现在,您可以在自己的类中使用上述代码:

class Foo():
  def toJSON(self):
        return json.loads(
            json.dumps(self, sort_keys=True, indent=4, separators=(',', ': '), default=json_default_format), cls=DateTimeDecoder)


Foo().toJSON() 

PYTHON 3:

class DateTimeDecoder(json.JSONDecoder):

   def __init__(self, *args, **kargs):
        JSONDecoder.__init__(self, object_hook=self.dict_to_object,
                         *args, **kargs)

   def dict_to_object(self, d):
       if '__type__' not in d:
          return d

       type = d.pop('__type__')
       try:
          dateobj = datetime(**d)
          return dateobj
       except:
          d['__type__'] = type
          return d

def json_default_format(value):
    try:
        if isinstance(value, datetime):
            return {
                '__type__': 'datetime',
                'year': value.year,
                'month': value.month,
                'day': value.day,
                'hour': value.hour,
                'minute': value.minute,
                'second': value.second,
                'microsecond': value.microsecond,
            }
        if isinstance(value, decimal.Decimal):
            return float(value)
        if isinstance(value, Enum):
            return value.name
        else:
            return vars(value)
    except Exception as e:
        raise ValueError

Now you can use above code inside your own class :

class Foo():
  def toJSON(self):
        return json.loads(
            json.dumps(self, sort_keys=True, indent=4, separators=(',', ': '), default=json_default_format), cls=DateTimeDecoder)


Foo().toJSON() 

回答 11

vars() 很棒,但是不适用于对象的嵌套对象

将对象的嵌套对象转换为dict:

def to_dict(self):
    return json.loads(json.dumps(self, default=lambda o: o.__dict__))

vars() is great, but doesn’t work for nested objects of objects

Convert nested object of objects to dict:

def to_dict(self):
    return json.loads(json.dumps(self, default=lambda o: o.__dict__))

如何从字典中获取值列表?

问题:如何从字典中获取值列表?

如何在Python中获取字典中的值列表?

在Java中,将Map的值作为List变得容易list = map.values();。我想知道Python中是否有类似的简单方法可以从字典中获取值列表。

How can I get a list of the values in a dict in Python?

In Java, getting the values of a Map as a List is as easy as doing list = map.values();. I’m wondering if there is a similarly simple way in Python to get a list of values from a dict.


回答 0

是的,这与Python 2完全相同:

d.values()

Python 3中(在其中dict.values返回字典值的视图):

list(d.values())

Yes it’s the exact same thing in Python 2:

d.values()

In Python 3 (where dict.values returns a view of the dictionary’s values instead):

list(d.values())

回答 1

您可以使用*运算符解压缩dict_values:

>>> d = {1: "a", 2: "b"}
>>> [*d.values()]
['a', 'b']

或列出对象

>>> d = {1: "a", 2: "b"}
>>> list(d.values())
['a', 'b']

You can use * operator to unpack dict_values:

>>> d = {1: "a", 2: "b"}
>>> [*d.values()]
['a', 'b']

or list object

>>> d = {1: "a", 2: "b"}
>>> list(d.values())
['a', 'b']

回答 2

应该有一种方法,最好只有一种方法。

因此list(dictionary.values())一种方法

但是,考虑到Python3,更快的方法是什么?

[*L]vs. [].extend(L)vs.list(L)

small_ds = {x: str(x+42) for x in range(10)}
small_df = {x: float(x+42) for x in range(10)}

print('Small Dict(str)')
%timeit [*small_ds.values()]
%timeit [].extend(small_ds.values())
%timeit list(small_ds.values())

print('Small Dict(float)')
%timeit [*small_df.values()]
%timeit [].extend(small_df.values())
%timeit list(small_df.values())

big_ds = {x: str(x+42) for x in range(1000000)}
big_df = {x: float(x+42) for x in range(1000000)}

print('Big Dict(str)')
%timeit [*big_ds.values()]
%timeit [].extend(big_ds.values())
%timeit list(big_ds.values())

print('Big Dict(float)')
%timeit [*big_df.values()]
%timeit [].extend(big_df.values())
%timeit list(big_df.values())
Small Dict(str)
256 ns ± 3.37 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
338 ns ± 0.807 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
336 ns ± 1.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Small Dict(float)
268 ns ± 0.297 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
343 ns ± 15.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
336 ns ± 0.68 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Big Dict(str)
17.5 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.5 ms ± 338 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.2 ms ± 19.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Big Dict(float)
13.2 ms ± 41 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
13.1 ms ± 919 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
12.8 ms ± 578 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

在Intel®Core™i7-8650U CPU @ 1.90GHz上完成。

# Name                    Version                   Build
ipython                   7.5.0            py37h24bf2e0_0

结果

  1. 对于小词典* operator更快
  2. 对于重要的大字典来说,list()可能会更快

There should be one ‒ and preferably only one ‒ obvious way to do it.

Therefore list(dictionary.values()) is the one way.

Yet, considering Python3, what is quicker?

[*L] vs. [].extend(L) vs. list(L)

small_ds = {x: str(x+42) for x in range(10)}
small_df = {x: float(x+42) for x in range(10)}

print('Small Dict(str)')
%timeit [*small_ds.values()]
%timeit [].extend(small_ds.values())
%timeit list(small_ds.values())

print('Small Dict(float)')
%timeit [*small_df.values()]
%timeit [].extend(small_df.values())
%timeit list(small_df.values())

big_ds = {x: str(x+42) for x in range(1000000)}
big_df = {x: float(x+42) for x in range(1000000)}

print('Big Dict(str)')
%timeit [*big_ds.values()]
%timeit [].extend(big_ds.values())
%timeit list(big_ds.values())

print('Big Dict(float)')
%timeit [*big_df.values()]
%timeit [].extend(big_df.values())
%timeit list(big_df.values())
Small Dict(str)
256 ns ± 3.37 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
338 ns ± 0.807 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
336 ns ± 1.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Small Dict(float)
268 ns ± 0.297 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
343 ns ± 15.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
336 ns ± 0.68 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Big Dict(str)
17.5 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.5 ms ± 338 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.2 ms ± 19.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Big Dict(float)
13.2 ms ± 41 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
13.1 ms ± 919 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
12.8 ms ± 578 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Done on Intel(R) Core(TM) i7-8650U CPU @ 1.90GHz.

# Name                    Version                   Build
ipython                   7.5.0            py37h24bf2e0_0

The result

  1. For small dictionaries * operator is quicker
  2. For big dictionaries where it matters list() is maybe slightly quicker

回答 3

请按照以下示例-

songs = [
{"title": "happy birthday", "playcount": 4},
{"title": "AC/DC", "playcount": 2},
{"title": "Billie Jean", "playcount": 6},
{"title": "Human Touch", "playcount": 3}
]

print("====================")
print(f'Songs --> {songs} \n')
title = list(map(lambda x : x['title'], songs))
print(f'Print Title --> {title}')

playcount = list(map(lambda x : x['playcount'], songs))
print(f'Print Playcount --> {playcount}')
print (f'Print Sorted playcount --> {sorted(playcount)}')

# Aliter -
print(sorted(list(map(lambda x: x['playcount'],songs))))

Follow the below example —

songs = [
{"title": "happy birthday", "playcount": 4},
{"title": "AC/DC", "playcount": 2},
{"title": "Billie Jean", "playcount": 6},
{"title": "Human Touch", "playcount": 3}
]

print("====================")
print(f'Songs --> {songs} \n')
title = list(map(lambda x : x['title'], songs))
print(f'Print Title --> {title}')

playcount = list(map(lambda x : x['playcount'], songs))
print(f'Print Playcount --> {playcount}')
print (f'Print Sorted playcount --> {sorted(playcount)}')

# Aliter -
print(sorted(list(map(lambda x: x['playcount'],songs))))

回答 4

out: dict_values([{1:a, 2:b}])

in:  str(dict.values())[14:-3]    
out: 1:a, 2:b

纯粹出于视觉目的。不会产生有用的产品…仅在您希望长字典以段落类型形式打印时才有用。

out: dict_values([{1:a, 2:b}])

in:  str(dict.values())[14:-3]    
out: 1:a, 2:b

Purely for visual purposes. Does not produce a useful product… Only useful if you want a long dictionary to print in a paragraph type form.


将Python字典转换为kwargs?

问题:将Python字典转换为kwargs?

我想使用类继承构建一个针对sunburnt(solr interface)的查询,因此将键-值对加在一起。sunburnt接口带有关键字参数。如何将字典({'type':'Event'})转换为关键字参数(type='Event')

I want to build a query for sunburnt(solr interface) using class inheritance and therefore adding key – value pairs together. The sunburnt interface takes keyword arguments. How can I transform a dict ({'type':'Event'}) into keyword arguments (type='Event')?


回答 0

使用双星运算符(又名double-splat?):

func(**{'type':'Event'})

相当于

func(type='Event')

Use the double-star (aka double-splat?) operator:

func(**{'type':'Event'})

is equivalent to

func(type='Event')

回答 1

** 操作员在这里会有所帮助。

**操作员将解开dict元素的包装,因此**{'type':'Event'}将被视为type='Event'

func(**{'type':'Event'}) 与…相同 func(type='Event') dict元素将转换为相同keyword arguments

费耶

* 将解压缩列表元素,它们将被视为 positional arguments

func(*['one', 'two']) 与…相同 func('one', 'two')

** operator would be helpful here.

** operator will unpack the dict elements and thus **{'type':'Event'} would be treated as type='Event'

func(**{'type':'Event'}) is same as func(type='Event') i.e the dict elements would be converted to the keyword arguments.

FYI

* will unpack the list elements and they would be treated as positional arguments.

func(*['one', 'two']) is same as func('one', 'two')


回答 2

这是一个完整的示例,显示了如何使用**运算符将字典中的值作为关键字参数传递。

>>> def f(x=2):
...     print(x)
... 
>>> new_x = {'x': 4}
>>> f()        #    default value x=2
2
>>> f(x=3)     #   explicit value x=3
3
>>> f(**new_x) # dictionary value x=4 
4

Here is a complete example showing how to use the ** operator to pass values from a dictionary as keyword arguments.

>>> def f(x=2):
...     print(x)
... 
>>> new_x = {'x': 4}
>>> f()        #    default value x=2
2
>>> f(x=3)     #   explicit value x=3
3
>>> f(**new_x) # dictionary value x=4 
4

通过字典中的值获取键

问题:通过字典中的值获取键

我制作了一个函数,该函数将查询年龄Dictionary并显示匹配的名称:

dictionary = {'george' : 16, 'amber' : 19}
search_age = raw_input("Provide age")
for age in dictionary.values():
    if age == search_age:
        name = dictionary[age]
        print name

我知道如何比较和查找年龄,但我不知道如何显示此人的名字。另外,KeyError由于第5行,我得到了提示。我知道这是不正确的,但我不知道如何使它向后搜索。

I made a function which will look up ages in a Dictionary and show the matching name:

dictionary = {'george' : 16, 'amber' : 19}
search_age = raw_input("Provide age")
for age in dictionary.values():
    if age == search_age:
        name = dictionary[age]
        print name

I know how to compare and find the age I just don’t know how to show the name of the person. Additionally, I am getting a KeyError because of line 5. I know it’s not correct but I can’t figure out how to make it search backwards.


