Getting a list of indices where pandas boolean series is True

I have a pandas series with boolean entries. I would like to get a list of indices where the values are True.

For example the input pd.Series([True, False, True, True, False, False, False, True])

should yield the output [0,2,3,7].

I can do it with a list comprehension, but is there something cleaner or faster?

Using Boolean Indexing

>>> s = pd.Series([True, False, True, True, False, False, False, True])
>>> s[s].index
Int64Index([0, 2, 3, 7], dtype='int64')

If need a np.array object, get the .values

>>> s[s].index.values
array([0, 2, 3, 7])

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Using np.nonzero

>>> np.nonzero(s)
(array([0, 2, 3, 7]),)

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Using np.flatnonzero

>>> np.flatnonzero(s)
array([0, 2, 3, 7])

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Using np.where

>>> np.where(s)[0]
array([0, 2, 3, 7])

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Using np.argwhere

>>> np.argwhere(s).ravel()
array([0, 2, 3, 7])

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Using pd.Series.index

>>> s.index[s]
array([0, 2, 3, 7])

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Using python's built-in filter

>>> [*filter(s.get, s.index)]
[0, 2, 3, 7]

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Using list comprehension

>>> [i for i in s.index if s[i]]
[0, 2, 3, 7]

As an addition to rafaelc's answer, here are the according times (from quickest to slowest) for the following setup

import numpy as np
import pandas as pd
s = pd.Series([x > 0.5 for x in np.random.random(size=1000)])

Using np.where

>>> timeit np.where(s)[0]
12.7 µs ± 77.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

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Using np.flatnonzero

>>> timeit np.flatnonzero(s)
18 µs ± 508 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

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Using pd.Series.index

The time difference to boolean indexing was really surprising to me, since the boolean indexing is usually more used.

>>> timeit s.index[s]
82.2 µs ± 38.9 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

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Using Boolean Indexing

>>> timeit s[s].index
1.75 ms ± 2.16 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

If you need a np.array object, get the .values

>>> timeit s[s].index.values
1.76 ms ± 3.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

If you need a slightly easier to read version <-- not in original answer

>>> timeit s[s==True].index
1.89 ms ± 3.52 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

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Using pd.Series.where <-- not in original answer

>>> timeit s.where(s).dropna().index
2.22 ms ± 3.32 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

>>> timeit s.where(s == True).dropna().index
2.37 ms ± 2.19 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

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Using pd.Series.mask <-- not in original answer

>>> timeit s.mask(s).dropna().index
2.29 ms ± 1.43 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

>>> timeit s.mask(s == True).dropna().index
2.44 ms ± 5.82 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

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Using list comprehension

>>> timeit [i for i in s.index if s[i]]
13.7 ms ± 40.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

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Using python's built-in filter

>>> timeit [*filter(s.get, s.index)]
14.2 ms ± 28.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

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Using np.nonzero <-- did not work out of the box for me

>>> timeit np.nonzero(s)
ValueError: Length of passed values is 1, index implies 1000.

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Using np.argwhere <-- did not work out of the box for me

>>> timeit np.argwhere(s).ravel()
ValueError: Length of passed values is 1, index implies 1000.

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Also works: s.where(lambda x: x).dropna().index, and it has the advantage of being easy to chain pipe - if your series is being computed on the fly, you don't need to assign it to a variable.

Note that if s is computed from r: s = cond(r) than you can also use: r.where(lambda x: cond(x)).dropna().index.