Intersect two numpy index arrays

Question

Suppose I have two images that are thresholded differently, and I want to find the set intersection of indexes over both threshold, in other words all the indexes that are in both lists. Assume the dimensions are equal, so there is nothing to worry about for that.

img1 = #something
img2 = #something slightly different
indexes1 = np.nonzero(img1)
indexes2 = np.nonzero(img2)
index_intersection = #???

How can I do this in a way that is easy to understand and in a way that is efficient?

Assume sample data for img1 and img2 and show us the expected output? — Divakar
– Divakar, Commented Sep 22, 2015 at 18:42
@AGML that question is about geometric (also falls under infinite set) intersection, while this one is about finite set intersection. Similar, but different! — Andrew Hundt
– Andrew Hundt, Commented Sep 22, 2015 at 23:04

Divakar · Accepted Answer · 2015-09-22 19:42:54Z

2

If you are looking for matching nonzero XY indices pairs, you can use boolean ANDing between the nonzero masks of the input arrays and then use np.nonzero, like so -

out = np.nonzero((img1!=0) & (img2!=0))

You can verify these results with np.intersect1d after getting the linear indices of matches from img1 and img2 giving us a second approach to solve the problem at hand , like so -

l_intsct = np.intersect1d(np.nonzero(img1.ravel())[0],np.nonzero(img2.ravel())[0])
out = np.unravel_index(l_intsct,img1.shape)

Sample run -

In [127]: img1
Out[127]: 
array([[3, 2, 3, 1, 0],
       [3, 1, 1, 2, 2],
       [0, 2, 3, 2, 1],
       [0, 0, 0, 4, 2]])

In [128]: img2
Out[128]: 
array([[1, 1, 4, 0, 0],
       [0, 0, 0, 0, 2],
       [4, 1, 0, 3, 1],
       [1, 0, 4, 1, 4]])

In [129]: np.nonzero(img1)
Out[129]: 
(array([0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3]),
 array([0, 1, 2, 3, 0, 1, 2, 3, 4, 1, 2, 3, 4, 3, 4]))

In [130]: np.nonzero(img2)
Out[130]: 
(array([0, 0, 0, 1, 2, 2, 2, 2, 3, 3, 3, 3]),
 array([0, 1, 2, 4, 0, 1, 3, 4, 0, 2, 3, 4]))

In [131]: np.nonzero((img1!=0) & (img2!=0))
Out[131]: (array([0, 0, 0, 1, 2, 2, 2, 3, 3]), array([0, 1, 2, 4, 1, 3, 4, 3, 4]))

In [132]: l_intsct = np.intersect1d(np.nonzero(img1.ravel())[0],np.nonzero(img2.ravel())[0])

In [133]: np.unravel_index(l_intsct,img1.shape)
Out[133]: (array([0, 0, 0, 1, 2, 2, 2, 3, 3]), array([0, 1, 2, 4, 1, 3, 4, 3, 4]))

edited Sep 22, 2015 at 19:42

answered Sep 22, 2015 at 19:32

Divakar

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3 Comments

Andrew Hundt Over a year ago

Interesting! What if (0,0) is in each?

Divakar Over a year ago

@AndrewHundt If R1,C1 = np.nonzero(img1) and R2,C2 = np.nonzero(img2) and you meant (0,0) as a pair of R and C, which is already happening in the listed sample, then looking at the output: R,C = np.nonzero((img1!=0) & (img2!=0)) from the listed solution gives you that pair, right? Or what else do you mean?

Andrew Hundt Over a year ago

In my use case I tried that command and it didn't seem to work. I'm not sure why.

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