回答 0

空无一人。 dict不打算以此方式使用。

dictionary = {'george': 16, 'amber': 19}
search_age = input("Provide age")
for name, age in dictionary.items():  # for name, age in dictionary.iteritems():  (for Python 2.x)
    if age == search_age:
        print(name)

There is none. dict is not intended to be used this way.

dictionary = {'george': 16, 'amber': 19}
search_age = input("Provide age")
for name, age in dictionary.items():  # for name, age in dictionary.iteritems():  (for Python 2.x)
    if age == search_age:
        print(name)

回答 1

mydict = {'george': 16, 'amber': 19}
print mydict.keys()[mydict.values().index(16)]  # Prints george

或在Python 3.x中:

mydict = {'george': 16, 'amber': 19}
print(list(mydict.keys())[list(mydict.values()).index(16)])  # Prints george

基本上,它将字典中的值分隔在一个列表中,找到您拥有的值的位置,并在该位置获取键。

有关Python 3的更多信息keys()如何从dict中获取值列表?.values()

mydict = {'george': 16, 'amber': 19}
print mydict.keys()[mydict.values().index(16)]  # Prints george

Or in Python 3.x:

mydict = {'george': 16, 'amber': 19}
print(list(mydict.keys())[list(mydict.values()).index(16)])  # Prints george

Basically, it separates the dictionary’s values in a list, finds the position of the value you have, and gets the key at that position.

More about keys() and .values() in Python 3: How can I get list of values from dict?


回答 2

如果你想同时姓名年龄,你应该用.items()它给你的关键(key, value)元组:

for name, age in mydict.items():
    if age == search_age:
        print name

您可以在for循环中将元组解压缩为两个单独的变量,然后匹配年龄。

如果您通常要按年龄查找,并且没有两个人具有相同的年龄,则还应该考虑反转字典。

{16: 'george', 19: 'amber'}

这样您就可以通过做

mydict[search_age]

我一直称其为,mydict而不是list因为它list是内置类型的名称,因此您不应将此名称用作其他名称。

您甚至可以在一行中获得所有给定年龄的人的列表:

[name for name, age in mydict.items() if age == search_age]

或每个年龄段只有一个人:

next((name for name, age in mydict.items() if age == search_age), None)

这只会给你 None如果没有那个年龄的人,。

最后,如果dict较长,并且您使用的是Python 2,则应考虑使用.iteritems()而不是.items()Cat Plus Plus在其答案中所做的操作,因为它不需要复制列表。

If you want both the name and the age, you should be using .items() which gives you key (key, value) tuples:

for name, age in mydict.items():
    if age == search_age:
        print name

You can unpack the tuple into two separate variables right in the for loop, then match the age.

You should also consider reversing the dictionary if you’re generally going to be looking up by age, and no two people have the same age:

{16: 'george', 19: 'amber'}

so you can look up the name for an age by just doing

mydict[search_age]

I’ve been calling it mydict instead of list because list is the name of a built-in type, and you shouldn’t use that name for anything else.

You can even get a list of all people with a given age in one line:

[name for name, age in mydict.items() if age == search_age]

or if there is only one person with each age:

next((name for name, age in mydict.items() if age == search_age), None)

which will just give you None if there isn’t anyone with that age.

Finally, if the dict is long and you’re on Python 2, you should consider using .iteritems() instead of .items() as Cat Plus Plus did in his answer, since it doesn’t need to make a copy of the list.


回答 3

我认为指出哪种方法最快,在哪种情况下会很有趣:

这是我运行的一些测试(在2012年的MacBook Pro上)

>>> def method1(list,search_age):
...     for name,age in list.iteritems():
...             if age == search_age:
...                     return name
... 
>>> def method2(list,search_age):
...     return [name for name,age in list.iteritems() if age == search_age]
... 
>>> def method3(list,search_age):
...     return list.keys()[list.values().index(search_age)]

profile.run()每种方法的结果100000次:

方法1:

>>> profile.run("for i in range(0,100000): method1(list,16)")
     200004 function calls in 1.173 seconds

方法2:

>>> profile.run("for i in range(0,100000): method2(list,16)")
     200004 function calls in 1.222 seconds

方法3:

>>> profile.run("for i in range(0,100000): method3(list,16)")
     400004 function calls in 2.125 seconds

因此,这表明对于小的字典而言,方法1最快。这很可能是因为它返回第一个匹配项,而不是像方法2那样的所有匹配项(请参见下面的注释)。


有趣的是,对我拥有2700个条目的字典执行相同的测试,我得到的结果却截然不同(这次运行10000次):

方法1:

>>> profile.run("for i in range(0,10000): method1(UIC_CRS,'7088380')")
     20004 function calls in 2.928 seconds

方法2:

>>> profile.run("for i in range(0,10000): method2(UIC_CRS,'7088380')")
     20004 function calls in 3.872 seconds

方法3:

>>> profile.run("for i in range(0,10000): method3(UIC_CRS,'7088380')")
     40004 function calls in 1.176 seconds

所以在这里,方法3是快。只是显示您的字典大小会影响您选择的方法。

注意:方法2返回所有名称的列表,而方法1和3仅返回第一个匹配项。我没有考虑内存使用情况。我不确定方法3是否创建2个额外的列表(keys()和values())并将其存储在内存中。

I thought it would be interesting to point out which methods are the quickest, and in what scenario:

Here’s some tests I ran (on a 2012 MacBook Pro)

>>> def method1(list,search_age):
...     for name,age in list.iteritems():
...             if age == search_age:
...                     return name
... 
>>> def method2(list,search_age):
...     return [name for name,age in list.iteritems() if age == search_age]
... 
>>> def method3(list,search_age):
...     return list.keys()[list.values().index(search_age)]

Results from profile.run() on each method 100000 times:

Method 1:

>>> profile.run("for i in range(0,100000): method1(list,16)")
     200004 function calls in 1.173 seconds

Method 2:

>>> profile.run("for i in range(0,100000): method2(list,16)")
     200004 function calls in 1.222 seconds

Method 3:

>>> profile.run("for i in range(0,100000): method3(list,16)")
     400004 function calls in 2.125 seconds

So this shows that for a small dict, method 1 is the quickest. This is most likely because it returns the first match, as opposed to all of the matches like method 2 (see note below).


Interestingly, performing the same tests on a dict I have with 2700 entries, I get quite different results (this time run 10000 times):

Method 1:

>>> profile.run("for i in range(0,10000): method1(UIC_CRS,'7088380')")
     20004 function calls in 2.928 seconds

Method 2:

>>> profile.run("for i in range(0,10000): method2(UIC_CRS,'7088380')")
     20004 function calls in 3.872 seconds

Method 3:

>>> profile.run("for i in range(0,10000): method3(UIC_CRS,'7088380')")
     40004 function calls in 1.176 seconds

So here, method 3 is much faster. Just goes to show the size of your dict will affect which method you choose.

Notes: Method 2 returns a list of all names, whereas methods 1 and 3 return only the first match. I have not considered memory usage. I’m not sure if method 3 creates 2 extra lists (keys() and values()) and stores them in memory.


回答 4

一线版本:(i是旧字典,p是反向字典)

说明:i.keys()i.values()返回两个分别具有字典键和值的列表。zip函数具有将列表捆绑在一起以生成字典的能力。

p = dict(zip(i.values(),i.keys()))

警告:仅当值是可哈希值且唯一时,此选项才起作用。

one line version: (i is an old dictionary, p is a reversed dictionary)

explanation : i.keys() and i.values() returns two lists with keys and values of the dictionary respectively. The zip function has the ability to tie together lists to produce a dictionary.

p = dict(zip(i.values(),i.keys()))

Warning : This will work only if the values are hashable and unique.


回答 5

a = {'a':1,'b':2,'c':3}
{v:k for k, v in a.items()}[1]

或更好

{k:v for k, v in a.items() if v == 1}
a = {'a':1,'b':2,'c':3}
{v:k for k, v in a.items()}[1]

or better

{k:v for k, v in a.items() if v == 1}

回答 6

key = next((k for k in my_dict if my_dict[k] == val), None)
key = next((k for k in my_dict if my_dict[k] == val), None)

回答 7

尝试以下这种单线来反向字典:

reversed_dictionary = dict(map(reversed, dictionary.items()))

Try this one-liner to reverse a dictionary:

reversed_dictionary = dict(map(reversed, dictionary.items()))

回答 8

我找到了这个答案非常有效,但对我来说却不太容易理解。

为了更加清晰,您可以反转字典的键和值。这是使键的值和值的键,因为看到这里

mydict = {'george':16,'amber':19}
res = dict((v,k) for k,v in mydict.iteritems())
print(res[16]) # Prints george

要么

mydict = {'george':16,'amber':19}
dict((v,k) for k,v in mydict.iteritems())[16]

这与其他答案基本相同。

I found this answer very effective but not very easy to read for me.

To make it more clear you can invert the key and the value of a dictionary. This is make the keys values and the values keys, as seen here.

mydict = {'george':16,'amber':19}
res = dict((v,k) for k,v in mydict.iteritems())
print(res[16]) # Prints george

or

mydict = {'george':16,'amber':19}
dict((v,k) for k,v in mydict.iteritems())[16]

which is essentially the same that this other answer.


回答 9

如果要通过值查找键,则可以使用字典理解来创建查找字典,然后使用该字典从值中查找键。

lookup = {value: key for key, value in self.data}
lookup[value]

If you want to find the key by the value, you can use a dictionary comprehension to create a lookup dictionary and then use that to find the key from the value.

lookup = {value: key for key, value in self.data}
lookup[value]

回答 10

您可以通过使用获取密钥dict.keys()dict.values()list.index()方法,请参见下面的代码示例:

names_dict = {'george':16,'amber':19}
search_age = int(raw_input("Provide age"))
key = names_dict.keys()[names_dict.values().index(search_age)]

You can get key by using dict.keys(), dict.values() and list.index() methods, see code samples below:

names_dict = {'george':16,'amber':19}
search_age = int(raw_input("Provide age"))
key = names_dict.keys()[names_dict.values().index(search_age)]

回答 11

这是我对这个问题的看法。:)我刚刚开始学习Python,所以我这样称呼:

“初学者的理解”解决方案。

#Code without comments.

list1 = {'george':16,'amber':19, 'Garry':19}
search_age = raw_input("Provide age: ")
print
search_age = int(search_age)

listByAge = {}

for name, age in list1.items():
    if age == search_age:
        age = str(age)
        results = name + " " +age
        print results

        age2 = int(age)
        listByAge[name] = listByAge.get(name,0)+age2

print
print listByAge

#Code with comments.
#I've added another name with the same age to the list.
list1 = {'george':16,'amber':19, 'Garry':19}
#Original code.
search_age = raw_input("Provide age: ")
print
#Because raw_input gives a string, we need to convert it to int,
#so we can search the dictionary list with it.
search_age = int(search_age)

#Here we define another empty dictionary, to store the results in a more 
#permanent way.
listByAge = {}

#We use double variable iteration, so we get both the name and age 
#on each run of the loop.
for name, age in list1.items():
    #Here we check if the User Defined age = the age parameter 
    #for this run of the loop.
    if age == search_age:
        #Here we convert Age back to string, because we will concatenate it 
        #with the person's name. 
        age = str(age)
        #Here we concatenate.
        results = name + " " +age
        #If you want just the names and ages displayed you can delete
        #the code after "print results". If you want them stored, don't...
        print results

        #Here we create a second variable that uses the value of
        #the age for the current person in the list.
        #For example if "Anna" is "10", age2 = 10,
        #integer value which we can use in addition.
        age2 = int(age)
        #Here we use the method that checks or creates values in dictionaries.
        #We create a new entry for each name that matches the User Defined Age
        #with default value of 0, and then we add the value from age2.
        listByAge[name] = listByAge.get(name,0)+age2

#Here we print the new dictionary with the users with User Defined Age.
print
print listByAge

#Results
Running: *\test.py (Thu Jun 06 05:10:02 2013)

Provide age: 19

amber 19
Garry 19

{'amber': 19, 'Garry': 19}

Execution Successful!

