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

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?

just saw that stackoverflow.com/questions/7546285/… is a duplicate question. how can I close this one or shall I delete it? — Stephan Kulla
– Stephan Kulla, Commented Jul 6, 2012 at 16:49
@tampis: Just vote to close – the moderators will eventually decide whether this question is merged or deleted. — Sven Marnach
– Sven Marnach, Commented Jul 6, 2012 at 16:51
Actually, I don't think that that even does what you think it does. Try adding del var after the first loop. — cha0site
– cha0site, Commented Jul 6, 2012 at 16:53
@cha0site del var gives the exception lobal name 'var' is not defined — Stephan Kulla
– Stephan Kulla, Commented Jul 6, 2012 at 16:56

Andrew Clark · Accepted Answer · 2012-07-06 16:55:27Z

11

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

edited Jul 6, 2012 at 16:55

answered Jul 6, 2012 at 16:50

Andrew Clark

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Sven Marnach · Accepted Answer · 2012-07-06 16:56:21Z

6

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)

edited Jul 6, 2012 at 16:56

answered Jul 6, 2012 at 16:49

Sven Marnach

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1 Comment

user166390 Over a year ago

Thanks for the edit, that is okay for a +1 now.

Jon Clements · Accepted Answer · 2012-07-06 16:51:51Z

1

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).

answered Jul 6, 2012 at 16:51

Jon Clements

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