Function object in Python

Question

I want to create a list of functions consisting of function objects created by a fixed base function but with different input parameters. I have written such code:

def f_base(name):
    print "I am %s" % name

f_list = []
for name in ['A', 'B']:
    f_list.append(
        lambda : f_base(name)
    )

f_list[0]()
f_list[1]()

The result is:

I am B
I am B

But I want to get:

I am A
I am B

Can someone explain why Python produces such output and what is the easiest way to get the result I need?

Jared · Accepted Answer · 2015-11-15 17:04:24Z

4

All you're doing with the lambda functions now is calling f_base on the variable name in the body. The name variable is determined at the time you call those lambda functions and not at the time you create each function.

The variable name refers to 'B' after you finish looping since python keeps variables defined in for loops around after the loop is done.

This is particularly obvious if you do,

name = 'C'

after the loop and before calling the functions. Then you get,

I am C
I am C

To do what you want you need to create a closure function or, probably better, use functools.partial.

import functools

def f_base(name):
    print "I am %s" % name

f_list = []
for name in ['A', 'B']:
    f_list.append(
        functools.partial(f_base, name)
    )

f_list[0]()
f_list[1]()

Which produces,

I am A
I am B

answered Nov 15, 2015 at 17:04

Jared

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2 Comments

bereal Over a year ago

Also known as infamous 'late binding gotcha'

Fomalhaut Over a year ago

Great! Thanks for the answer!

hjpotter92 · Accepted Answer · 2015-11-15 16:59:42Z

1

You can create closure:

def f_base(name):
    def closure():
        print "I am %s" % name
    return closure

f_list = []
for name in ['A', 'B']:
    f_list.append( f_base(name) )

answered Nov 15, 2015 at 16:59

hjpotter92

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