Here is my take on this problem. :) I have just started learning Python, so I call this:

“The Understandable for beginners” solution.

#Code without comments.

list1 = {'george':16,'amber':19, 'Garry':19}
search_age = raw_input("Provide age: ")
print
search_age = int(search_age)

listByAge = {}

for name, age in list1.items():
    if age == search_age:
        age = str(age)
        results = name + " " +age
        print results

        age2 = int(age)
        listByAge[name] = listByAge.get(name,0)+age2

print
print listByAge

.

#Code with comments.
#I've added another name with the same age to the list.
list1 = {'george':16,'amber':19, 'Garry':19}
#Original code.
search_age = raw_input("Provide age: ")
print
#Because raw_input gives a string, we need to convert it to int,
#so we can search the dictionary list with it.
search_age = int(search_age)

#Here we define another empty dictionary, to store the results in a more 
#permanent way.
listByAge = {}

#We use double variable iteration, so we get both the name and age 
#on each run of the loop.
for name, age in list1.items():
    #Here we check if the User Defined age = the age parameter 
    #for this run of the loop.
    if age == search_age:
        #Here we convert Age back to string, because we will concatenate it 
        #with the person's name. 
        age = str(age)
        #Here we concatenate.
        results = name + " " +age
        #If you want just the names and ages displayed you can delete
        #the code after "print results". If you want them stored, don't...
        print results

        #Here we create a second variable that uses the value of
        #the age for the current person in the list.
        #For example if "Anna" is "10", age2 = 10,
        #integer value which we can use in addition.
        age2 = int(age)
        #Here we use the method that checks or creates values in dictionaries.
        #We create a new entry for each name that matches the User Defined Age
        #with default value of 0, and then we add the value from age2.
        listByAge[name] = listByAge.get(name,0)+age2

#Here we print the new dictionary with the users with User Defined Age.
print
print listByAge

.

#Results
Running: *\test.py (Thu Jun 06 05:10:02 2013)

Provide age: 19

amber 19
Garry 19

{'amber': 19, 'Garry': 19}

Execution Successful!

回答 12

get_key = lambda v, d: next(k for k in d if d[k] is v)
get_key = lambda v, d: next(k for k in d if d[k] is v)

回答 13

考虑使用熊猫。正如William McKinney的“ Python for Data Analysis”中所述

考虑序列的另一种方法是定长排序的字典,因为它是索引值到数据值的映射。它可以在可能使用字典的许多情况下使用。

import pandas as pd
list = {'george':16,'amber':19}
lookup_list = pd.Series(list)

要查询您的系列,请执行以下操作:

lookup_list[lookup_list.values == 19]

生成:

Out[1]: 
amber    19
dtype: int64

如果您需要对输出执行其他任何操作,将答案转换为列表可能会很有用:

answer = lookup_list[lookup_list.values == 19].index
answer = pd.Index.tolist(answer)

Consider using Pandas. As stated in William McKinney’s “Python for Data Analysis’

Another way to think about a Series is as a fixed-length, ordered dict, as it is a mapping of index values to data values. It can be used in many contexts where you might use a dict.

import pandas as pd
list = {'george':16,'amber':19}
lookup_list = pd.Series(list)

To query your series do the following:

lookup_list[lookup_list.values == 19]

Which yields:

Out[1]: 
amber    19
dtype: int64

If you need to do anything else with the output transforming the answer into a list might be useful:

answer = lookup_list[lookup_list.values == 19].index
answer = pd.Index.tolist(answer)

回答 14

在这里,recover_key将使用字典和要在字典中找到的值。然后,我们遍历字典中的键,并与value的键进行比较,然后返回该特定键。

def recover_key(dicty,value):
    for a_key in dicty.keys():
        if (dicty[a_key] == value):
            return a_key

Here, recover_key takes dictionary and value to find in dictionary. We then loop over the keys in dictionary and make a comparison with that of value and return that particular key.

def recover_key(dicty,value):
    for a_key in dicty.keys():
        if (dicty[a_key] == value):
            return a_key

回答 15

for name in mydict:
    if mydict[name] == search_age:
        print(name) 
        #or do something else with it. 
        #if in a function append to a temporary list, 
        #then after the loop return the list
for name in mydict:
    if mydict[name] == search_age:
        print(name) 
        #or do something else with it. 
        #if in a function append to a temporary list, 
        #then after the loop return the list

回答 16

我们可以得到Keydict是:

def getKey(dct,value):
     return [key for key in dct if (dct[key] == value)]

we can get the Key of dict by :

def getKey(dct,value):
     return [key for key in dct if (dct[key] == value)]

回答 17

它得到了回答,但是可以通过使用“ map / reduce”来完成,例如:

def find_key(value, dictionary):
    return reduce(lambda x, y: x if x is not None else y,
                  map(lambda x: x[0] if x[1] == value else None, 
                      dictionary.iteritems()))

it’s answered, but it could be done with a fancy ‘map/reduce’ use, e.g.:

def find_key(value, dictionary):
    return reduce(lambda x, y: x if x is not None else y,
                  map(lambda x: x[0] if x[1] == value else None, 
                      dictionary.iteritems()))

回答 18

Cat Plus Plus提到,这不是打算使用字典的方式。原因如下:

字典的定义类似于数学中的映射。在这种情况下,字典是K(键集)到V(值)的映射-但反之亦然。如果取消引用dict,则期望返回的值恰好是一个。但是,将不同的键映射到相同的值是完全合法的,例如:

d = { k1 : v1, k2 : v2, k3 : v1}

当您通过键的对应值查找键时,实际上就是在反转字典。但是,映射不一定是可逆的!在此示例中,请求对应于v1的密钥可能会产生k1或k3。您应该同时退货吗?只是找到第一个?这就是为什么字典未定义indexof()的原因。

如果您知道自己的数据,则可以这样做。但是API不能假定任意字典都是可逆的,因此缺少这种操作。

Cat Plus Plus mentioned that this isn’t how a dictionary is intended to be used. Here’s why:

The definition of a dictionary is analogous to that of a mapping in mathematics. In this case, a dict is a mapping of K (the set of keys) to V (the values) – but not vice versa. If you dereference a dict, you expect to get exactly one value returned. But, it is perfectly legal for different keys to map onto the same value, e.g.:

d = { k1 : v1, k2 : v2, k3 : v1}

When you look up a key by it’s corresponding value, you’re essentially inverting the dictionary. But a mapping isn’t necessarily invertible! In this example, asking for the key corresponding to v1 could yield k1 or k3. Should you return both? Just the first one found? That’s why indexof() is undefined for dictionaries.

If you know your data, you could do this. But an API can’t assume that an arbitrary dictionary is invertible, hence the lack of such an operation.


回答 19

这是我的看法。万一您需要一个结果,这对于显示多个结果很有用。所以我也添加了列表

myList = {'george':16,'amber':19, 'rachel':19, 
           'david':15 }                         #Setting the dictionary
result=[]                                       #Making ready of the result list
search_age = int(input('Enter age '))

for keywords in myList.keys():
    if myList[keywords] ==search_age:
    result.append(keywords)                    #This part, we are making list of results

for res in result:                             #We are now printing the results
    print(res)

就是这样…

here is my take on it. This is good for displaying multiple results just in case you need one. So I added the list as well

myList = {'george':16,'amber':19, 'rachel':19, 
           'david':15 }                         #Setting the dictionary
result=[]                                       #Making ready of the result list
search_age = int(input('Enter age '))

for keywords in myList.keys():
    if myList[keywords] ==search_age:
    result.append(keywords)                    #This part, we are making list of results

for res in result:                             #We are now printing the results
    print(res)

And that’s it…


回答 20

d= {'george':16,'amber':19}

dict((v,k) for k,v in d.items()).get(16)

输出如下:

-> prints george
d= {'george':16,'amber':19}

dict((v,k) for k,v in d.items()).get(16)

The output is as follows:

-> prints george

回答 21

没有一种简单的方法可以通过“查找”值在列表中找到键。但是,如果您知道该值(通过键进行迭代),则可以通过该元素在字典中查找值。如果D [element](其中D是字典对象)等于您要查找的键,则可以执行一些代码。

D = {'Ali': 20, 'Marina': 12, 'George':16}
age = int(input('enter age:\t'))  
for element in D.keys():
    if D[element] == age:
        print(element)

There is no easy way to find a key in a list by ‘looking up’ the value. However, if you know the value, iterating through the keys, you can look up values in the dictionary by the element. If D[element] where D is a dictionary object, is equal to the key you’re trying to look up, you can execute some code.

D = {'Ali': 20, 'Marina': 12, 'George':16}
age = int(input('enter age:\t'))  
for element in D.keys():
    if D[element] == age:
        print(element)

回答 22

您需要使用字典,然后反向使用该字典。这意味着您需要另一个数据结构。如果您使用的是python 3,请使用enum模块,但如果使用的是python 2.7,请使用enum34python 2的反向移植。

例:

from enum import Enum

class Color(Enum): 
    red = 1 
    green = 2 
    blue = 3

>>> print(Color.red) 
Color.red

>>> print(repr(Color.red)) 
<color.red: 1=""> 

>>> type(Color.red) 
<enum 'color'=""> 
>>> isinstance(Color.green, Color) 
True 

>>> member = Color.red 
>>> member.name 
'red' 
>>> member.value 
1 

You need to use a dictionary and reverse of that dictionary. It means you need another data structure. If you are in python 3, use enum module but if you are using python 2.7 use enum34 which is back ported for python 2.

