Using an existing function, how to define a new function with different default optional arguments?

Question

Given a function fxn1()

def fxn1(a,b=1):
    return a*b

I would like to do something like:

fxn2=fxn1(.,b=2)

In words, I would like to make a new function fxn2() which is identical to fxn1() but with different default values of optional arguments. I have not been able to find an answer/direction on web.

You may want to read about [`functools.partial`](https://docs.python.org/3/library/functools.html#functools.partial) — DeepSpace, May 27 '19 at 19:55

score 2 · Answer 1 · answered May 27 '19 at 19:56

2

You can use partial function:

from functools import partial
fxn2 = partial(fxn1, b=2)

fxn2(1)

answered May 27 '19 at 19:56

Nikita Ryanov

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Andrej Kesely · Accepted Answer · 2019-05-27T20:03:36.603

2

You can use functools.partial (official doc):

from functools import partial

def fxn1(a,b=1):
    return a*b

fxn2 = partial(fxn1, b=2)

print(fxn2(4))

Prints:

edited May 27 '19 at 20:03

answered May 27 '19 at 19:56

Andrej Kesely

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score 1 · Answer 3 · answered May 27 '19 at 19:52

1

def fxn1(a,b=1):
    return a*b

def fxn2(a,b=2):
    return fxn1(a,b)

answered May 27 '19 at 19:52

Victor Hugo Borges

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score 0 · Answer 4 · edited May 27 '19 at 19:54

0

The easiest way will be

def fxn2(a, b=2):
    return fxn1(a, b)

edited May 27 '19 at 19:54

Arnav Poddar

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answered May 27 '19 at 19:52

JohnO

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It should be `return fxn1(a, b)` – Tianbo Ji May 27 '19 at 19:52
You are of course right, thanks for catching it :-) – JohnO May 27 '19 at 19:55
Actually, the easiest way will be [`functools.partial`](https://docs.python.org/3/library/functools.html#functools.partial) – DeepSpace May 27 '19 at 19:56
Partial is arguably more *elegant*. I figured that for a beginner, the other option would be easier. – JohnO May 27 '19 at 20:07

Using an existing function, how to define a new function with different default optional arguments?

4 Answers4