What will be the output of the code below? Explain your answer.
def multipliers(): return [lambda x : i * x for i in range(4)] print [m(2) for m in multipliers()]
How would you modify the definition of
multipliers to produce the presumably desired behavior?
The output of the above code will be
[6, 6, 6, 6] (not
[0, 2, 4, 6]).
The reason for this is that Python’s closures are late binding. This means that the values of variables used in closures are looked up at the time the inner function is called. So as a result, when any of the functions returned by
multipliers() are called, the value of
i is looked up in the surrounding scope at that time. By then, regardless of which of the returned functions is called, the
for loop has completed and
i is left with its final value of 3. Therefore, every returned function multiplies the value it is passed by 3, so since a value of 2 is passed in the above code, they all return a value of 6 (i.e., 3 x 2).
(Incidentally, as pointed out in The Hitchhiker’s Guide to Python, there is a somewhat widespread misconception that this has something to do with lambdas, which is not the case. Functions created with a
lambda expression are in no way special and the same behavior is exhibited by functions created using an ordinary
Below are a few examples of ways to circumvent this issue.
One solution would be use a Python generator as follows:
def multipliers(): for i in range(4): yield lambda x : i * x
Another solution is to create a closure that binds immediately to its arguments by using a default argument. For example:
def multipliers(): return [lambda x, i=i : i * x for i in range(4)]
Or alternatively, you can use the
from functools import partial from operator import mul def multipliers(): return [partial(mul, i) for i in range(4)]