How can I force Python to create a new variable / new scope inside a loop? [duplicate]

问题

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?

回答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

回答2:

You can't create a new variable in the local scope inside the loop. Whatever name you choose, your function will always be a closure over that name and use its most recent value.

The easiest way around this is to use a keyword argument:

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

回答3:

In your example, var is always bound to the the last value of the loop. You need to bind it inside the lambda using closures.append(lambda var=var: var).

来源：https://stackoverflow.com/questions/11366375/how-can-i-force-python-to-create-a-new-variable-new-scope-inside-a-loop

标签

python

variables

for-loop

parameters

closures