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python - Filtering a list based on a list of booleans

I have a list of values which I need to filter given the values in a list of booleans:

list_a = [1, 2, 4, 6]
filter = [True, False, True, False]

I generate a new filtered list with the following line:

filtered_list = [i for indx,i in enumerate(list_a) if filter[indx] == True]

which results in:

print filtered_list
[1,4]

The line works but looks (to me) a bit overkill and I was wondering if there was a simpler way to achieve the same.


Advices

Summary of two good advices given in the answers below:

1- Don't name a list filter like I did because it is a built-in function.

2- Don't compare things to True like I did with if filter[idx]==True.. since it's unnecessary. Just using if filter[idx] is enough.

See Question&Answers more detail:os

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You're looking for itertools.compress:

>>> from itertools import compress
>>> list_a = [1, 2, 4, 6]
>>> fil = [True, False, True, False]
>>> list(compress(list_a, fil))
[1, 4]

Timing comparisons(py3.x):

>>> list_a = [1, 2, 4, 6]
>>> fil = [True, False, True, False]
>>> %timeit list(compress(list_a, fil))
100000 loops, best of 3: 2.58 us per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v]  #winner
100000 loops, best of 3: 1.98 us per loop

>>> list_a = [1, 2, 4, 6]*100
>>> fil = [True, False, True, False]*100
>>> %timeit list(compress(list_a, fil))              #winner
10000 loops, best of 3: 24.3 us per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v]
10000 loops, best of 3: 82 us per loop

>>> list_a = [1, 2, 4, 6]*10000
>>> fil = [True, False, True, False]*10000
>>> %timeit list(compress(list_a, fil))              #winner
1000 loops, best of 3: 1.66 ms per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v] 
100 loops, best of 3: 7.65 ms per loop

Don't use filter as a variable name, it is a built-in function.


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