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Lambdas inside list comprehensions and for loops

Asked 2012-04-09T07:57:57.827
10

I wanted to have a list of lambdas that act as sort of a cache to some heavy computation and noticed this:

>>> [j() for j in [lambda:i for i in range(10)]]
[9, 9, 9, 9, 9, 9, 9, 9, 9, 9]

Although

>>> list([lambda:i for i in range(10)])
[<function <lambda> at 0xb6f9d1ec>, <function <lambda> at 0xb6f9d22c>, <function <lambda> at 0xb6f9d26c>, <function <lambda> at 0xb6f9d2ac>, <function <lambda> at 0xb6f9d2ec>, <function <lambda> at 0xb6f9d32c>, <function <lambda> at 0xb6f9d36c>, <function <lambda> at 0xb6f9d3ac>, <function <lambda> at 0xb6f9d3ec>, <function <lambda> at 0xb6f9d42c>]

Meaning that the lambdas are unique functions but they somehow all share the same index value.

Is this a bug or a feature? How do I avoid this problem? It's not limited to list comprehensions...

>>> funcs = []
... for i in range(10):
...     funcs.append(lambda:i)
... [j() for j in funcs]
[9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
python lambda functional-programming list-comprehension
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1 Answer

18

The lambda returns the value of i at the time you call it. Since you call the lambda after the loop has finished running, the value of i will always be 9.

You can create a local i variable in the lambda to hold the value at the time the lambda was defined:

>>> [j() for j in [lambda i=i:i for i in range(10)]]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Another solution is to create a function that returns the lambda:

def create_lambda(i):
    return lambda:i
>>> [j() for j in [create_lambda(i) for i in range(10)]]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

This works because there is a different closure (holding a different value of i) created for each invocation of create_lambda.

answered 2012-04-09T08:08:10.817

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