Ignore additional keyword arguments in python

Question

Imagine I have a function like

def foo(x):
    ...

When I call it with the dictionary { 'x': 42, 'y': 23 } as keyword arguments I get an TypeError:

>>> foo(**{ 'x': 42, 'y': 23 })
...
TypeError: foo() got an unexpected keyword argument 'y'

Is there a good way to make a function call with keyword arguments where additional keyword arguments are just ignored?

My solution so far: I can define a helper function:

import inspect

def call_with_kwargs(func, kwargs):
    params = inspect.getargspec(func).args

    return func(**{ k: v for k,v in kwargs.items() if k in params})

Now I can do

>>> call_with_kwargs(foo, { 'x': 42, 'y': 23 })
42

Is there a better way?

So altering `func1()` to use a catch-all is not an option, it has to work for *all* functions? e.g. `def func1(x, **kwargs):`? — Martijn Pieters, Apr 16 '15 at 16:31
not a solution but you can make a decorator out of your `call_with_kwargs` function — Julien Spronck, Apr 16 '15 at 16:32
@jonrsharpe Yes, this would work fine in my case. Thanks a lot. If you want, you can add an answer to this question (so that I can accept it...) — Stephan Kulla, Apr 16 '15 at 16:36

score 5 · Accepted Answer · answered Apr 16 '15 at 16:37

5

If altering your functions is fine, then just add a catch-all **kw argument to it:

def foo(x, **kw):
    # ...

and ignore whatever kw captured in the function.

answered Apr 16 '15 at 16:37

Martijn Pieters

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Ignore additional keyword arguments in python

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