How to make python built-in functions support keyword arguments?

Question

I want to use partial() to build a function that only takes one argument,
so that I can pass it to some high-order functions (ex: map()/filter()):

>>> from operator import sub
>>> from functools import partial

>>> map(lambda x:sub(x, 5), [1,2,3])
[-4, -3, -2]

>>> help(sub)
Help on built-in function sub in module operator:

sub(...)
    sub(a, b) -- Same as a - b.

>>> map(partial(sub, b=5), [1,2,3])
TypeError: sub() takes no keyword arguments

Is there some way to make sub()(or any other built-in functions) support keyword arguments?

Why do you need to use a keyword argument as opposed to a static argument. Why not 5 instead of b=5? — CorbinMc, May 30 '14 at 08:07
@CorbinMc: because then you cannot use a `functools.partial()` wrapper, that's why. — Martijn Pieters, May 31 '14 at 10:15

score 2 · Accepted Answer · answered May 30 '14 at 08:04

2

If a function defined in C does not take keyword arguments then there is no way to force it to do so. Either use lamdba and fill the arguments in the hard way, or wrap the function in a Python function that can take keyword arguments.

answered May 30 '14 at 08:04

Ignacio Vazquez-Abrams

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How to make python built-in functions support keyword arguments?

1 Answers1