Python create function in a loop capturing the loop variable

Question

What's going on here? I'm trying to create a list of functions:

def f(a,b):
    return a*b

funcs = []

for i in range(0,10):
    funcs.append(lambda x:f(i,x))

This isn't doing what I expect. I would expect the list to act like this:

funcs[3](3) = 9
funcs[0](5) = 0

But all the functions in the list seem to be identical, and be setting the fixed value to be 9:

funcs[3](3) = 27
funcs[3](1) = 9

funcs[2](6) = 54

Any ideas?

Answer 1

lambdas in python are closures.... the arguments you give it aren't going to be evaluated until the lambda is evaluated. At that time, i=9 regardless, because your iteration is finished.

The behavior you're looking for can be achieved with functools.partial

import functools

def f(a,b):
    return a*b

funcs = []

for i in range(0,10):
    funcs.append(functools.partial(f,i))

Answer 2

Yep, the usual "scoping problem" (actually a binding-later-than-you want problem, but it's often called by that name). You've already gotten the two best (because simplest) answers -- the "fake default" i=i solution, and functools.partial, so I'm only giving the third one of the classic three, the "factory lambda":

for i in range(0,10):
    funcs.append((lambda i: lambda x: f(i, x))(i))

Personally I'd go with i=i if there's no risk of the functions in funcs being accidentally called with 2 parameters instead of just 1, but the factory function approach is worth considering when you need something a little bit richer than just pre-binding one arg.

Answer 3

There's only one i which is bound to each lambda, contrary to what you think. This is a common mistake.

One way to get what you want is:

for i in range(0,10):
    funcs.append(lambda x, i=i: f(i, x))

Now you're creating a default parameter i in each lambda closure and binding to it the current value of the looping variable i.