One liner to get an array of positions form NUMPY array in python

Question

I have the following numpy array named histarr with the shape 1, 13

array([0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0], dtype=uint32)

I want to get an array that gives me positions where 1's are, hence I used np.where

where_are_ones_arr = np.where(histarr == 1)

The output is:

(array([1, 2, 4, 5, 6], dtype=int32),)

I was confused for a while but than I checked the type and I realized that where_are_ones_arr is not an array but it is actually a tuple, so if I wanted to get an array I used:

where_are_ones_arr[0]

Result:

array([1, 2, 4, 5, 6], dtype=int32)

Now that is all fine but I found it unbelievable that I cannot get that in one line, so I looked around and tried:

where_are_ones_give_me_only_array = histarr[np.where(histarr == 1)]

But it spits out:

array([1, 1, 1, 1, 1], dtype=uint32)

Which is not what I want and what I can explain? What is it that I do not get?

np.where(histarr == 1)[0] or np.flatnonzero(histarr == 1)? — Divakar
– Divakar, Commented May 27, 2020 at 7:45
Does this answer your question? output of numpy.where(condition) is not an array, but a tuple of arrays: why? — StupidWolf
– StupidWolf, Commented May 27, 2020 at 8:00
Thanks Divakar, that works. @StupidWolf I read that answer and an answer from paulj line 280 where he uses technique that I tried to reproduce in my question does not work for me. — rioZg
– rioZg, Commented May 27, 2020 at 8:14
I still do not get how I get the result that I get in my question. — rioZg
– rioZg, Commented May 27, 2020 at 8:17
Also: np.where() with only one argument is just calling under the hood just np.nonzero(), so calling np.nonzero() directly should be preferred. — norok2
– norok2, Commented May 27, 2020 at 8:30

CHRD · Accepted Answer · 2020-05-27 07:48:13Z

1

You should be able to do it in one line:

np.where(histarr == 1)[0]

answered May 27, 2020 at 7:48

CHRD

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1 Comment

rioZg Over a year ago

Thanks that works NOW magically. I tried that before and got some error that I can not reproduce anymore.

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