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

## 09 July 2019 - 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.

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]],
...