Numpy where returns empty array

Question

I have an array e.g.

a = [5,1,3,0,2]

I apply the where function:

np.where(a == 2)

The output is an empty array

(array([], dtype=int64),)

I found kind of the same problem here, but in my case it really dosen't make any sense or dose it?

Btw. i'm on a Mac using Python 2.7.10

The interpreter first evaluates a==2, and then passes the result to the where function. Does that a==2 make sense? — hpaulj
– hpaulj, Commented Jan 21, 2018 at 17:41

Kasravnd · Accepted Answer · 2018-01-21 17:18:21Z

13

You're passing a list to where() function not a Numpy array. Use an array instead:

In [20]: a = np.array([5,1,3,0,2])

In [21]: np.where(a == 2)
Out[21]: (array([4]),)

Also as mentioned in comments, in this case the value of a == 2 is False, and this is the value passed to where. If a is a numpy array, then the value of a == 2 is a numpy array of bools, and the where function will give you the desire results.

edited Jan 21, 2018 at 17:18

answered Jan 21, 2018 at 16:39

Kasravnd

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3 Comments

Warren Weckesser Over a year ago

Actually, the function is not being applied to a list. The value of a == 2 in the question is simply False, and this is the value passed to where. But you are correct in that the problem is that a is a list and not a numpy array. If a is a numpy array, then the value of a == 2 is a numpy array of bools, and then where works as expected.

Kasravnd Over a year ago

@WarrenWeckesser Indeed! thanks for point that out, it could cause a miss understanding.

Sebastian N Over a year ago

Ah ok now I understand! Thank you very much for your quick and well explained help!

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