CODEWITHSUNDEEP
pythonarraysnumpy
zlpython
zlpython

Reputation: 29

python numpy array indexing

I think I understand the indexing of array in python/numpy correctly. But today I met a problem as follows:

I have a 6-d array e.g. A and A.shape = (11,1,9,1,5,7). Then I use the indexing as follows:

B = A[:,0,0,0,[3,4,2],0]

and B.shape = (11,3) as expected;

C = A[:,0,0,0,[3,4,2],:]

and C.shape = (11,3,7) as expected;

But when I say:

D = A[:,0,:,0,[3,4,2],0]

and D.shape should be (11,9,3) as I can expect, however, python returned the D.shape = (3, 11, 9).

And I'm really confused about the shape of array D.

Is there any one can give me a brief explanation? Thanks a lot!

Upvotes: 1

Views: 189

Answers (1)

hpaulj
hpaulj

Reputation: 231325

As discussed in https://docs.scipy.org/doc/numpy-1.12.0/reference/arrays.indexing.html#combining-advanced-and-basic-indexing

A[:,0,:,0,[3,4,2],0]

indexes with the 'advanced' list, [3,4,2] producing the size 3 dimension. And the 1st and 3rd dimensions are added on after, resulting in the (3,11,9) shape.

This behavior is somewhat controversial, especially when the other indices are scalars. The justification given in the docs is clearer when there are two indexing lists.

Numpy sub-array assignment with advanced, mixed indexing

Upvotes: 2

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