Example:

from enum import Enum

class Color(Enum): 
    red = 1 
    green = 2 
    blue = 3

>>> print(Color.red) 
Color.red

>>> print(repr(Color.red)) 
<color.red: 1=""> 

>>> type(Color.red) 
<enum 'color'=""> 
>>> isinstance(Color.green, Color) 
True 

>>> member = Color.red 
>>> member.name 
'red' 
>>> member.value 
1 

回答 23

def get_Value(dic,value):
    for name in dic:
        if dic[name] == value:
            del dic[name]
            return name
def get_Value(dic,value):
    for name in dic:
        if dic[name] == value:
            del dic[name]
            return name

回答 24

只是我的回答lambdafilter

filter( lambda x, dictionary=dictionary, search_age=int(search_age): dictionary[x] == search_age  , dictionary )

Just my answer in lambda and filter.

filter( lambda x, dictionary=dictionary, search_age=int(search_age): dictionary[x] == search_age  , dictionary )

回答 25

已经回答了,但是由于有几个人提到反转字典,因此这是您在一行中进行操作的方式(假设1:1映射)和一些各种性能数据:

python 2.6:

reversedict = dict([(value, key) for key, value in mydict.iteritems()])

2.7+:

reversedict = {value:key for key, value in mydict.iteritems()}

如果您认为不是1:1,则仍然可以用几行代码创建合理的反向映射:

reversedict = defaultdict(list)
[reversedict[value].append(key) for key, value in mydict.iteritems()]

速度有多慢:比简单的搜索要慢,但是却没有您想像的慢-在“直” 100000条目字典上,“快速”搜索(即在键中寻找一个早的值)比反转整个字典快约10倍,“慢速”搜索(快到结尾)约快4-5倍。因此,经过最多约10次查询后,它是自付费用的。

第二个版本(每个项目都有列表)的花费约为简单版本的2.5倍。

largedict = dict((x,x) for x in range(100000))

# Should be slow, has to search 90000 entries before it finds it
In [26]: %timeit largedict.keys()[largedict.values().index(90000)]
100 loops, best of 3: 4.81 ms per loop

# Should be fast, has to only search 9 entries to find it. 
In [27]: %timeit largedict.keys()[largedict.values().index(9)]
100 loops, best of 3: 2.94 ms per loop

# How about using iterkeys() instead of keys()?
# These are faster, because you don't have to create the entire keys array.
# You DO have to create the entire values array - more on that later.

In [31]: %timeit islice(largedict.iterkeys(), largedict.values().index(90000))
100 loops, best of 3: 3.38 ms per loop

In [32]: %timeit islice(largedict.iterkeys(), largedict.values().index(9))
1000 loops, best of 3: 1.48 ms per loop

In [24]: %timeit reversedict = dict([(value, key) for key, value in largedict.iteritems()])
10 loops, best of 3: 22.9 ms per loop

In [23]: %%timeit
....: reversedict = defaultdict(list)
....: [reversedict[value].append(key) for key, value in largedict.iteritems()]
....:
10 loops, best of 3: 53.6 ms per loop

使用ifilter也有一些有趣的结果。从理论上讲,ifilter应该更快,因为我们可以使用itervalues(),而不必创建/遍历整个值列表。实际上,结果是…很奇怪…

In [72]: %%timeit
....: myf = ifilter(lambda x: x[1] == 90000, largedict.iteritems())
....: myf.next()[0]
....:
100 loops, best of 3: 15.1 ms per loop

In [73]: %%timeit
....: myf = ifilter(lambda x: x[1] == 9, largedict.iteritems())
....: myf.next()[0]
....:
100000 loops, best of 3: 2.36 us per loop

因此,对于较小的偏移量,它比以前的任何版本都快得多(2.36 * u * S,而以前的版本最低为1.48 * m * S)。但是,对于列表末尾附近的较大偏移量,它的速度要慢得多(15.1ms与相同的1.48mS)。恕我直言,低端的少量节省不值得高端的成本。

already been answered, but since several people mentioned reversing the dictionary, here’s how you do it in one line (assuming 1:1 mapping) and some various perf data:

python 2.6:

reversedict = dict([(value, key) for key, value in mydict.iteritems()])

2.7+:

reversedict = {value:key for key, value in mydict.iteritems()}

if you think it’s not 1:1, you can still create a reasonable reverse mapping with a couple lines:

reversedict = defaultdict(list)
[reversedict[value].append(key) for key, value in mydict.iteritems()]

how slow is this: slower than a simple search, but not nearly as slow as you’d think – on a ‘straight’ 100000 entry dictionary, a ‘fast’ search (i.e. looking for a value that should be early in the keys) was about 10x faster than reversing the entire dictionary, and a ‘slow’ search (towards the end) about 4-5x faster. So after at most about 10 lookups, it’s paid for itself.

the second version (with lists per item) takes about 2.5x as long as the simple version.

largedict = dict((x,x) for x in range(100000))

# Should be slow, has to search 90000 entries before it finds it
In [26]: %timeit largedict.keys()[largedict.values().index(90000)]
100 loops, best of 3: 4.81 ms per loop

# Should be fast, has to only search 9 entries to find it. 
In [27]: %timeit largedict.keys()[largedict.values().index(9)]
100 loops, best of 3: 2.94 ms per loop

# How about using iterkeys() instead of keys()?
# These are faster, because you don't have to create the entire keys array.
# You DO have to create the entire values array - more on that later.

In [31]: %timeit islice(largedict.iterkeys(), largedict.values().index(90000))
100 loops, best of 3: 3.38 ms per loop

In [32]: %timeit islice(largedict.iterkeys(), largedict.values().index(9))
1000 loops, best of 3: 1.48 ms per loop

In [24]: %timeit reversedict = dict([(value, key) for key, value in largedict.iteritems()])
10 loops, best of 3: 22.9 ms per loop

In [23]: %%timeit
....: reversedict = defaultdict(list)
....: [reversedict[value].append(key) for key, value in largedict.iteritems()]
....:
10 loops, best of 3: 53.6 ms per loop

Also had some interesting results with ifilter. Theoretically, ifilter should be faster, in that we can use itervalues() and possibly not have to create/go through the entire values list. In practice, the results were… odd…

In [72]: %%timeit
....: myf = ifilter(lambda x: x[1] == 90000, largedict.iteritems())
....: myf.next()[0]
....:
100 loops, best of 3: 15.1 ms per loop

In [73]: %%timeit
....: myf = ifilter(lambda x: x[1] == 9, largedict.iteritems())
....: myf.next()[0]
....:
100000 loops, best of 3: 2.36 us per loop

So, for small offsets, it was dramatically faster than any previous version (2.36 *u*S vs. a minimum of 1.48 *m*S for previous cases). However, for large offsets near the end of the list, it was dramatically slower (15.1ms vs. the same 1.48mS). The small savings at the low end is not worth the cost at the high end, imho.


回答 26

有时可能需要int():

titleDic = {'Фильмы':1, 'Музыка':2}

def categoryTitleForNumber(self, num):
    search_title = ''
    for title, titleNum in self.titleDic.items():
        if int(titleNum) == int(num):
            search_title = title
    return search_title

Sometimes int() may be needed:

titleDic = {'Фильмы':1, 'Музыка':2}

def categoryTitleForNumber(self, num):
    search_title = ''
    for title, titleNum in self.titleDic.items():
        if int(titleNum) == int(num):
            search_title = title
    return search_title

回答 27

这是一个在Python 2和Python 3中都可以使用的解决方案。

dict((v, k) for k, v in list.items())[search_age]

直到[search_age]构造反向字典为止(其中值是键,反之亦然)。您可以创建一个辅助方法,该方法将缓存此反向字典,如下所示:

def find_name(age, _rev_lookup=dict((v, k) for k, v in ages_by_name.items())):
    return _rev_lookup[age]

甚至更一般的工厂将为您的一个或多个列表创建按年龄查找名称的方法

def create_name_finder(ages_by_name):
    names_by_age = dict((v, k) for k, v in ages_by_name.items())
    def find_name(age):
      return names_by_age[age]

这样您就可以执行以下操作:

find_teen_by_age = create_name_finder({'george':16,'amber':19})
...
find_teen_by_age(search_age)

请注意,由于前者是预定义类型,因此我将其重命名listages_by_name

Here is a solution which works both in Python 2 and Python 3:

dict((v, k) for k, v in list.items())[search_age]

The part until [search_age] constructs the reverse dictionary (where values are keys and vice-versa). You could create a helper method which will cache this reversed dictionary like so:

def find_name(age, _rev_lookup=dict((v, k) for k, v in ages_by_name.items())):
    return _rev_lookup[age]

or even more generally a factory which would create a by-age name lookup method for one or more of you lists

def create_name_finder(ages_by_name):
    names_by_age = dict((v, k) for k, v in ages_by_name.items())
    def find_name(age):
      return names_by_age[age]

so you would be able to do:

find_teen_by_age = create_name_finder({'george':16,'amber':19})
...
find_teen_by_age(search_age)

Note that I renamed list to ages_by_name since the former is a predefined type.


回答 28

这是您访问字典以执行所需操作的方式:

list = {'george': 16, 'amber': 19}
search_age = raw_input("Provide age")
for age in list:
    if list[age] == search_age:
        print age

当然,您的名字太不正确了,看起来好像要打印一个年龄,但确实可以打印出名字。由于您是按名称访问的,因此如果您输入以下内容,它将变得更加容易理解:

list = {'george': 16, 'amber': 19}
search_age = raw_input("Provide age")
for name in list:
    if list[name] == search_age:
        print name

更好的是:

people = {'george': {'age': 16}, 'amber': {'age': 19}}
search_age = raw_input("Provide age")
for name in people:
    if people[name]['age'] == search_age:
        print name

This is how you access the dictionary to do what you want:

list = {'george': 16, 'amber': 19}
search_age = raw_input("Provide age")
for age in list:
    if list[age] == search_age:
        print age

of course, your names are so off it looks like it would be printing an age, but it DOES print the name. Since you are accessing by name, it becomes more understandable if you write:

list = {'george': 16, 'amber': 19}
search_age = raw_input("Provide age")
for name in list:
    if list[name] == search_age:
        print name

Better yet:

people = {'george': {'age': 16}, 'amber': {'age': 19}}
search_age = raw_input("Provide age")
for name in people:
    if people[name]['age'] == search_age:
        print name

回答 29

dictionary = {'george' : 16, 'amber' : 19}
search_age = raw_input("Provide age")
key = [filter( lambda x: dictionary[x] == k  , dictionary ),[None]][0] 
# key = None from [None] which is a safeguard for not found.

对于多次出现,请使用:

keys = [filter( lambda x: dictionary[x] == k  , dictionary )]
dictionary = {'george' : 16, 'amber' : 19}
search_age = raw_input("Provide age")
key = [filter( lambda x: dictionary[x] == k  , dictionary ),[None]][0] 
# key = None from [None] which is a safeguard for not found.

For multiple occurrences use:

keys = [filter( lambda x: dictionary[x] == k  , dictionary )]

将嵌套的Python字典转换为对象?

问题:将嵌套的Python字典转换为对象?

我正在寻找一种优雅的方式来获取数据,该数据使用具有一些嵌套的字典和列表的字典的属性访问(即javascript样式的对象语法)。

例如:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

应该以这种方式访问​​:

>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
bar

我认为,没有递归是不可能的,但是获得字典对象样式的一种好方法是什么?

I’m searching for an elegant way to get data using attribute access on a dict with some nested dicts and lists (i.e. javascript-style object syntax).

For example:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

Should be accessible in this way:

>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
bar

I think, this is not possible without recursion, but what would be a nice way to get an object style for dicts?


