获取 boolean 列表中真值的索引

Question

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"

Answer 1

Use enumerate , list.index returns the index of first match found.

>>> t = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]
>>> [i for i, x in enumerate(t) if x]
[4, 5, 7]

For huge lists, it'd be better to use itertools.compress :

>>> from itertools import compress
>>> list(compress(xrange(len(t)), t))
[4, 5, 7]
>>> t = t*1000
>>> %timeit [i for i, x in enumerate(t) if x]
100 loops, best of 3: 2.55 ms per loop
>>> %timeit list(compress(xrange(len(t)), t))
1000 loops, best of 3: 696 µs per loop

Answer 2

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])

Answer 3

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)

Answer 4

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))))

Answer 5

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}

Answer 6

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}

Answer 7

Simply do this:

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

Answer 8

I got different benchmark result compared to @meysham answer . In this test, compress seems the fastest (python 3.7).

from itertools import compress
import numpy as np

t = [True, False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]

%timeit [i for i, x in enumerate(t) if x]
%timeit list(compress(range(len(t)), t))
%timeit list(filter(lambda x: t[x], range(len(t))))
%timeit np.where(t)[0]

# 2.54 µs ± 400 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# 2.67 µs ± 600 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# 6.22 µs ± 624 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# 6.52 µs ± 768 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

t = 1000*t
%timeit [i for i, x in enumerate(t) if x]
%timeit list(compress(range(len(t)), t))
%timeit list(filter(lambda x: t[x], range(len(t))))
%timeit np.where(t)[0]

# 1.68 ms ± 112 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# 947 µs ± 105 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# 3.96 ms ± 97 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# 2.14 ms ± 45.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 9

You can filter by using boolean mask array with square bracket, it's faster than np.where

>>> states = [True, False, False, True]
>>> np.arange(len(states))[states]
array([0, 3])

>>> size = 1_000_000
>>> states = np.arange(size) % 2 == 0
>>> states 
array([ True, False,  True, ..., False,  True, False])
>>> true_index = np.arange(size)[states]
>>> len(true_index)
500000
>>> true_index
array([     0,      2,      4, ..., 999994, 999996, 999998])

Answer 10

index_of_element=list_name.index(True)