Get (column, row) index from NumPy array that meets a boolean condition

Question

I am working with a 2D NumPy array. I would like to get the (column, row) index, or (x, y) coordinate, if you prefer thinking that way, from my 2D array that meets a boolean condition.

The best way I can explain what I am trying to do is via a trivial example:

>>> a = np.arange(9).reshape(3, 3)
>>> b = a > 4
>>> b
>>> array([[False, False, False],
           [False, False,  True],
           [ True,  True,  True]], dtype=bool)

At this point I now have a boolean array, indicating where a > 4.

My goal at this point is grab the indexes of the boolean array where the value is True. For example, the indexes (1, 2), (2, 0), (2, 1), and (2, 2) all have a value of True.

My end goal is to end up with a list of indexes:

>>> indexes = [(1, 2), (2, 0), (2, 1), (2, 2)]

Again, I stress the point that the code above is a trivial example, but the application of what I'm trying to do could have arbitrary indexes where a > 4 and not something based on arange and reshape.

Note that the indices you gave are actually (y,x) order, which is also what the answer delivers. — letmaik
– letmaik, Commented May 10, 2014 at 15:55

Ashwini Chaudhary · Accepted Answer · 2014-01-28 14:11:01Z

18

Use numpy.where with numpy.column_stack:

>>> np.column_stack(np.where(b))
array([[1, 2],
       [2, 0],
       [2, 1],
       [2, 2]])

answered Jan 28, 2014 at 14:11

Ashwini Chaudhary

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ev-br · Accepted Answer · 2014-01-29 02:16:50Z

An alternative to the answer of @Ashwini Chaudhary, is numpy.nonzero

>>> a = np.arange(9).reshape(3,3)
>>> b = a > 4
>>> np.nonzero(b)
(array([1, 2, 2, 2]), array([2, 0, 1, 2]))

>>> np.transpose(np.nonzero(b))
array([[1, 2],
       [2, 0],
       [2, 1],
       [2, 2]])

EDIT: What is faster. nonzero and where are essentially equivalent, but transpose turns out to be the wrong one here (even though it's mentioned in the docs):

In [15]: N = 5000

In [16]: a = np.random.random((N, N))

In [17]: %timeit np.nonzero(a > 0.5)
1 loops, best of 3: 470 ms per loop

In [18]: %timeit np.transpose(np.nonzero(a > 0.5))     # ooops
1 loops, best of 3: 2.56 s per loop

In [19]: %timeit np.where(a > 0.5)
1 loops, best of 3: 467 ms per loop

In [20]: %timeit np.column_stack(np.where(a > 0.5))
1 loops, best of 3: 653 ms per loop

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