回答 0

更新:在Python 2.6及更高版本中,请考虑namedtuple数据结构是否满足您的需求:

>>> from collections import namedtuple
>>> MyStruct = namedtuple('MyStruct', 'a b d')
>>> s = MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s
MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s.a
1
>>> s.b
{'c': 2}
>>> s.c
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'MyStruct' object has no attribute 'c'
>>> s.d
['hi']

备选方案(原始答案内容)为:

class Struct:
    def __init__(self, **entries):
        self.__dict__.update(entries)

然后,您可以使用:

>>> args = {'a': 1, 'b': 2}
>>> s = Struct(**args)
>>> s
<__main__.Struct instance at 0x01D6A738>
>>> s.a
1
>>> s.b
2

Update: In Python 2.6 and onwards, consider whether the namedtuple data structure suits your needs:

>>> from collections import namedtuple
>>> MyStruct = namedtuple('MyStruct', 'a b d')
>>> s = MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s
MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s.a
1
>>> s.b
{'c': 2}
>>> s.c
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'MyStruct' object has no attribute 'c'
>>> s.d
['hi']

The alternative (original answer contents) is:

class Struct:
    def __init__(self, **entries):
        self.__dict__.update(entries)

Then, you can use:

>>> args = {'a': 1, 'b': 2}
>>> s = Struct(**args)
>>> s
<__main__.Struct instance at 0x01D6A738>
>>> s.a
1
>>> s.b
2

回答 1

class obj(object):
    def __init__(self, d):
        for a, b in d.items():
            if isinstance(b, (list, tuple)):
               setattr(self, a, [obj(x) if isinstance(x, dict) else x for x in b])
            else:
               setattr(self, a, obj(b) if isinstance(b, dict) else b)

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = obj(d)
>>> x.b.c
2
>>> x.d[1].foo
'bar'
class obj(object):
    def __init__(self, d):
        for a, b in d.items():
            if isinstance(b, (list, tuple)):
               setattr(self, a, [obj(x) if isinstance(x, dict) else x for x in b])
            else:
               setattr(self, a, obj(b) if isinstance(b, dict) else b)

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = obj(d)
>>> x.b.c
2
>>> x.d[1].foo
'bar'

回答 2

令人惊讶的是,没有人提到邦奇。该库专门用于提供对dict对象的属性样式访问,并且完全符合OP的要求。演示:

>>> from bunch import bunchify
>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = bunchify(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'

可以从https://github.com/Infinidat/munch获得Python 3库- 版权归codyzu所有

Surprisingly no one has mentioned Bunch. This library is exclusively meant to provide attribute style access to dict objects and does exactly what the OP wants. A demonstration:

>>> from bunch import bunchify
>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = bunchify(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'

A Python 3 library is available at https://github.com/Infinidat/munch – Credit goes to codyzu


回答 3

x = type('new_dict', (object,), d)

然后向其中添加递归就可以了。

编辑这是我将如何实现:

>>> d
{'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> def obj_dic(d):
    top = type('new', (object,), d)
    seqs = tuple, list, set, frozenset
    for i, j in d.items():
        if isinstance(j, dict):
            setattr(top, i, obj_dic(j))
        elif isinstance(j, seqs):
            setattr(top, i, 
                type(j)(obj_dic(sj) if isinstance(sj, dict) else sj for sj in j))
        else:
            setattr(top, i, j)
    return top

>>> x = obj_dic(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'
x = type('new_dict', (object,), d)

then add recursion to this and you’re done.

edit this is how I’d implement it:

>>> d
{'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> def obj_dic(d):
    top = type('new', (object,), d)
    seqs = tuple, list, set, frozenset
    for i, j in d.items():
        if isinstance(j, dict):
            setattr(top, i, obj_dic(j))
        elif isinstance(j, seqs):
            setattr(top, i, 
                type(j)(obj_dic(sj) if isinstance(sj, dict) else sj for sj in j))
        else:
            setattr(top, i, j)
    return top

>>> x = obj_dic(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'

回答 4

有一个名为的收集助手namedtuple,可以为您完成此操作:

from collections import namedtuple

d_named = namedtuple('Struct', d.keys())(*d.values())

In [7]: d_named
Out[7]: Struct(a=1, b={'c': 2}, d=['hi', {'foo': 'bar'}])

In [8]: d_named.a
Out[8]: 1

There’s a collection helper called namedtuple, that can do this for you:

from collections import namedtuple

d_named = namedtuple('Struct', d.keys())(*d.values())

In [7]: d_named
Out[7]: Struct(a=1, b={'c': 2}, d=['hi', {'foo': 'bar'}])

In [8]: d_named.a
Out[8]: 1

回答 5

class Struct(object):
    """Comment removed"""
    def __init__(self, data):
        for name, value in data.iteritems():
            setattr(self, name, self._wrap(value))

    def _wrap(self, value):
        if isinstance(value, (tuple, list, set, frozenset)): 
            return type(value)([self._wrap(v) for v in value])
        else:
            return Struct(value) if isinstance(value, dict) else value

可以与任何深度的任何序列/字典/值结构一起使用。

class Struct(object):
    """Comment removed"""
    def __init__(self, data):
        for name, value in data.iteritems():
            setattr(self, name, self._wrap(value))

    def _wrap(self, value):
        if isinstance(value, (tuple, list, set, frozenset)): 
            return type(value)([self._wrap(v) for v in value])
        else:
            return Struct(value) if isinstance(value, dict) else value

Can be used with any sequence/dict/value structure of any depth.


回答 6

以我认为是前面示例的最佳方面,这是我想到的:

class Struct:
  '''The recursive class for building and representing objects with.'''
  def __init__(self, obj):
    for k, v in obj.iteritems():
      if isinstance(v, dict):
        setattr(self, k, Struct(v))
      else:
        setattr(self, k, v)
  def __getitem__(self, val):
    return self.__dict__[val]
  def __repr__(self):
    return '{%s}' % str(', '.join('%s : %s' % (k, repr(v)) for
      (k, v) in self.__dict__.iteritems()))

Taking what I feel are the best aspects of the previous examples, here’s what I came up with:

class Struct:
  '''The recursive class for building and representing objects with.'''
  def __init__(self, obj):
    for k, v in obj.iteritems():
      if isinstance(v, dict):
        setattr(self, k, Struct(v))
      else:
        setattr(self, k, v)
  def __getitem__(self, val):
    return self.__dict__[val]
  def __repr__(self):
    return '{%s}' % str(', '.join('%s : %s' % (k, repr(v)) for
      (k, v) in self.__dict__.iteritems()))

回答 7

如果您的字典来自json.loads(),则可以一行将其变成一个对象(而不是字典):

import json
from collections import namedtuple

json.loads(data, object_hook=lambda d: namedtuple('X', d.keys())(*d.values()))

另请参阅如何将JSON数据转换为Python对象

If your dict is coming from json.loads(), you can turn it into an object instead (rather than a dict) in one line:

import json
from collections import namedtuple

json.loads(data, object_hook=lambda d: namedtuple('X', d.keys())(*d.values()))

See also How to convert JSON data into a Python object.


回答 8

如果要将字典键作为对象(或作为困难键的字典)访问,请递归地进行操作,并且还能够更新原始字典,则可以执行以下操作:

class Dictate(object):
    """Object view of a dict, updating the passed in dict when values are set
    or deleted. "Dictate" the contents of a dict...: """

    def __init__(self, d):
        # since __setattr__ is overridden, self.__dict = d doesn't work
        object.__setattr__(self, '_Dictate__dict', d)

    # Dictionary-like access / updates
    def __getitem__(self, name):
        value = self.__dict[name]
        if isinstance(value, dict):  # recursively view sub-dicts as objects
            value = Dictate(value)
        return value

    def __setitem__(self, name, value):
        self.__dict[name] = value
    def __delitem__(self, name):
        del self.__dict[name]

    # Object-like access / updates
    def __getattr__(self, name):
        return self[name]

    def __setattr__(self, name, value):
        self[name] = value
    def __delattr__(self, name):
        del self[name]

    def __repr__(self):
        return "%s(%r)" % (type(self).__name__, self.__dict)
    def __str__(self):
        return str(self.__dict)

用法示例:

d = {'a': 'b', 1: 2}
dd = Dictate(d)
assert dd.a == 'b'  # Access like an object
assert dd[1] == 2  # Access like a dict
# Updates affect d
dd.c = 'd'
assert d['c'] == 'd'
del dd.a
del dd[1]
# Inner dicts are mapped
dd.e = {}
dd.e.f = 'g'
assert dd['e'].f == 'g'
assert d == {'c': 'd', 'e': {'f': 'g'}}

If you want to access dict keys as an object (or as a dict for difficult keys), do it recursively, and also be able to update the original dict, you could do:

class Dictate(object):
    """Object view of a dict, updating the passed in dict when values are set
    or deleted. "Dictate" the contents of a dict...: """

    def __init__(self, d):
        # since __setattr__ is overridden, self.__dict = d doesn't work
        object.__setattr__(self, '_Dictate__dict', d)

    # Dictionary-like access / updates
    def __getitem__(self, name):
        value = self.__dict[name]
        if isinstance(value, dict):  # recursively view sub-dicts as objects
            value = Dictate(value)
        return value

    def __setitem__(self, name, value):
        self.__dict[name] = value
    def __delitem__(self, name):
        del self.__dict[name]

    # Object-like access / updates
    def __getattr__(self, name):
        return self[name]

    def __setattr__(self, name, value):
        self[name] = value
    def __delattr__(self, name):
        del self[name]

    def __repr__(self):
        return "%s(%r)" % (type(self).__name__, self.__dict)
    def __str__(self):
        return str(self.__dict)

Example usage:

d = {'a': 'b', 1: 2}
dd = Dictate(d)
assert dd.a == 'b'  # Access like an object
assert dd[1] == 2  # Access like a dict
# Updates affect d
dd.c = 'd'
assert d['c'] == 'd'
del dd.a
del dd[1]
# Inner dicts are mapped
dd.e = {}
dd.e.f = 'g'
assert dd['e'].f == 'g'
assert d == {'c': 'd', 'e': {'f': 'g'}}

回答 9

>>> def dict2obj(d):
        if isinstance(d, list):
            d = [dict2obj(x) for x in d]
        if not isinstance(d, dict):
            return d
        class C(object):
            pass
        o = C()
        for k in d:
            o.__dict__[k] = dict2obj(d[k])
        return o


>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'
>>> def dict2obj(d):
        if isinstance(d, list):
            d = [dict2obj(x) for x in d]
        if not isinstance(d, dict):
            return d
        class C(object):
            pass
        o = C()
        for k in d:
            o.__dict__[k] = dict2obj(d[k])
        return o


>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'

回答 10

我最终都尝试了AttrDictBunch库,发现它们对于我的使用而言太慢了。经过一个朋友和我的研究,我们发现编写这些库的主要方法导致该库通过嵌套对象积极地递归并在整个字典对象中进行复制。考虑到这一点,我们进行了两个关键更改。1)我们使属性延迟加载2)我们创建轻量级代理对象的副本,而不是创建字典对象的副本。这是最终的实现。使用此代码的性能提升令人难以置信。当使用AttrDict或Bunch时,仅这两个库分别消耗了我的请求时间的1/2和1/3(什么!?)。这段代码将时间减少到几乎没有(在0.5ms范围内)。当然这取决于您的需求,

class DictProxy(object):
    def __init__(self, obj):
        self.obj = obj

    def __getitem__(self, key):
        return wrap(self.obj[key])

    def __getattr__(self, key):
        try:
            return wrap(getattr(self.obj, key))
        except AttributeError:
            try:
                return self[key]
            except KeyError:
                raise AttributeError(key)

    # you probably also want to proxy important list properties along like
    # items(), iteritems() and __len__

class ListProxy(object):
    def __init__(self, obj):
        self.obj = obj

    def __getitem__(self, key):
        return wrap(self.obj[key])

