[python] Getting indices of True values in a boolean list

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 "off" equal False. So now I just want to return a list of all the True values and their position. This is all I have but it only return the position of the first occurrence of True (this is just a portion of my code):

self.states = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]

def which_switch(self):
    x = [self.states.index(i) for i in self.states if i == True]

This only returns "4"

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

• Method 1: Using 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)

• Method 2: Using itertools.compress

>>> %timeit list(compress(range(len(t)), t))
210 µs ± 704 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

• Method 3 (the fastest method): Using numpy.where

>>> %timeit np.where(t)
179 µs ± 593 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

You can use filter for it:

filter(lambda x: self.states[x], range(len(self.states)))

The range here enumerates elements of your list and since we want only those where self.states is True, we are applying a filter based on this condition.

For Python > 3.0:

list(filter(lambda x: self.states[x], range(len(self.states))))

Using element-wise multiplication and a set:

>>> states = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]
>>> set(multiply(states,range(1,len(states)+1))-1).difference({-1})

Output: {4, 5, 7}

If you have numpy available:

>>> import numpy as np
>>> states = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]
>>> np.where(states)[0]
array([4, 5, 7])

Simply do this:

def which_index(self):
    return [
        i for i in range(len(self.states))
        if self.states[i] == True
    ]

Use dictionary comprehension way,

x = {k:v for k,v in enumerate(states) if v == True}

Input:

states = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]

Output:

{4: True, 5: True, 7: True}