Creating Python function with partial parameters

Question

I want to pass a Python function to another function with some of its parameters "filled out" ahead of time.

This is simplification what I am doing:

def add(x, y):
    return x + y

def increment_factory(i):  # create a function that increments by i
    return (lambda y: add(i, y))

inc2 = increment_factory(2)

print inc2(3) # prints 5

I don't want to use some sort of passing of args and later exploding it with *args because the function I am passing inc2 into doesn't know to pass args to it.

This feels a bit too clever for a group project... is there a more straightforward or pythonic way to do this?

Thanks!

David · Accepted Answer · 2010-07-15 18:41:57Z

37

This is called currying, or partial application. You can use the built-in functools.partial(). Something like the following would do what you want.

import functools
def add(x,y):
    return x + y

inc2 = functools.partial(add, 2)
print inc2(3)

answered Jul 15, 2010 at 18:41

David

7,1931 gold badge44 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Donald Miner Over a year ago

Thanks! One of my favorite features of Haskell is the built-in currying. I should have just used the term :).

João Bravo · Accepted Answer · 2022-04-06 16:14:00Z

4

You could also accomplish the same with a lambda function:

inc2 = lambda y: add(2, y)

print inc2(3)

answered Apr 6, 2022 at 16:14

João Bravo

2676 silver badges19 bronze badges

1 Comment

Jacob Waters Over a year ago

This is a way better answer. Accomplishes currying with simpler code and no import.

Collectives™ on Stack Overflow

Creating Python function with partial parameters

2 Answers 2

1 Comment

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related