如何忽略传递给函数的意外关键字参数? [英] How does one ignore unexpected keyword arguments passed to a function?

问题描述

假设我有一些功能，f:

def f (a=None):
    print a

现在，如果我有像dct = {"a":"Foo"}这样的字典，我可以调用f(**dct)并得到打印结果Foo.

Now, if I have a dictionary such as dct = {"a":"Foo"}, I may call f(**dct) and get the result Foo printed.

但是，假设我有一本字典dct2 = {"a":"Foo", "b":"Bar"}.如果我拨打f(**dct2)我会得到

However, suppose I have a dictionary dct2 = {"a":"Foo", "b":"Bar"}. If I call f(**dct2) I get a

TypeError: f() got an unexpected keyword argument 'b'

足够公平.但是，无论如何，在f的定义中或在调用它时，是否要告诉Python仅忽略不是参数名称的任何键?最好是一种可以指定默认值的方法.

Fair enough. However, is there anyway to, in the definition of f or in the calling of it, tell Python to just ignore any keys that are not parameter names? Preferable a method that allows defaults to be specified.

推荐答案

作为@Bas发布答案的扩展，我建议将kwargs参数(可变长度关键字参数)添加为函数的第二个参数

As an extension to the answer posted by @Bas, I would suggest to add the kwargs arguments (variable length keyword arguments) as the second parameter to the function

>>> def f (a=None, **kwargs):
    print a


>>> dct2 = {"a":"Foo", "b":"Bar"}
>>> f(**dct2)
Foo

这足以满足

1. 只忽略不是参数名称的任何键
2. 但是，它缺少参数的默认值，这是一个很好的功能，可以很好地保留

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