arguments in functions python

Question

This is a subtle question about notation. I want to call a function with specific arguments, but without having to redefine it.

For example, min() with a key function on the second argument key = itemgetter(1) would look like:

min_arg2 = lambda p,q = min(p,q, key = itemgetter(1))

I'm hoping to just call it as something like min( *itemgetter(1) )...

Does anyone know how to do this? Thank you.

"but without having to redefine it." How can this possibly be avoided? What are you trying to do? If min() doesn't mean __builtins__.min() people will hate you. — S.Lott
– S.Lott, Commented Sep 23, 2011 at 17:34

Duncan · Accepted Answer · 2011-09-23 17:20:55Z

10

You want to use functools.partial():

min_arg2 = functools.partial(min, key=itemgetter(1))

Example:

>>> import functools
>>> from operator import itemgetter
>>> min_arg2 = functools.partial(min, key=itemgetter(1))
>>> min_arg2(vals)
('b', 0)

answered Sep 23, 2011 at 17:20

Duncan

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1 Comment

Where do you define vals?

Community · Accepted Answer · 2017-05-23 12:20:34Z

2

Using functools (as in Duncan's answer) is a better approach, however you can use a lambda expression, you just didn't get the syntax correct:

min_arg2 = lambda p,q: min(p,q, key=itemgetter(1))

CommunityBot

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answered Sep 23, 2011 at 17:34

Zach Kelling

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thanks, that mistake wasa typo, although functools.partial was what I wanted.