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name

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name Punditsdkoslkdosdkoskdo

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.

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    • Some style notes: if filter[indx] == True Do not use == if you want to check for identity with True, use is. Anyway in this case the whole comparison is useless, you could simply use if filter[indx]. Lastly: never use the name of a built-in as a variable/module name(I'm referring to the name filter). Using something like included, so that the if reads nicely (if included[indx]).

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.

  • 183
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Like so:

filtered_list = [i for (i, v) in zip(list_a, filter) if v]

Using zip is the pythonic way to iterate over multiple sequences in parallel, without needing any indexing. This assumes both sequences have the same length (zip stops after the shortest runs out). Using itertools for such a simple case is a bit overkill ...

One thing you do in your example you should really stop doing is comparing things to True, this is usually not necessary. Instead of if filter[idx]==True: ..., you can simply write if filter[idx]: ....

  • 42
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With numpy:

In [128]: list_a = np.array([1, 2, 4, 6])
In [129]: filter = np.array([True, False, True, False])
In [130]: list_a[filter]

Out[130]: array([1, 4])

or see Alex Szatmary's answer if list_a can be a numpy array but not filter

Numpy usually gives you a big speed boost as well

In [133]: list_a = [1, 2, 4, 6]*10000
In [134]: fil = [True, False, True, False]*10000
In [135]: list_a_np = np.array(list_a)
In [136]: fil_np = np.array(fil)

In [139]: %timeit list(itertools.compress(list_a, fil))
1000 loops, best of 3: 625 us per loop

In [140]: %timeit list_a_np[fil_np]
10000 loops, best of 3: 173 us per loop
  • 40
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      • 2
    • Good point, I prefer using NumPy over list where possible. But if you need to use list anyway, you have (using NumPy solution) create np.array from both lists, use boolean indexing and finally converting array back to list with tolist() method. To be precise, you should include those objects creation into time comparison. Then, using itertools.compress will be still the fastest solution.

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