Changing keyword defaults of a function in Python

Question

Let's say I have a function like this:

def func_1(a, b=1, c=None):
     code here

Now I want to make another function that has the same internals but different keyword arguments.

def func_2(a, b=2, c='asdf'):
    code here

One option would be a closure like this:

 def make_func(b, c):
     def func(a, b=b, c=c):
         code here
     return func

 func_1 = make_func(1, None)
 func_2 = make_func(2, 'asdf')

Is there a more concise/Pythonic way to go about this?

Why use a closure? Just call one from the other: def func_2(a, b=2, c='asdf'): func_1(a, b, c) — jonrsharpe
– jonrsharpe, Commented Mar 15, 2016 at 22:43
@jonrsharpe: the closure acts as a factory to produce more such functions. I had made the same comment, then deleted it again once I realised that. — Martijn Pieters
– Martijn Pieters, Commented Mar 15, 2016 at 22:45
@MartijnPieters true, but generating the functions dynamically seems unnecessary for a small number (not to mention likely makes life harder for IDEs). — jonrsharpe
– jonrsharpe, Commented Mar 15, 2016 at 22:47
Or does the OP actually mean changing, i.e. altering the defaults in an existing function object, which isn't what the example does. — jonrsharpe
– jonrsharpe, Commented Mar 15, 2016 at 22:48
@jonrsharpe I meant making another function with same behavior, different keywords - sorry for ambiguous wording. — Andrew
– Andrew, Commented Mar 16, 2016 at 0:04

Martijn Pieters · Accepted Answer · 2016-03-15 22:43:32Z

3

Use a functools.partial() object:

from functools import partial

func_2 = partial(func_1, b=2, c='asdf')

The defaults supplied to the partial can be overridden again when calling it:

>>> from functools import partial
>>> def func_1(a, b=1, c=None):
...     return a, b, c
...
>>> func_2 = partial(func_1, b=2, c='asdf')
>>> func_2(42)
(42, 2, 'asdf')
>>> func_2(42, b=5, c='spam')
(42, 5, 'spam')

answered Mar 15, 2016 at 22:43

Martijn Pieters

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