1

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.

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  • 1
    You may want to read about functools.partial Commented May 27, 2019 at 19:55
  • It's called early binding. Use partial from functools. Commented May 27, 2019 at 19:55

4 Answers 4

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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:

8
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2

You can use partial function:

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

fxn2(1)
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1 
def fxn1(a,b=1):
    return a*b

def fxn2(a,b=2):
    return fxn1(a,b)
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0

The easiest way will be

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

It should be return fxn1(a, b)
You are of course right, thanks for catching it :-)
Actually, the easiest way will be functools.partial
Partial is arguably more elegant. I figured that for a beginner, the other option would be easier.

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