Index confusion in numpy arrays

Question

I'm really confused by the index logic of numpy arrays with several dimensions. Here is an example:

import numpy as np
A = np.arange(18).reshape(3,2,3)
[[[ 0,  1,  2],
  [ 3,  4,  5]],

 [[ 6,  7,  8],
  [ 9, 10, 11]],

 [[12, 13, 14],
  [15, 16, 17]]])

this gives me an array of shape (3,2,3), call them (x,y,z) for sake of argument. Now I want an array B with the elements from A corresponding to x = 0,2 y =0,1 and z = 1,2. Like

array([[[ 1,  2],
        [4,  5]],

       [[13, 14],
        [16, 17]]])

Naively I thought that

B=A[[0,2],[0,1],[1,2]]

would do the job. But it gives

array([  2, 104])

and does not work.

A[[0,2],:,:][:,:,[1,2]]

does the job. But I still wonder whats wrong with my first try. And what is the best way to do what I want to do?

My personal take: both of your ways trigger Numpy's advanced indexing. In advanced indexing context, A[[0,2],[0,1],[1,2]] is interpreted as "selected each indexing number in each of your dimension(a somewhat greedy approach). — JP Zhang
– JP Zhang, Commented Oct 17, 2017 at 15:02

Community · Accepted Answer · 2017-05-23 12:25:01Z

18

There are two types of indexing in NumPy basic and advanced. Basic indexing uses tuples of slices for indexing, and does not copy the array, but rather creates a view with adjusted strides. Advanced indexing in contrast also uses lists or arrays of indices and copies the array.

Your first attempt

B = A[[0, 2], [0, 1], [1, 2]]

uses advanced indexing. In advanced indexing, all index lists are first broadcasted to the same shape, and this shape is used for the output array. In this case, they already have the same shape, so the broadcasting does not do anything. The output array will also have this shape of two entries. The first entry of the output array is obtained by using all first indices of the three lists, and the second by using all second indices:

B = numpy.array([A[0, 0, 1], A[2, 1, 2]])

Your second approach

B = A[[0,2],:,:][:,:,[1,2]]

does work, but it is inefficient. It uses advanced indexing twice, so your data will be copied twice.

To get what you actually want with advanced indexing, you can use

A[np.ix_([0,2],[0,1],[1,2])]

as pointed out by nikow. This will copy the data only once.

In your example, you can get away without copying the data at all, using only basic indexing:

B = A[::2, :, 1:2]

edited May 23, 2017 at 12:25

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answered Jan 27, 2011 at 13:59

Sven Marnach

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3 Comments

jonalm Over a year ago

Thanks Sven, but is there a better way to extract indices which arent contius (or regularly spaced) from an array, than using advanced indexing. (My example was too simple, as all the indices im extracting are continously spaced)

Sven Marnach Over a year ago

If your indices aren't regularly spaced, you will of course have to use advanced indexing. The way to go in this case is the ix_ syntax, as already pointed out by nikow. I just wanted to add some more details, such as that your own approach is inefficient for large arrays.

nikow Over a year ago

Excellent answer, and thanks for pointing out my errors. I should have thought about the efficiency implications of advanced indexing.

nikow · Accepted Answer · 2011-01-27 20:33:19Z

7

I recommend the following advanced tutorial, which explains the various indexing methods: NumPy MedKit

Once you understand the powerful ways to index arrays (and how they can be combined) it will make sense. If your first try was valid then this would collide with some of the other indexing techniques (reducing your options in other use cases).

In your example you can exploit that the third index covers a continuous range:

 A[[0,2],:,1:]

You could also use

A[np.ix_([0,2],[0,1],[1,2])]

which is handy in more general cases, when the latter indices are not continuous. np.ix_ simply constructs three index arrays.

As Sven pointed out in his answer, there is a more efficient way in this specific case (using a view instead of a copied version).

Edit: As pointed out by Sven my answer contained some errors, which I have removed. I still think that his answer is better, but unfortunately I can't delete mine now.

edited Jan 27, 2011 at 20:33

answered Jan 27, 2011 at 10:46

nikow

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1 Comment

Sven Marnach Over a year ago

Some remarks: 1. You first two code snippets are not equivalent -- they yield arrays of different shapes. The first line is not interpreted like the second. 2. The solution by jonalm is not advisable since it copies the array twice. I try to explain these issues a bit more in another answer.

user2290820 · Accepted Answer · 2014-06-30 22:36:32Z

0

A[(0,2),:,1:]

If you wanted

array([[[ 1,  2],
        [ 4,  5]],

       [[13, 14],
        [16, 17]]])

A[indices you want,rows you want, col you want]

answered Jun 30, 2014 at 22:36

user2290820

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