    # you probably also want to proxy important list properties along like
    # __iter__ and __len__

def wrap(value):
    if isinstance(value, dict):
        return DictProxy(value)
    if isinstance(value, (tuple, list)):
        return ListProxy(value)
    return value

通过https://stackoverflow.com/users/704327/michael-merickel查看此处的原始实现。

还要注意的另一件事是,此实现非常简单,并没有实现您可能需要的所有方法。您需要根据需要在DictProxy或ListProxy对象上编写这些内容。

I ended up trying BOTH the AttrDict and the Bunch libraries and found them to be way too slow for my uses. After a friend and I looked into it, we found that the main method for writing these libraries results in the library aggressively recursing through a nested object and making copies of the dictionary object throughout. With this in mind, we made two key changes. 1) We made attributes lazy-loaded 2) instead of creating copies of a dictionary object, we create copies of a light-weight proxy object. This is the final implementation. The performance increase of using this code is incredible. When using AttrDict or Bunch, these two libraries alone consumed 1/2 and 1/3 respectively of my request time(what!?). This code reduced that time to almost nothing(somewhere in the range of 0.5ms). This of course depends on your needs, but if you are using this functionality quite a bit in your code, definitely go with something simple like this.

class DictProxy(object):
    def __init__(self, obj):
        self.obj = obj

    def __getitem__(self, key):
        return wrap(self.obj[key])

    def __getattr__(self, key):
        try:
            return wrap(getattr(self.obj, key))
        except AttributeError:
            try:
                return self[key]
            except KeyError:
                raise AttributeError(key)

    # you probably also want to proxy important list properties along like
    # items(), iteritems() and __len__

class ListProxy(object):
    def __init__(self, obj):
        self.obj = obj

    def __getitem__(self, key):
        return wrap(self.obj[key])

    # you probably also want to proxy important list properties along like
    # __iter__ and __len__

def wrap(value):
    if isinstance(value, dict):
        return DictProxy(value)
    if isinstance(value, (tuple, list)):
        return ListProxy(value)
    return value

See the original implementation here by https://stackoverflow.com/users/704327/michael-merickel.

The other thing to note, is that this implementation is pretty simple and doesn’t implement all of the methods you might need. You’ll need to write those as required on the DictProxy or ListProxy objects.


回答 11

x.__dict__.update(d) 应该做的很好。

x.__dict__.update(d) should do fine.


回答 12

您可以通过自定义对象挂钩来利用标准库的json模块

import json

class obj(object):
    def __init__(self, dict_):
        self.__dict__.update(dict_)

def dict2obj(d):
    return json.loads(json.dumps(d), object_hook=obj)

用法示例:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> o = dict2obj(d)
>>> o.a
1
>>> o.b.c
2
>>> o.d[0]
u'hi'
>>> o.d[1].foo
u'bar'

而且它不是严格的只读方式,例如namedtuple,您可以更改值,而不是结构:

>>> o.b.c = 3
>>> o.b.c
3

You can leverage the json module of the standard library with a custom object hook:

import json

class obj(object):
    def __init__(self, dict_):
        self.__dict__.update(dict_)

def dict2obj(d):
    return json.loads(json.dumps(d), object_hook=obj)

Example usage:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> o = dict2obj(d)
>>> o.a
1
>>> o.b.c
2
>>> o.d[0]
u'hi'
>>> o.d[1].foo
u'bar'

And it is not strictly read-only as it is with namedtuple, i.e. you can change values – not structure:

>>> o.b.c = 3
>>> o.b.c
3

回答 13

这应该使您开始:

class dict2obj(object):
    def __init__(self, d):
        self.__dict__['d'] = d

    def __getattr__(self, key):
        value = self.__dict__['d'][key]
        if type(value) == type({}):
            return dict2obj(value)

        return value

d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

x = dict2obj(d)
print x.a
print x.b.c
print x.d[1].foo

它不适用于列表。您必须将列表包装在UserList中,并重载__getitem__以包装字典。

This should get your started:

class dict2obj(object):
    def __init__(self, d):
        self.__dict__['d'] = d

    def __getattr__(self, key):
        value = self.__dict__['d'][key]
        if type(value) == type({}):
            return dict2obj(value)

        return value

d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

x = dict2obj(d)
print x.a
print x.b.c
print x.d[1].foo

It doesn’t work for lists, yet. You’ll have to wrap the lists in a UserList and overload __getitem__ to wrap dicts.


回答 14

我知道这里已经有很多答案了,我参加聚会很晚,但是这种方法将递归地将“字典”转换成类似对象的结构…在3.xx中有效

def dictToObject(d):
    for k,v in d.items():
        if isinstance(v, dict):
            d[k] = dictToObject(v)
    return namedtuple('object', d.keys())(*d.values())

# Dictionary created from JSON file
d = {
    'primaryKey': 'id', 
    'metadata': 
        {
            'rows': 0, 
            'lastID': 0
        }, 
    'columns': 
        {
            'col2': {
                'dataType': 'string', 
                'name': 'addressLine1'
            }, 
            'col1': {
                'datatype': 'string', 
                'name': 'postcode'
            }, 
            'col3': {
                'dataType': 'string', 
                'name': 'addressLine2'
            }, 
            'col0': {
                'datatype': 'integer', 
                'name': 'id'
            }, 
            'col4': {
                'dataType': 'string', 
                'name': 'contactNumber'
            }
        }, 
        'secondaryKeys': {}
}

d1 = dictToObject(d)
d1.columns.col1 # == object(datatype='string', name='postcode')
d1.metadata.rows # == 0

I know there’s already a lot of answers here already and I’m late to the party but this method will recursively and ‘in place’ convert a dictionary to an object-like structure… Works in 3.x.x

def dictToObject(d):
    for k,v in d.items():
        if isinstance(v, dict):
            d[k] = dictToObject(v)
    return namedtuple('object', d.keys())(*d.values())

# Dictionary created from JSON file
d = {
    'primaryKey': 'id', 
    'metadata': 
        {
            'rows': 0, 
            'lastID': 0
        }, 
    'columns': 
        {
            'col2': {
                'dataType': 'string', 
                'name': 'addressLine1'
            }, 
            'col1': {
                'datatype': 'string', 
                'name': 'postcode'
            }, 
            'col3': {
                'dataType': 'string', 
                'name': 'addressLine2'
            }, 
            'col0': {
                'datatype': 'integer', 
                'name': 'id'
            }, 
            'col4': {
                'dataType': 'string', 
                'name': 'contactNumber'
            }
        }, 
        'secondaryKeys': {}
}

d1 = dictToObject(d)
d1.columns.col1 # == object(datatype='string', name='postcode')
d1.metadata.rows # == 0

回答 15

from mock import Mock
d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
my_data = Mock(**d)

# We got
# my_data.a == 1
from mock import Mock
d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
my_data = Mock(**d)

# We got
# my_data.a == 1

回答 16

让我解释一下我的解决方案几乎使用前一段时间。但是首先,以下代码说明了我没有这样做的原因:

d = {'from': 1}
x = dict2obj(d)

print x.from

给出此错误:

  File "test.py", line 20
    print x.from == 1
                ^
SyntaxError: invalid syntax

由于“ from”是Python关键字,因此某些字典关键字是您不允许的。


现在,我的解决方案允许直接使用字典项的名称来访问字典项。但它也允许您使用“字典语义”。这是带有示例用法的代码:

class dict2obj(dict):
    def __init__(self, dict_):
        super(dict2obj, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj(it)
            elif isinstance(item, dict):
                self[key] = dict2obj(item)

    def __getattr__(self, key):
        return self[key]

d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

x = dict2obj(d)

assert x.a == x['a'] == 1
assert x.b.c == x['b']['c'] == 2
assert x.d[1].foo == x['d'][1]['foo'] == "bar"

Let me explain a solution I almost used some time ago. But first, the reason I did not is illustrated by the fact that the following code:

d = {'from': 1}
x = dict2obj(d)

print x.from

gives this error:

  File "test.py", line 20
    print x.from == 1
                ^
SyntaxError: invalid syntax

Because “from” is a Python keyword there are certain dictionary keys you cannot allow.


Now my solution allows access to the dictionary items by using their names directly. But it also allows you to use “dictionary semantics”. Here is the code with example usage:

class dict2obj(dict):
    def __init__(self, dict_):
        super(dict2obj, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj(it)
            elif isinstance(item, dict):
                self[key] = dict2obj(item)

    def __getattr__(self, key):
        return self[key]

d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

x = dict2obj(d)

assert x.a == x['a'] == 1
assert x.b.c == x['b']['c'] == 2
assert x.d[1].foo == x['d'][1]['foo'] == "bar"

回答 17

过去的问答,但我还有话要说。似乎没有人谈论递归字典。这是我的代码:

#!/usr/bin/env python

class Object( dict ):
    def __init__( self, data = None ):
        super( Object, self ).__init__()
        if data:
            self.__update( data, {} )

    def __update( self, data, did ):
        dataid = id(data)
        did[ dataid ] = self

        for k in data:
            dkid = id(data[k])
            if did.has_key(dkid):
                self[k] = did[dkid]
            elif isinstance( data[k], Object ):
                self[k] = data[k]
            elif isinstance( data[k], dict ):
                obj = Object()
                obj.__update( data[k], did )
                self[k] = obj
                obj = None
            else:
                self[k] = data[k]

    def __getattr__( self, key ):
        return self.get( key, None )

    def __setattr__( self, key, value ):
        if isinstance(value,dict):
            self[key] = Object( value )
        else:
            self[key] = value

    def update( self, *args ):
        for obj in args:
            for k in obj:
                if isinstance(obj[k],dict):
                    self[k] = Object( obj[k] )
                else:
                    self[k] = obj[k]
        return self

    def merge( self, *args ):
        for obj in args:
            for k in obj:
                if self.has_key(k):
                    if isinstance(self[k],list) and isinstance(obj[k],list):
                        self[k] += obj[k]
                    elif isinstance(self[k],list):
                        self[k].append( obj[k] )
                    elif isinstance(obj[k],list):
                        self[k] = [self[k]] + obj[k]
                    elif isinstance(self[k],Object) and isinstance(obj[k],Object):
                        self[k].merge( obj[k] )
                    elif isinstance(self[k],Object) and isinstance(obj[k],dict):
                        self[k].merge( obj[k] )
                    else:
                        self[k] = [ self[k], obj[k] ]
                else:
                    if isinstance(obj[k],dict):
                        self[k] = Object( obj[k] )
                    else:
                        self[k] = obj[k]
        return self

def test01():
    class UObject( Object ):
        pass
    obj = Object({1:2})
    d = {}
    d.update({
        "a": 1,
        "b": {
            "c": 2,
            "d": [ 3, 4, 5 ],
            "e": [ [6,7], (8,9) ],
            "self": d,
        },
        1: 10,
        "1": 11,
        "obj": obj,
    })
    x = UObject(d)


    assert x.a == x["a"] == 1
    assert x.b.c == x["b"]["c"] == 2
    assert x.b.d[0] == 3
    assert x.b.d[1] == 4
    assert x.b.e[0][0] == 6
    assert x.b.e[1][0] == 8
    assert x[1] == 10
    assert x["1"] == 11
    assert x[1] != x["1"]
    assert id(x) == id(x.b.self.b.self) == id(x.b.self)
    assert x.b.self.a == x.b.self.b.self.a == 1

    x.x = 12
    assert x.x == x["x"] == 12
    x.y = {"a":13,"b":[14,15]}
    assert x.y.a == 13
    assert x.y.b[0] == 14

def test02():
    x = Object({
        "a": {
            "b": 1,
            "c": [ 2, 3 ]
        },
        1: 6,
        2: [ 8, 9 ],
        3: 11,
    })
    y = Object({
        "a": {
            "b": 4,
            "c": [ 5 ]
        },
        1: 7,
        2: 10,
        3: [ 12 , 13 ],
    })
    z = {
        3: 14,
        2: 15,
        "a": {
            "b": 16,
            "c": 17,
        }
    }
    x.merge( y, z )
    assert 2 in x.a.c
    assert 3 in x.a.c
    assert 5 in x.a.c
    assert 1 in x.a.b
    assert 4 in x.a.b
    assert 8 in x[2]
    assert 9 in x[2]
    assert 10 in x[2]
    assert 11 in x[3]
    assert 12 in x[3]
    assert 13 in x[3]
    assert 14 in x[3]
    assert 15 in x[2]
    assert 16 in x.a.b
    assert 17 in x.a.c

if __name__ == '__main__':
    test01()
    test02()

