Question

Today I explored a weird behavior of Python. An example:

closures = []
for x in [1, 2, 3]:
    # store `x' in a "new" local variable
    var = x

    # store a closure which returns the value of `var'
    closures.append(lambda: var)

for c in closures:
    print(c())

The above code prints

3
3
3

But I want it to print

1
2
3

I explain this behavior for myself that var is always the same local variable (and python does not create a new one like in other languages). How can I fix the above code, so that each closure will return another value?

Answer 1

The easiest way to do this is to use a default argument for your lambda, this way the current value of x is bound as the default argument of the function, instead of var being looked up in a containing scope on each call:

closures = []
for x in [1, 2, 3]:
    closures.append(lambda var=x: var)

for c in closures:
    print(c())

Alternatively you can create a closure (what you have is not a closure, since each function is created in the global scope):

make_closure = lambda var: lambda: var
closures = []
for x in [1, 2, 3]:
    closures.append(make_closure(x))

for c in closures:
    print(c())

make_closure() could also be written like this, which may make it more readable:

def make_closure(var):
    return lambda: var