using numpy 'where' with an array subset

Question

I'm using python 2.7.13, and I'm perplexed by a behavior that I'm seeing when using array slicing with numpy.

import numpy as np
a=np.array([1,1,2,3,4,5,6,7])
print a
print a[1:]
print a[1:] > 3

print np.where( a[1:]  > 1 )

I was expecting that I would see for the final output [2 3 4 5 6 7 8] i.e. the indices of the array that were found within the slice.

My goal was to apply the Boolean mask to all elements of the array except the first element. Then to grab value in the array that corresponded to the first 'True' value in the index array. Is this possible?

where just tells you where a[1:]>1 is True. It does nothing with a. — hpaulj
– hpaulj, Commented Aug 16, 2017 at 6:43

Landmaster · Accepted Answer · 2017-08-16 06:45:30Z

3

Your code is doing exactly what you want it to do. When you slice a[1:], you get np.array([1, 2, 3, 4, 5, 6, 7]). The output I got:

(array([1, 2, 3, 4, 5, 6], dtype=int64),)

is giving you the indices of 2, 3, 4, 5, 6, 7, in the slice of a you specified.

answered Aug 16, 2017 at 6:45

Landmaster

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Comments

Ramon van der Werf · Accepted Answer · 2017-08-16 06:56:53Z

1

I guess this solves your problem:

import numpy as np
a=np.array([1,1,2,3,4,5,6,7])
b=a[1:]
print np.where( b  > 1 )

In your case, the print statement returns the element indices in the array a, while skipping the first elememnt. to get the indices in array a, print np.where(a>1) I hope this helps.

answered Aug 16, 2017 at 6:56

Ramon van der Werf

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