Python numpy array indexing. How is this working?

Question

I came across this python code (which works) and to me it seems amazing. However, I am unable to figure out what this code is doing. To replicate it, I sort of wrote a test code:

import numpy as np

# Create a random array which represent the 6 unique coeff. 
# of a symmetric 3x3 matrix
x = np.random.rand(10, 10, 6)

So, I have 100 symmetric 3x3 matrices and I am only storing the unique components. Now, I want to generate the full 3x3 matrix and this is where the magic happens.

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

I see what this is doing. This is how the 0-5 index components should be arranged in the 3x3 matrix to have a symmetric matrix.

mat = x[..., indices]

This line has me lost. So, it is working on the last dimension of the x array but it is not at all clear to me how the rearrangement and reshaping is done but this indeed returns an array of shape (10, 10, 3, 3). I am amazed and confused!

My answer to this question: stackoverflow.com/questions/10921893/… may help. — user545424
– user545424, Commented Jan 8, 2015 at 23:38
I started to write an answer but found myself largely repeating this reference. If the reference is not clear, maybe update your question with the part you don't understand and we can help you more. — Bi Rico
– Bi Rico, Commented Jan 9, 2015 at 1:12

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

From the advanced indexing documentation - bi rico's link.

Example

Suppose x.shape is (10,20,30) and ind is a (2,3,4)-shaped indexing intp array, thenresult = x[...,ind,:] has shape (10,2,3,4,30) because the (20,)-shaped subspace has been replaced with a (2,3,4)-shaped broadcasted indexing subspace. If we let i, j, kloop over the (2,3,4)-shaped subspace then result[...,i,j,k,:] =x[...,ind[i,j,k],:]. This example produces the same result as x.take(ind, axis=-2).

edited Jun 20, 2020 at 9:12

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answered Jan 9, 2015 at 8:49

hpaulj

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