Python：kwargs.pop() 和 kwargs.get() 之间的区别

Python: Difference between kwargs.pop() and kwargs.get()

提问by Aliquis

I have seen both ways but I do not understand what the difference is and what I should use as "best practice":

我已经看到了两种方式，但我不明白有什么区别以及我应该将什么用作“最佳实践”：

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

def custom_function2(**kwargs):
    foo = kwargs.get('foo')
    bar = kwargs.get('bar')
    ...

回答by DhiaTN

get(key[, default]): return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError.

get(key[, default])：如果键在字典中，则返回键的值，否则返回默认值。如果未给出默认值，则默认为 None，因此此方法永远不会引发KeyError.

d = {'a' :1, 'c' :2}
print(d.get('b', 0)) # return 0
print(d.get('c', 0)) # return 2

pop(key[, default])if key is in the dictionary, remove it and return its value, else return default. If default is not given and key is not in the dictionary, a KeyErroris raised.

pop(key[, default])如果键在字典中，删除它并返回它的值，否则返回默认值。如果未给出默认值并且键不在字典中，KeyError则引发 a。

d = {'a' :1, 'c' :2}
print(d.pop('c', 0)) # return 2
print(d) # returns {'a': 1}
print(d.get('c', 0)) # return 0

NB:Regarding best practice question, I would say it depends on your use case but I would go by default for .getunless I have a real need to .pop

注意：关于最佳实践问题，我会说这取决于您的用例，但我会默认为.get除非我真的需要.pop

回答by shx2

The difference is popalso removes the item from the dict.

不同之处pop还在于从字典中删除了该项目。

There is no best practice. Use the one which is more convenient for your particular use case.

没有最佳实践。使用对您的特定用例更方便的一种。

Most times, all you need is getting the value.

大多数时候，您所需要的只是get调整价值。

Other times, you want to make sure no extra/unexpected kwargs are provided. In this case, it is convenient to use pop. E.g.:

其他时候，你想确保没有提供额外/意外的 kwargs。在这种情况下，使用pop. 例如：

a = kw.pop('a')
b = kw.pop('b')
if kw:
    raise TypeError('Unepxected kwargs provided: %s' % list(kw.keys()))

回答by ssoler

Consider the next example, where the use of getor popmakes a difference:

考虑下一个示例，其中使用getorpop有所不同：

Let's begin with get:

让我们开始get：

class Foo(object):
    def __init__(self, foo_param=None):
        print("In Foo: {}".format(foo_param))

class Bar(Foo):
    def __init__(self, **kwargs):
        bar_param = kwargs.get('bar_param')
        print("In Bar: {}".format(bar_param))
        super(Bar, self).__init__(**kwargs)

bar = Bar(foo_param='F', bar_param='B')

This code snippet raises TypeErrorexception:

此代码片段引发TypeError异常：

TypeError: __init__() got an unexpected keyword argument 'bar_param'

When Bar executes super(Bar, self).__init__(**kwargs)it is forwarding to Foo the same dict he has recived: {foo_param='F', bar_param='B'}. Then Foo raises TypeErrorbecause input paramteres doesn't respect its interface.

当 Bar 执行时，super(Bar, self).__init__(**kwargs)它会向 Foo 转发他收到的同一个 dict: {foo_param='F', bar_param='B'}。然后 Foo 引发，TypeError因为输入参数不尊重其接口。

If you popbar_parambefore executing the call to super, Foo only recives its required input parameter foo_param, and all goes fine.

如果您popbar_param在执行对 的调用之前super， Foo 只接收其所需的输入参数foo_param，并且一切正常。

class Foo(object):
    def __init__(self, foo_param=None):
        print("In Foo: {}".format(foo_param))

class Bar(Foo):
    def __init__(self, **kwargs):
        bar_param = kwargs.pop('bar_param')
        print("In Bar: {}".format(bar_param))
        super(Bar, self).__init__(**kwargs)

bar = Bar(foo_param='F', bar_param='B')

Output is:

输出是：

In Bar: B
In Foo: F

回答by hwhite4

So the get and pop functions do very different things

所以 get 和 pop 函数做了非常不同的事情

get is used to return a value for a given key in the dictionary

get 用于返回字典中给定键的值

pop removes the value from the dictionary and returns the removed value

pop 从字典中移除值并返回移除的值

All of the dictionary functions are documented here (for python3): https://docs.python.org/3/library/stdtypes.html#mapping-types-dict

此处记录了所有字典函数（对于 python3）：https: //docs.python.org/3/library/stdtypes.html#mapping-types-dict