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pythonarraysnumpy
invis
invis

Reputation: 1100

Python: intersection indices numpy array

How can I get the indices of intersection points between two numpy arrays? I can get intersecting values with intersect1d:

import numpy as np

a = np.array(xrange(11))
b = np.array([2, 7, 10])
inter = np.intersect1d(a, b)
# inter == array([ 2,  7, 10])

But how can I get the indices into a of the values in inter?

Upvotes: 45

Views: 44697

Answers (6)

Sambid Wasti
Sambid Wasti

Reputation: 1

This is a really old post but the numpy.intersect1d() has a return_indices flag.

common, inda, indb = numpy.intersect1d(a,b, return_indices=True)

would return with inda for indices/position with common values for a, and indb for b.

However, it returns the 1st intersection point. For example, if a is not unique and has 4 similar values, the indices returned for a are the first.

Upvotes: 0

Michael Yahalom
Michael Yahalom

Reputation: 11

As of numpy version 1.15.0 intersect1d has a return_indices option :

numpy.intersect1d(ar1, ar2, assume_unique=False, return_indices=False)

Upvotes: 1

Will
Will

Reputation: 161

indices = np.argwhere(np.in1d(a,b))

Upvotes: 2

WY Hsu
WY Hsu

Reputation: 1905

For Python >= 3.5, there's another solution to do so

Other Solution

Let we go through this step by step.

Based on the original code from the question

import numpy as np

a = np.array(range(11))
b = np.array([2, 7, 10])
inter = np.intersect1d(a, b)

First, we create a numpy array with zeros

c = np.zeros(len(a))
print (c)

output

>>> [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]

Second, change array value of c using intersect index. Hence, we have

c[inter] = 1
print (c)

output

>>>[ 0.  0.  1.  0.  0.  0.  0.  1.  0.  0.  1.]

The last step, use the characteristic of np.nonzero(), it will return exactly the index of the non-zero term you want.

inter_with_idx = np.nonzero(c)
print (inter_with_idx)

Final output

array([ 2, 7, 10])

Reference

[1] numpy.nonzero

Upvotes: 1

Rik
Rik

Reputation: 1987

If you need to get unique values as given by intersect1d:

import numpy as np

a = np.array([range(11,21), range(11,21)]).reshape(20)
b = np.array([12, 17, 20])
print(np.intersect1d(a,b))
#unique values

inter = np.in1d(a, b)
print(a[inter])
#you can see these values are not unique

indices=np.array(range(len(a)))[inter]
#These are the non-unique indices

_,unique=np.unique(a[inter], return_index=True)

uniqueIndices=indices[unique]
#this grabs the unique indices

print(uniqueIndices)
print(a[uniqueIndices])
#now they are unique as you would get from np.intersect1d()

Output:

[12 17 20]
[12 17 20 12 17 20]
[1 6 9]
[12 17 20]

Upvotes: 2

senderle
senderle

Reputation: 150957

You could use the boolean array produced by in1d to index an arange. Reversing a so that the indices are different from the values:

>>> a[::-1]
array([10,  9,  8,  7,  6,  5,  4,  3,  2,  1,  0])
>>> a = a[::-1]

intersect1d still returns the same values...

>>> numpy.intersect1d(a, b)
array([ 2,  7, 10])

But in1d returns a boolean array:

>>> numpy.in1d(a, b)
array([ True, False, False,  True, False, False, False, False,  True,
       False, False], dtype=bool)

Which can be used to index a range:

>>> numpy.arange(a.shape[0])[numpy.in1d(a, b)]
array([0, 3, 8])
>>> indices = numpy.arange(a.shape[0])[numpy.in1d(a, b)]
>>> a[indices]
array([10,  7,  2])

To simplify the above, though, you could use nonzero -- this is probably the most correct approach, because it returns a tuple of uniform lists of X, Y... coordinates:

>>> numpy.nonzero(numpy.in1d(a, b))
(array([0, 3, 8]),)

Or, equivalently:

>>> numpy.in1d(a, b).nonzero()
(array([0, 3, 8]),)

The result can be used as an index to arrays of the same shape as a with no problems. 

>>> a[numpy.nonzero(numpy.in1d(a, b))]
array([10,  7,  2])

But note that under many circumstances, it makes sense just to use the boolean array itself, rather than converting it into a set of non-boolean indices.

Finally, you can also pass the boolean array to argwhere, which produces a slightly differently-shaped result that's not as suitable for indexing, but might be useful for other purposes.

>>> numpy.argwhere(numpy.in1d(a, b))
array([[0],
       [3],
       [8]])

Upvotes: 48

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