Numpy 'where' returns empty array for all 0s array for condition >= 0

Question

I am using np.where function to get indices where the value matches 0. But the following code returns me empty array, which is not expected.

import numpy as np
a = np.array([0,0,0,0])
np.where(a[a>=0])

This gives: (array([], dtype=int64),).

Could anyone please point out what I am missing?

https://numpy.org/doc/stable/reference/generated/numpy.where.html

Quang Hoang · Accepted Answer · 2021-10-21 18:44:04Z

2

That's expected. a>=0 gives you all True, i.e a[a>=0] gives you a, which does not contain any non-zero element. So np.where(a) returns empty array.

Are you looking for np.where(a>=0)?

answered Oct 21, 2021 at 18:44

Quang Hoang

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Ulises Bussi · Accepted Answer · 2021-10-21 18:45:32Z

1

you are telling to find index where a[ a>=0] a>=0 returns a [True,True,True,True] then a[a>=0] returns [0,0,0,0] then there's no True condition to return.

Maybe you want to do

 np.where(a>=0)

answered Oct 21, 2021 at 18:45

Ulises Bussi

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1 Comment

Ulises Bussi Over a year ago

my response is the same as Quang Hoang I think he explains it better: np.where returns the arguments where the list have a non-zero element so you were using the original list (for your example a[a>=0] it's a )

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