Indexes of elements in NumPy array that satisfy conditions on the value and the index

Question

I have a NumPy array, A. I want to know the indexes of the elements in A equal to a value and which indexes satisfy some condition:

import numpy as np
A = np.array([1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4])

value = 2
ind = np.array([0, 1, 5, 10])  # Index belongs to ind

Here is what I did:

B = np.where(A==value)[0]  # Gives the indexes in A for which value = 2
print(B)

[1 5 9]

mask = np.in1d(B, ind)  # Gives the index values that belong to the ind array
print(mask)
array([ True, True, False], dtype=bool)

print B[mask]  # Is the solution
[1 5]

The solution works, but I find it complicated. Also, in1d does a sort which is slow. Is there a better way of achieving this?

Robb · Accepted Answer · 2014-08-22 02:28:58Z

22

If you flip the operation order around you can do it in one line:

B = ind[A[ind]==value]
print B
[1 5]

Breaking that down:

#subselect first
print A[ind]
[1 2 2 3]

#create a mask for the indices
print A[ind]==value
[False  True  True False]

print ind
[ 0  1  5 10]

print ind[A[ind]==value]
[1 5]

answered Aug 22, 2014 at 2:28

Robb

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John Zwinck · Accepted Answer · 2014-08-22 02:02:22Z

14

B = np.where(A==value)[0]  #gives the indexes in A for which value = 2
print np.intersect1d(B, ind)
[1 5]

answered Aug 22, 2014 at 2:02

John Zwinck

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Paul · Accepted Answer · 2014-08-22 02:01:25Z

2

The second two steps can be replaced by intersect1D. Probably also does a sort. Don't know how you'd avoid that unless you can guarantee your ind array is ordered.

answered Aug 22, 2014 at 2:01

Paul

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Praveen · Accepted Answer · 2014-08-22 02:11:34Z

2

How about deferring the np.where until the end, like so:

res = (A == value)
mask = np.zeros(A.size)
mask[ind] = 1
print np.where(res * z)[0]

That shouldn't require any sorting.

answered Aug 22, 2014 at 2:11

Praveen

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wwii · Accepted Answer · 2014-08-22 03:14:09Z

1

This is a bit different - I didn't do any timing tests.

>>> 
>>> A = np.array([1,2,3,4,1,2,3,4,1,2,3,4])
>>> ind = np.array([0,1,5,10])
>>> b = np.ix_(A==2)
>>> np.intersect1d(ind, *b)
array([1, 5])
>>>

Although after looking at @Robb's solution, that's probably the way to do it.

answered Aug 22, 2014 at 3:14

wwii

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