Getting argsort of numpy array

Question

I have a numpy array as follows:

array([ True,  True,  True,  True,  True, False,  True,  True, False,
        True, False,  True,  True,  True,  True,  True,  True, False,
       False, False, False, False,  True,  True, False, False, False,
        True,  True,  True,  True,  True,  True,  True, False,  True,
        True,  True,  True, False,  True,  True, False, False,  True,
        True,  True, False,  True,  True,  True, False],

I want to get the indices of all the True elements. There is no get_loc method in numpy like Pandas Series and similarly no index method like a list. I don't want to convert it into a list and then use .index.

Any idea?

wim · Accepted Answer · 2015-04-30 00:12:47Z

3

Use ndarray.nonzero:

>>> a.nonzero()
(array([ 0,  1,  2,  3,  4,  6,  7,  9, 11, 12, 13, 14, 15, 16, 22, 23, 27,
        28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 44, 45, 46, 48, 49,
        50]),)

answered Apr 30, 2015 at 0:12

wim

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EdChum · Accepted Answer · 2015-04-30 08:45:04Z

To do this in pandas:

In [255]:

s[s==True].index
Out[255]:
Int64Index([0, 1, 2, 3, 4, 6, 7, 9, 11, 12, 13, 14, 15, 16, 22, 23, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 44, 45, 46, 48, 49, 50], dtype='int64')

Update

Actually you can use the fact that the values are already boolean values to mask the series:

In [256]:

s[s].index
Out[256]:
Int64Index([0, 1, 2, 3, 4, 6, 7, 9, 11, 12, 13, 14, 15, 16, 22, 23, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 44, 45, 46, 48, 49, 50], dtype='int64')

Similarly for numpy arrays you can use the boolean values to mask the array and get the index values using np.where:

In [261]:

np.where(a)

Out[261]:
(array([ 0,  1,  2,  3,  4,  6,  7,  9, 11, 12, 13, 14, 15, 16, 22, 23, 27,
        28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 44, 45, 46, 48, 49,
        50], dtype=int64),)

fixxxer · Accepted Answer · 2015-04-30 09:09:58Z

0

The np.ix_ way seems to be the slowest.

In [846]: % timeit a.nonzero()
1000000 loops, best of 3: 707 ns per loop

In [845]: % timeit np.where(a)
1000000 loops, best of 3: 883 ns per loop

In [849]: %timeit np.ix_(a==True)
100000 loops, best of 3: 9.21 µs per loop

answered Apr 30, 2015 at 9:09

fixxxer

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