Find 2d array inside a 3d array

Question

Is there a fast way to find all indices where a 2d array is inside a 3d array?

I have this 3d numpy array:

arr = np.array([
        [[0,1],[0,2],[0,3],[0,4],[0,4],[0,5],[0,5],[0,5],[0,5],[0,5]],
        [[0,1],[0,2],[0,2],[0,2],[0,3],[0,4],[0,4],[0,4],[0,5],[0,5]],
        [[0,1],[0,2],[0,3],[0,3],[0,3],[0,4],[0,4],[0,5],[0,5],[0,5]]
       ])

And I would like to find all indices where [0,4] occurs. I've tried this one:

whereInd = np.argwhere(arr == np.array([0,4]))

but it doesn't work. The expected result is:

[[0 3],[0 4],[1 5],[1 6],[1 7],[2 5],[2 6]]

Another question is, will this be fast? Because I would like to use it for a (10000,100,2) array.

Alex Riley · Accepted Answer · 2015-06-07 11:31:54Z

2

Using argwhere() is a good idea, but you also need to use all() to get your desired output:

>>> np.argwhere((arr == [0, 4]).all(axis=2))
array([[0, 3],
       [0, 4],
       [1, 5],
       [1, 6],
       [1, 7],
       [2, 5],
       [2, 6]])

Here all() is used to check each row is [True, True] following the comparison (that is, the row is equal to [0, 4]). In a 3D array, axis=2 points along rows.

This reduces the number of dimensions to two, and argwhere() returns the desired array of indices.

Regarding performance, this method should handle arrays of the size you specify fairly quickly:

In [20]: arr = np.random.randint(0, 10, size=(10000, 100, 2))
In [21]: %timeit np.argwhere((arr == [0, 4]).all(axis=2))
10 loops, best of 3: 44.9 ms per loop

edited Jun 7, 2015 at 11:31

answered Jun 7, 2015 at 11:16

Alex Riley

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AvidLearner · Accepted Answer · 2015-06-07 11:30:22Z

0

The simplest solution I can think of is:

import numpy as np
arr = np.array([
        [[0,1],[0,2],[0,3],[0,4],[0,4],[0,5],[0,5],[0,5],[0,5],[0,5]],
        [[0,1],[0,2],[0,2],[0,2],[0,3],[0,4],[0,4],[0,4],[0,5],[0,5]],
        [[0,1],[0,2],[0,3],[0,3],[0,3],[0,4],[0,4],[0,5],[0,5],[0,5]]
       ])

whereInd = []
for i,row in enumerate(arr):
    for j,elem in enumerate(row):
        if all(elem == [0,4]):
            whereInd.append((i,j))

print whereInd
#prints [(0, 3), (0, 4), (1, 5), (1, 6), (1, 7), (2, 5), (2, 6)]

Although any solution with np.argwhere should work around 10 times faster.

edited Jun 7, 2015 at 11:30

answered Jun 7, 2015 at 11:16

AvidLearner

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