在 Python 中使用 **kwargs 的正确方法

当涉及到默认值时,在 Python 中使用 **kwargs 的正确方法是什么?

kwargs 返回一个字典,但是设置默认值的最佳方法是什么,或者有没有?我应该将它作为字典访问吗?使用get函数?

class ExampleClass:
    def __init__(self, **kwargs):
        self.val = kwargs['val']
        self.val2 = kwargs.get('val2')

一个简单的问题,但我找不到好的资源。人们在我见过的代码中以不同的方式进行操作,很难知道该使用什么。

stack overflow Proper way to use **kwargs in Python
原文答案
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接受的答案

You can pass a default value to get() for keys that are not in the dictionary:

self.val2 = kwargs.get('val2',"default value")

However, if you plan on using a particular argument with a particular default value, why not use named arguments in the first place?

def __init__(self, val2="default value", **kwargs):

答案:

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While most answers are saying that, e.g.,

def f(**kwargs):
    foo = kwargs.pop('foo')
    bar = kwargs.pop('bar')
    ...etc...

is "the same as"

def f(foo=None, bar=None, **kwargs):
    ...etc...

this is not true. In the latter case, f can be called as f(23, 42), while the former case accepts named arguments only -- no positional calls. Often you want to allow the caller maximum flexibility and therefore the second form, as most answers assert, is preferable: but that is not always the case. When you accept many optional parameters of which typically only a few are passed, it may be an excellent idea (avoiding accidents and unreadable code at your call sites!) to force the use of named arguments -- threading.Thread is an example. The first form is how you implement that in Python 2.

The idiom is so important that in Python 3 it now has special supporting syntax: every argument after a single * in the def signature is keyword-only, that is, cannot be passed as a positional argument, but only as a named one. So in Python 3 you could code the above as:

def f(*, foo=None, bar=None, **kwargs):
    ...etc...

Indeed, in Python 3 you can even have keyword-only arguments that aren't optional (ones without a default value).

However, Python 2 still has long years of productive life ahead, so it's better to not forget the techniques and idioms that let you implement in Python 2 important design ideas that are directly supported in the language in Python 3!

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I suggest something like this

def testFunc( **kwargs ):
    options = {
            'option1' : 'default_value1',
            'option2' : 'default_value2',
            'option3' : 'default_value3', }

    options.update(kwargs)
    print options

testFunc( option1='new_value1', option3='new_value3' )
# {'option2': 'default_value2', 'option3': 'new_value3', 'option1': 'new_value1'}

testFunc( option2='new_value2' )
# {'option1': 'default_value1', 'option3': 'default_value3', 'option2': 'new_value2'}

And then use the values any way you want

dictionaryA.update(dictionaryB) adds the contents of dictionaryB to dictionaryA overwriting any duplicate keys.

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You'd do

self.attribute = kwargs.pop('name', default_value)

or

self.attribute = kwargs.get('name', default_value)

If you use pop, then you can check if there are any spurious values sent, and take the appropriate action (if any).

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Using **kwargs and default values is easy. Sometimes, however, you shouldn't be using **kwargs in the first place.

In this case, we're not really making best use of **kwargs.

class ExampleClass( object ):
    def __init__(self, **kwargs):
        self.val = kwargs.get('val',"default1")
        self.val2 = kwargs.get('val2',"default2")

The above is a "why bother?" declaration. It is the same as

class ExampleClass( object ):
    def __init__(self, val="default1", val2="default2"):
        self.val = val
        self.val2 = val2

When you're using **kwargs, you mean that a keyword is not just optional, but conditional. There are more complex rules than simple default values.

When you're using **kwargs, you usually mean something more like the following, where simple defaults don't apply.

class ExampleClass( object ):
    def __init__(self, **kwargs):
        self.val = "default1"
        self.val2 = "default2"
        if "val" in kwargs:
            self.val = kwargs["val"]
            self.val2 = 2*self.val
        elif "val2" in kwargs:
            self.val2 = kwargs["val2"]
            self.val = self.val2 / 2
        else:
            raise TypeError( "must provide val= or val2= parameter values" )
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Since **kwargs is used when the number of arguments is unknown, why not doing this?

class Exampleclass(object):
  def __init__(self, **kwargs):
    for k in kwargs.keys():
       if k in [acceptable_keys_list]:
          self.__setattr__(k, kwargs[k])
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I think the proper way to use **kwargs in Python when it comes to default values is to use the dictionary method setdefault, as given below:

class ExampleClass:
    def __init__(self, **kwargs):
        kwargs.setdefault('val', value1)
        kwargs.setdefault('val2', value2)

In this way, if a user passes 'val' or 'val2' in the keyword args, they will be used; otherwise, the default values that have been set will be used.

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Here's another approach:

def my_func(arg1, arg2, arg3):
    ... so something ...

kwargs = {'arg1': 'Value One', 'arg2': 'Value Two', 'arg3': 'Value Three'}
# Now you can call the function with kwargs like this:

my_func(**kwargs)
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Following up on @srhegde suggestion of using setattr:

class ExampleClass(object):
    __acceptable_keys_list = ['foo', 'bar']

    def __init__(self, **kwargs):
        [self.__setattr__(key, kwargs.get(key)) for key in self.__acceptable_keys_list]

This variant is useful when the class is expected to have all of the items in our acceptable list.

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You could do something like this

class ExampleClass:
    def __init__(self, **kwargs):
        arguments = {'val':1, 'val2':2}
        arguments.update(kwargs)
        self.val = arguments['val']
        self.val2 = arguments['val2']