Old Q&A, but I get something more to talk. Seems no one talk about recursive dict. This is my code:

#!/usr/bin/env python

class Object( dict ):
    def __init__( self, data = None ):
        super( Object, self ).__init__()
        if data:
            self.__update( data, {} )

    def __update( self, data, did ):
        dataid = id(data)
        did[ dataid ] = self

        for k in data:
            dkid = id(data[k])
            if did.has_key(dkid):
                self[k] = did[dkid]
            elif isinstance( data[k], Object ):
                self[k] = data[k]
            elif isinstance( data[k], dict ):
                obj = Object()
                obj.__update( data[k], did )
                self[k] = obj
                obj = None
            else:
                self[k] = data[k]

    def __getattr__( self, key ):
        return self.get( key, None )

    def __setattr__( self, key, value ):
        if isinstance(value,dict):
            self[key] = Object( value )
        else:
            self[key] = value

    def update( self, *args ):
        for obj in args:
            for k in obj:
                if isinstance(obj[k],dict):
                    self[k] = Object( obj[k] )
                else:
                    self[k] = obj[k]
        return self

    def merge( self, *args ):
        for obj in args:
            for k in obj:
                if self.has_key(k):
                    if isinstance(self[k],list) and isinstance(obj[k],list):
                        self[k] += obj[k]
                    elif isinstance(self[k],list):
                        self[k].append( obj[k] )
                    elif isinstance(obj[k],list):
                        self[k] = [self[k]] + obj[k]
                    elif isinstance(self[k],Object) and isinstance(obj[k],Object):
                        self[k].merge( obj[k] )
                    elif isinstance(self[k],Object) and isinstance(obj[k],dict):
                        self[k].merge( obj[k] )
                    else:
                        self[k] = [ self[k], obj[k] ]
                else:
                    if isinstance(obj[k],dict):
                        self[k] = Object( obj[k] )
                    else:
                        self[k] = obj[k]
        return self

def test01():
    class UObject( Object ):
        pass
    obj = Object({1:2})
    d = {}
    d.update({
        "a": 1,
        "b": {
            "c": 2,
            "d": [ 3, 4, 5 ],
            "e": [ [6,7], (8,9) ],
            "self": d,
        },
        1: 10,
        "1": 11,
        "obj": obj,
    })
    x = UObject(d)


    assert x.a == x["a"] == 1
    assert x.b.c == x["b"]["c"] == 2
    assert x.b.d[0] == 3
    assert x.b.d[1] == 4
    assert x.b.e[0][0] == 6
    assert x.b.e[1][0] == 8
    assert x[1] == 10
    assert x["1"] == 11
    assert x[1] != x["1"]
    assert id(x) == id(x.b.self.b.self) == id(x.b.self)
    assert x.b.self.a == x.b.self.b.self.a == 1

    x.x = 12
    assert x.x == x["x"] == 12
    x.y = {"a":13,"b":[14,15]}
    assert x.y.a == 13
    assert x.y.b[0] == 14

def test02():
    x = Object({
        "a": {
            "b": 1,
            "c": [ 2, 3 ]
        },
        1: 6,
        2: [ 8, 9 ],
        3: 11,
    })
    y = Object({
        "a": {
            "b": 4,
            "c": [ 5 ]
        },
        1: 7,
        2: 10,
        3: [ 12 , 13 ],
    })
    z = {
        3: 14,
        2: 15,
        "a": {
            "b": 16,
            "c": 17,
        }
    }
    x.merge( y, z )
    assert 2 in x.a.c
    assert 3 in x.a.c
    assert 5 in x.a.c
    assert 1 in x.a.b
    assert 4 in x.a.b
    assert 8 in x[2]
    assert 9 in x[2]
    assert 10 in x[2]
    assert 11 in x[3]
    assert 12 in x[3]
    assert 13 in x[3]
    assert 14 in x[3]
    assert 15 in x[2]
    assert 16 in x.a.b
    assert 17 in x.a.c

if __name__ == '__main__':
    test01()
    test02()

回答 18

想要上传我的这个小范例版本。

class Struct(dict):
  def __init__(self,data):
    for key, value in data.items():
      if isinstance(value, dict):
        setattr(self, key, Struct(value))
      else:   
        setattr(self, key, type(value).__init__(value))

      dict.__init__(self,data)

它保留导入到类中的类型的属性。我唯一关心的是从解析的字典中覆盖方法。但是否则看起来很稳固!

Wanted to upload my version of this little paradigm.

class Struct(dict):
  def __init__(self,data):
    for key, value in data.items():
      if isinstance(value, dict):
        setattr(self, key, Struct(value))
      else:   
        setattr(self, key, type(value).__init__(value))

      dict.__init__(self,data)

It preserves the attributes for the type that’s imported into the class. My only concern would be overwriting methods from within the dictionary your parsing. But otherwise seems solid!


回答 19

这也很好

class DObj(object):
    pass

dobj = Dobj()
dobj.__dict__ = {'a': 'aaa', 'b': 'bbb'}

print dobj.a
>>> aaa
print dobj.b
>>> bbb

This also works well too

class DObj(object):
    pass

dobj = Dobj()
dobj.__dict__ = {'a': 'aaa', 'b': 'bbb'}

print dobj.a
>>> aaa
print dobj.b
>>> bbb

回答 20

这是实现SilentGhost原始建议的另一种方法:

def dict2obj(d):
  if isinstance(d, dict):
    n = {}
    for item in d:
      if isinstance(d[item], dict):
        n[item] = dict2obj(d[item])
      elif isinstance(d[item], (list, tuple)):
        n[item] = [dict2obj(elem) for elem in d[item]]
      else:
        n[item] = d[item]
    return type('obj_from_dict', (object,), n)
  else:
    return d

Here is another way to implement SilentGhost’s original suggestion:

def dict2obj(d):
  if isinstance(d, dict):
    n = {}
    for item in d:
      if isinstance(d[item], dict):
        n[item] = dict2obj(d[item])
      elif isinstance(d[item], (list, tuple)):
        n[item] = [dict2obj(elem) for elem in d[item]]
      else:
        n[item] = d[item]
    return type('obj_from_dict', (object,), n)
  else:
    return d

回答 21

我偶然发现需要递归将字典列表转换为对象列表的情况,因此根据罗伯托的代码段,这里为我做了什么工作:

def dict2obj(d):
    if isinstance(d, dict):
        n = {}
        for item in d:
            if isinstance(d[item], dict):
                n[item] = dict2obj(d[item])
            elif isinstance(d[item], (list, tuple)):
                n[item] = [dict2obj(elem) for elem in d[item]]
            else:
                n[item] = d[item]
        return type('obj_from_dict', (object,), n)
    elif isinstance(d, (list, tuple,)):
        l = []
        for item in d:
            l.append(dict2obj(item))
        return l
    else:
        return d

注意,出于明显的原因,任何元组都将转换为其等效列表。

希望这对某人有帮助,就像您为我所做的所有回答一样。

I stumbled upon the case I needed to recursively convert a list of dicts to list of objects, so based on Roberto’s snippet here what did the work for me:

def dict2obj(d):
    if isinstance(d, dict):
        n = {}
        for item in d:
            if isinstance(d[item], dict):
                n[item] = dict2obj(d[item])
            elif isinstance(d[item], (list, tuple)):
                n[item] = [dict2obj(elem) for elem in d[item]]
            else:
                n[item] = d[item]
        return type('obj_from_dict', (object,), n)
    elif isinstance(d, (list, tuple,)):
        l = []
        for item in d:
            l.append(dict2obj(item))
        return l
    else:
        return d

Note that any tuple will be converted to its list equivalent, for obvious reasons.

Hope this helps someone as much as all your answers did for me, guys.


回答 22

仅将您dict的分配给__dict__空对象该怎么办?

class Object:
    """If your dict is "flat", this is a simple way to create an object from a dict

    >>> obj = Object()
    >>> obj.__dict__ = d
    >>> d.a
    1
    """
    pass

当然,这在您嵌套的dict示例中将失败,除非您递归遍历该dict:

# For a nested dict, you need to recursively update __dict__
def dict2obj(d):
    """Convert a dict to an object

    >>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
    >>> obj = dict2obj(d)
    >>> obj.b.c
    2
    >>> obj.d
    ["hi", {'foo': "bar"}]
    """
    try:
        d = dict(d)
    except (TypeError, ValueError):
        return d
    obj = Object()
    for k, v in d.iteritems():
        obj.__dict__[k] = dict2obj(v)
    return obj

您的示例list元素可能应该是Mapping,是(键,值)对的列表,如下所示:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': [("hi", {'foo': "bar"})]}
>>> obj = dict2obj(d)
>>> obj.d.hi.foo
"bar"

What about just assigning your dict to the __dict__ of an empty object?

class Object:
    """If your dict is "flat", this is a simple way to create an object from a dict

    >>> obj = Object()
    >>> obj.__dict__ = d
    >>> d.a
    1
    """
    pass

Of course this fails on your nested dict example unless you walk the dict recursively:

# For a nested dict, you need to recursively update __dict__
def dict2obj(d):
    """Convert a dict to an object

    >>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
    >>> obj = dict2obj(d)
    >>> obj.b.c
    2
    >>> obj.d
    ["hi", {'foo': "bar"}]
    """
    try:
        d = dict(d)
    except (TypeError, ValueError):
        return d
    obj = Object()
    for k, v in d.iteritems():
        obj.__dict__[k] = dict2obj(v)
    return obj

And your example list element was probably meant to be a Mapping, a list of (key, value) pairs like this:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': [("hi", {'foo': "bar"})]}
>>> obj = dict2obj(d)
>>> obj.d.hi.foo
"bar"

回答 23

这是另一个实现:

class DictObj(object):
    def __init__(self, d):
        self.__dict__ = d

def dict_to_obj(d):
    if isinstance(d, (list, tuple)): return map(dict_to_obj, d)
    elif not isinstance(d, dict): return d
    return DictObj(dict((k, dict_to_obj(v)) for (k,v) in d.iteritems()))

[编辑]关于还处理列表中的命令,而不仅仅是其他命令的遗漏之处。添加了修复程序。

Here’s another implementation:

class DictObj(object):
    def __init__(self, d):
        self.__dict__ = d

def dict_to_obj(d):
    if isinstance(d, (list, tuple)): return map(dict_to_obj, d)
    elif not isinstance(d, dict): return d
    return DictObj(dict((k, dict_to_obj(v)) for (k,v) in d.iteritems()))

[Edit] Missed bit about also handling dicts within lists, not just other dicts. Added fix.


