How does one ignore unexpected keyword arguments passed to a function?

Question

Suppose I have some function, f:

def f (a=None):     print a 

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

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' 

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.

Answer 1

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 

This would necessarily suffice the case of

1. to just ignore any keys that are not parameter names
2. However, it lacks the default values of parameters, which is a nice feature that it would be nice to keep