functions depending on iterables

Question

funcs = []
for i in range(3):
    a = i
    func = lambda x: a
    funcs.append(func)

print [func(0) for func in funcs]

I would like this to print [0,1,2], instead it prints [2,2,2]. I see what's going on, the question is how do I circumvent this?

Answer 1

You can get around this by binding a to the local scope of the lambda using a keyword argument:

funcs = []
for i in range(3):
    a = i
    func = lambda x, a=a: a
    funcs.append(func)
print [func(0) for func in funcs]

Output:

[0, 1, 2]

Answer 2

This line:

func = lambda x: a

states that you want the function to return a no matter what you pass it.

So when you call the functions with the argument 0, it returns the value of a. Which is 2 after it exited from the for loop.

To circumvent that you would need to change the lambda function to:

func = lambda x: x

and change the following line:

print [func(0) for func in funcs]

to:

print [func(x) for x in range(3)]

Or bind a to the local scope of lambda as suggested by @dano.