回答 24

class Struct(dict):
    def __getattr__(self, name):
        try:
            return self[name]
        except KeyError:
            raise AttributeError(name)

    def __setattr__(self, name, value):
        self[name] = value

    def copy(self):
        return Struct(dict.copy(self))

用法:

points = Struct(x=1, y=2)
# Changing
points['x'] = 2
points.y = 1
# Accessing
points['x'], points.x, points.get('x') # 2 2 2
points['y'], points.y, points.get('y') # 1 1 1
# Accessing inexistent keys/attrs 
points['z'] # KeyError: z
points.z # AttributeError: z
# Copying
points_copy = points.copy()
points.x = 2
points_copy.x # 1
class Struct(dict):
    def __getattr__(self, name):
        try:
            return self[name]
        except KeyError:
            raise AttributeError(name)

    def __setattr__(self, name, value):
        self[name] = value

    def copy(self):
        return Struct(dict.copy(self))

Usage:

points = Struct(x=1, y=2)
# Changing
points['x'] = 2
points.y = 1
# Accessing
points['x'], points.x, points.get('x') # 2 2 2
points['y'], points.y, points.get('y') # 1 1 1
# Accessing inexistent keys/attrs 
points['z'] # KeyError: z
points.z # AttributeError: z
# Copying
points_copy = points.copy()
points.x = 2
points_copy.x # 1

回答 25

这个怎么样:

from functools import partial
d2o=partial(type, "d2o", ())

然后可以这样使用:

>>> o=d2o({"a" : 5, "b" : 3})
>>> print o.a
5
>>> print o.b
3

How about this:

from functools import partial
d2o=partial(type, "d2o", ())

This can then be used like this:

>>> o=d2o({"a" : 5, "b" : 3})
>>> print o.a
5
>>> print o.b
3

回答 26

我认为一个字典由数字,字符串和字典组成,大多数时候就足够了。因此,我忽略了元组,列表和其他类型未出现在字典最终维度中的情况。

考虑到继承,再结合递归,可以方便地解决打印问题,并提供两种查询数据的方式,一种编辑数据的方式。

请参阅下面的示例,该字典描述了有关学生的一些信息:

group=["class1","class2","class3","class4",]
rank=["rank1","rank2","rank3","rank4","rank5",]
data=["name","sex","height","weight","score"]

#build a dict based on the lists above
student_dic=dict([(g,dict([(r,dict([(d,'') for d in data])) for r in rank ]))for g in group])

#this is the solution
class dic2class(dict):
    def __init__(self, dic):
        for key,val in dic.items():
            self.__dict__[key]=self[key]=dic2class(val) if isinstance(val,dict) else val


student_class=dic2class(student_dic)

#one way to edit:
student_class.class1.rank1['sex']='male'
student_class.class1.rank1['name']='Nan Xiang'

#two ways to query:
print student_class.class1.rank1
print student_class.class1['rank1']
print '-'*50
for rank in student_class.class1:
    print getattr(student_class.class1,rank)

结果:

{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
--------------------------------------------------
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}

I think a dict consists of number, string and dict is enough most time. So I ignore the situation that tuples, lists and other types not appearing in the final dimension of a dict.

Considering inheritance, combined with recursion, it solves the print problem conveniently and also provides two ways to query a data,one way to edit a data.

See the example below, a dict that describes some information about students:

group=["class1","class2","class3","class4",]
rank=["rank1","rank2","rank3","rank4","rank5",]
data=["name","sex","height","weight","score"]

#build a dict based on the lists above
student_dic=dict([(g,dict([(r,dict([(d,'') for d in data])) for r in rank ]))for g in group])

#this is the solution
class dic2class(dict):
    def __init__(self, dic):
        for key,val in dic.items():
            self.__dict__[key]=self[key]=dic2class(val) if isinstance(val,dict) else val


student_class=dic2class(student_dic)

#one way to edit:
student_class.class1.rank1['sex']='male'
student_class.class1.rank1['name']='Nan Xiang'

#two ways to query:
print student_class.class1.rank1
print student_class.class1['rank1']
print '-'*50
for rank in student_class.class1:
    print getattr(student_class.class1,rank)

Results:

{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
--------------------------------------------------
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}

回答 27

通常,您想将dict层次结构镜像到您的对象中,而不是通常位于最低级别的列表或元组。所以这就是我这样做的方式:

class defDictToObject(object):

    def __init__(self, myDict):
        for key, value in myDict.items():
            if type(value) == dict:
                setattr(self, key, defDictToObject(value))
            else:
                setattr(self, key, value)

因此,我们这样做:

myDict = { 'a': 1,
           'b': { 
              'b1': {'x': 1,
                    'y': 2} },
           'c': ['hi', 'bar'] 
         }

并获得:

x.b.b1.x 1个

x.c [‘hi’,’bar’]

Typically you want to mirror dict hierarchy into your object but not list or tuples which are typically at lowest level. So this is how I did this:

class defDictToObject(object):

    def __init__(self, myDict):
        for key, value in myDict.items():
            if type(value) == dict:
                setattr(self, key, defDictToObject(value))
            else:
                setattr(self, key, value)

So we do:

myDict = { 'a': 1,
           'b': { 
              'b1': {'x': 1,
                    'y': 2} },
           'c': ['hi', 'bar'] 
         }

and get:

x.b.b1.x 1

x.c [‘hi’, ‘bar’]


回答 28

我的字典是这样的格式:

addr_bk = {
    'person': [
        {'name': 'Andrew', 'id': 123, 'email': 'andrew@mailserver.com',
         'phone': [{'type': 2, 'number': '633311122'},
                   {'type': 0, 'number': '97788665'}]
        },
        {'name': 'Tom', 'id': 456,
         'phone': [{'type': 0, 'number': '91122334'}]}, 
        {'name': 'Jack', 'id': 7788, 'email': 'jack@gmail.com'}
    ]
}

可以看出,我有嵌套的字典字典列表。这是因为addr_bk是从使用lwpb.codec转换为python dict的协议缓冲区数据中解码的。有可选字段(例如,电子邮件=>,其中的密钥可能不可用)和重复字段(例如,电话=>转换为词典列表)。

我尝试了以上所有建议的解决方案。有些不能很好地处理嵌套字典。其他人无法轻松打印对象详细信息。

只有Dawie Strauss的解决方案dict2obj(dict)最有效。

当找不到密钥时,我对它进行了一些处理:

# Work the best, with nested dictionaries & lists! :)
# Able to print out all items.
class dict2obj_new(dict):
    def __init__(self, dict_):
        super(dict2obj_new, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj_new(it)
            elif isinstance(item, dict):
                self[key] = dict2obj_new(item)

    def __getattr__(self, key):
        # Enhanced to handle key not found.
        if self.has_key(key):
            return self[key]
        else:
            return None

然后,我用以下方法进行了测试:

# Testing...
ab = dict2obj_new(addr_bk)

for person in ab.person:
  print "Person ID:", person.id
  print "  Name:", person.name
  # Check if optional field is available before printing.
  if person.email:
    print "  E-mail address:", person.email

  # Check if optional field is available before printing.
  if person.phone:
    for phone_number in person.phone:
      if phone_number.type == codec.enums.PhoneType.MOBILE:
        print "  Mobile phone #:",
      elif phone_number.type == codec.enums.PhoneType.HOME:
        print "  Home phone #:",
      else:
        print "  Work phone #:",
      print phone_number.number

My dictionary is of this format:

addr_bk = {
    'person': [
        {'name': 'Andrew', 'id': 123, 'email': 'andrew@mailserver.com',
         'phone': [{'type': 2, 'number': '633311122'},
                   {'type': 0, 'number': '97788665'}]
        },
        {'name': 'Tom', 'id': 456,
         'phone': [{'type': 0, 'number': '91122334'}]}, 
        {'name': 'Jack', 'id': 7788, 'email': 'jack@gmail.com'}
    ]
}

As can be seen, I have nested dictionaries and list of dicts. This is because the addr_bk was decoded from protocol buffer data that converted to a python dict using lwpb.codec. There are optional field (e.g. email => where key may be unavailable) and repeated field (e.g. phone => converted to list of dict).

I tried all the above proposed solutions. Some doesn’t handle the nested dictionaries well. Others cannot print the object details easily.

Only the solution, dict2obj(dict) by Dawie Strauss, works best.

I have enhanced it a little to handle when the key cannot be found:

# Work the best, with nested dictionaries & lists! :)
# Able to print out all items.
class dict2obj_new(dict):
    def __init__(self, dict_):
        super(dict2obj_new, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj_new(it)
            elif isinstance(item, dict):
                self[key] = dict2obj_new(item)

    def __getattr__(self, key):
        # Enhanced to handle key not found.
        if self.has_key(key):
            return self[key]
        else:
            return None

Then, I tested it with:

# Testing...
ab = dict2obj_new(addr_bk)

for person in ab.person:
  print "Person ID:", person.id
  print "  Name:", person.name
  # Check if optional field is available before printing.
  if person.email:
    print "  E-mail address:", person.email

  # Check if optional field is available before printing.
  if person.phone:
    for phone_number in person.phone:
      if phone_number.type == codec.enums.PhoneType.MOBILE:
        print "  Mobile phone #:",
      elif phone_number.type == codec.enums.PhoneType.HOME:
        print "  Home phone #:",
      else:
        print "  Work phone #:",
      print phone_number.number

回答 29

建立我对“ python:如何动态地向类添加属性? ”的答案:

class data(object):
    def __init__(self,*args,**argd):
        self.__dict__.update(dict(*args,**argd))

def makedata(d):
    d2 = {}
    for n in d:
        d2[n] = trydata(d[n])
    return data(d2)

def trydata(o):
    if isinstance(o,dict):
        return makedata(o)
    elif isinstance(o,list):
        return [trydata(i) for i in o]
    else:
        return o

您调用makedata要转换的字典,或者trydata取决于您期望输入的内容,它会吐出一个数据对象。

笔记:

  • trydata如果需要更多功能,可以添加Elif 。
  • 显然,如果您想要x.a = {}或类似的方法将无法使用。
  • 如果需要只读版本,请使用原始答案中的类数据。

Building off my answer to “python: How to add property to a class dynamically?“:

class data(object):
    def __init__(self,*args,**argd):
        self.__dict__.update(dict(*args,**argd))

def makedata(d):
    d2 = {}
    for n in d:
        d2[n] = trydata(d[n])
    return data(d2)

def trydata(o):
    if isinstance(o,dict):
        return makedata(o)
    elif isinstance(o,list):
        return [trydata(i) for i in o]
    else:
        return o

You call makedata on the dictionary you want converted, or maybe trydata depending on what you expect as input, and it spits out a data object.

Notes:

  • You can add elifs to trydata if you need more functionality.
  • Obviously this won’t work if you want x.a = {} or similar.
  • If you want a readonly version, use the class data from the original answer.