lambda function in Python

Asked May 10 '18 at 17:17

Active May 10 '18 at 17:41

Viewed 261 times

I found a interesting behavior of lambda function in Python.

If I do,

lambda_funcs = []
for i in range(10):
    lambda_func = lambda x:x[i]
    lambda_funcs.append(lambda_func)

X = ['a', 'b', 'c']
print lambda_funcs[0](X)

It throws an error: IndexError: list index out of range.

This code aims to create 10 lambda functions each of which returns i-th element. But, all lambda_func in the above code attempts to retrieve 10th element.

To make it work, I found that the following definition is a right way:

lambda_func = (lambda j: lambda x: x[j])(i)

Can anyone tell me what is going on inside lambda evaluation in Python?

asked May 10 '18 at 17:17

DSKim

1

change to `lambda_func = lambda x, i=i:x[i]` – eyllanesc May 10 '18 at 17:21
classic lazy evaluation issue where `i` is the same for all lambdas – Jean-François Fabre May 10 '18 at 17:21
Try printing the value of `i` in your lambda function – phoxis May 10 '18 at 17:21
@eyllanesc they already showed a more "correct" solution, `(lambda j: lambda x: x[j])(i)` – juanpa.arrivillaga May 10 '18 at 17:41
This is due to the late-binding behavior of Python closures (so this has nothing to do with `lambda` per se). Here is a very good explanation: http://docs.python-guide.org/en/latest/writing/gotchas/#late-binding-closures – juanpa.arrivillaga May 10 '18 at 17:43

lambda function in Python

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