Advanced 3d numpy array slicing with alternation

Question

So, I want to slice my 3d array to skip the first 2 arrays and then return the next two arrays. And I want the slice to keep following this pattern, alternating skipping 2 and giving 2 arrays etc.. I have found a solution, but I was wondering if there is a more elegant way to go about this? Preferably without having to reshape?

arr = np.arange(1, 251).reshape((10, 5, 5))
sliced_array = np.concatenate((arr[2::4], arr[3::4]), axis=1).ravel().reshape((4, 5, 5))

Try using this notation `arr[[2, 3, 6, 7], :, :]`. // You can supply lists of indices to pull from specific axes of matrices. Further, these lists only really need to be made up of integers and be within range. So you could provide the list `[2, 2, 1, 3, 3]` and get multiple copies of an axis and not necessarily in order either! — KDecker, Jan 19 '23 at 15:24
Cool, I didn't know all that. I tried to make it a bit more generalized so that it can be any length: `slice = np.array([[i, i+1] for i in range(2, 10, 4)]).flatten()` and `sliced_array = arr[slice, :, :]`. Does that make sense? — Tea, Jan 19 '23 at 15:36
`np.arange(10).reshape(5,2)[1::2]` should produce `[[2,3],[6,7]]` — hpaulj, Jan 19 '23 at 16:33

score 1 · Accepted Answer · answered Jan 19 '23 at 16:02

You can use boolean indexing using a mask that repeats [False, False, True, True, ...]:

import numpy as np

arr = np.arange(1, 251).reshape((10, 5, 5))
mask = np.arange(arr.shape[0]) % 4 >= 2

out = arr[mask]

out:

array([[[ 51,  52,  53,  54,  55],
        [ 56,  57,  58,  59,  60],
        [ 61,  62,  63,  64,  65],
        [ 66,  67,  68,  69,  70],
        [ 71,  72,  73,  74,  75]],

       [[ 76,  77,  78,  79,  80],
        [ 81,  82,  83,  84,  85],
        [ 86,  87,  88,  89,  90],
        [ 91,  92,  93,  94,  95],
        [ 96,  97,  98,  99, 100]],

       [[151, 152, 153, 154, 155],
        [156, 157, 158, 159, 160],
        [161, 162, 163, 164, 165],
        [166, 167, 168, 169, 170],
        [171, 172, 173, 174, 175]],

       [[176, 177, 178, 179, 180],
        [181, 182, 183, 184, 185],
        [186, 187, 188, 189, 190],
        [191, 192, 193, 194, 195],
        [196, 197, 198, 199, 200]]])

score 0 · Answer 2 · answered Jan 19 '23 at 17:16

Since you want to select, and skip, the same numbers, reshaping works.

For a 1d array:

In [97]: np.arange(10).reshape(5,2)[1::2]
Out[97]: 
array([[2, 3],
       [6, 7]])

which can then be ravelled.

Generalizing to more dimensions:

In [98]: x = np.arange(100).reshape(10,10)    
In [99]: x.reshape(5,2,10)[1::2,...].reshape(-1,10)
Out[99]: 
array([[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
       [70, 71, 72, 73, 74, 75, 76, 77, 78, 79]])

I won't go on to 3d because the display will be longer, but it should be straight forward.

Advanced 3d numpy array slicing with alternation

2 Answers2