I have a piece of my code where I\'m supposed to create a switchboard. I want to return a list of all the switches that are on. Here \"on\" will equal True
and
TL; DR: use np.where
as it is the fastest option. Your options are np.where
, itertools.compress
, and list comprehension
.
See the detailed comparison below, where it can be seen np.where
outperforms both itertools.compress
and also list comprehension
.
>>> from itertools import compress
>>> import numpy as np
>>> t = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]`
>>> t = 1000*t
list comprehension
>>> %timeit [i for i, x in enumerate(t) if x]
457 µs ± 1.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
itertools.compress
>>> %timeit list(compress(range(len(t)), t))
210 µs ± 704 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
numpy.where
>>> %timeit np.where(t)
179 µs ± 593 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)