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Indexing Numpy Array With Index Array Of Lower Dim Yields Array Of Higher Dim Than Both

- 1 answer

a = np.zeros((5,4,3))
v = np.ones((5, 4), dtype=int)
data = a[v]
shp = data.shape

This code gives shp==(5,4,4,3)

I don't understand why. How can a larger array be output? makes no sense to me and would love an explanation.

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Answer

This is known as advanced indexing. Advanced indexing allows you to select arbitrary elements in the input array based on an N-dimensional index.

Let's use another example to make it clearer:

a = np.random.randint(1, 5, (5,4,3))
v = np.ones((5, 4), dtype=int)

Say in this case a is:

array([[[2, 1, 1],
        [3, 4, 4],
        [4, 3, 2],
        [2, 2, 2]],

       [[4, 4, 1],
        [3, 3, 4],
        [3, 4, 2],
        [1, 3, 1]],

       [[3, 1, 3],
        [4, 3, 1],
        [2, 1, 4],
        [1, 2, 2]],
        ...

By indexing with an array of np.ones:

print(v)

array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]])

You will simply be indexing a with 1 along the first axis as many times as v. Putting it in another way, when you do:

a[1]

[[4, 4, 1],
 [3, 3, 4],
 [3, 4, 2],
 [1, 3, 1]]

You're indexing along the first axis, as no indexing is specified along the additional axes. It is the same as doing a[1, ...], i.e taking a full slice along the remaining axes. Hence by indexing with a 2D array of ones, you will have the above 2D array (5, 4) times stacked together, resulting in an ndarray of shape (5, 4, 4, 3). Or in other words, a[1], of shape (4,3), stacked 5*4=20 times.

Hence, in this case you'd be getting:

array([[[[4, 4, 1],
         [3, 3, 4],
         [3, 4, 2],
         [1, 3, 1]],

        [[4, 4, 1],
         [3, 3, 4],
         [3, 4, 2],
         [1, 3, 1]],
         ...
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source: stackoverflow.com
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