Boolean Indexing in Numpy Arrays

Question

I was learning boolean indexing in numpy and came across this. How is the indexing below not producing a Index Error as for axis 0 as there are only two blocks?

x = np.arange(30).reshape(2, 3, 5)
x
array([[[ 0,  1,  2,  3,  4],
    [ 5,  6,  7,  8,  9],
    [10, 11, 12, 13, 14]],

   [[15, 16, 17, 18, 19],
    [20, 21, 22, 23, 24],
    [25, 26, 27, 28, 29]]])

x[[[True, True, False], [False, True, True]]]
array([[ 0,  1,  2,  3,  4],
   [ 5,  6,  7,  8,  9],
   [20, 21, 22, 23, 24],
   [25, 26, 27, 28, 29]])

The boolean indexing docs suggest using np.nonzero(...) to see the equivalent advanced indexing arrays, e.g. (array([0, 0, 1, 1]), array([0, 1, 1, 2])). So the equivalent operation is x[[0,0,1,1], [0,1,1,2], :] — hpaulj
– hpaulj, Commented Feb 27 at 18:41

mozway · Accepted Answer · 2025-02-27 12:40:01Z

You are performing boolean array indexing, which is fine.

You would have an indexing error with:

#  first dimension      second dimension
x[[True, True, False], [False, True, True]]
# IndexError: boolean index did not match indexed array along dimension 0;
# dimension is 2 but corresponding boolean dimension is 3

However, in your case, you have an extra set of brackets, which makes it index only the first dimension, using an array:

# [             first dimension            ]
#  [ second dimension], [ second dimension]
x[[[True, True, False], [False, True, True]]]

This means, that using a 2x3 array, you request in the second dimension, for the first "row" [True, True, False], and for the second "row" [False, True, True].

Since your array shape matches the first two dimensions, this is valid, and more or less equivalent to:

np.concatenate([x[0][[True, True, False]],
                x[1][[False, True, True]]])

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