An unusual resorting of a 3d numpy array

Question

Consider the 3d numpy array arr:

[[[ 0,  9,  0,  2], [ 6,  8,  0,  2]],
 [[ 0,  9,  1,  5], [ 4,  8,  1,  5]],
 [[ 1,  0,  1,  2], [ 6,  6,  1,  2]],
 [[ 0,  9,  1,  5], [ 5,  5,  1,  5]],
 [[ 0,  9,  1,  5], [ 6,  5,  1,  5]],
 [[ 1,  0,  1,  2], [ 4,  7,  1,  2]],  
 [[ 0,  9,  3,  4], [ 6,  4,  3,  4]]])

I want to create a new array, new_arr, of the same shape by resorting arr as follows:

Sort by arr[:, 0, 0], then arr[:, 1, 0], then arr[:, 0, 1], then arr[:, 1, 1], etc.

The 2d arrays themselves are unchanged, but they are presented in a different (lexicographic) order within the 3d array.

The desired result is:

new_arr =  np.array(
[[[ 0,  9,  1,  5], [ 4,  8,  1,  5]],
 [[ 0,  9,  1,  5], [ 5,  5,  1,  5]],
 [[ 0,  9,  3,  4], [ 6,  4,  3,  4]],
 [[ 0,  9,  1,  5], [ 6,  5,  1,  5]],
 [[ 0,  9,  0,  2], [ 6,  8,  0,  2]], 
 [[ 1,  0,  1,  2], [ 4,  7,  1,  2]], 
 [[ 1,  0,  1,  2], [ 6,  6,  1,  2]]])

Added answer using numpy.lexsort

LMC
– LMC

2025-06-26 02:31:06 +00:00
Commented Jun 26 at 2:31 — LMC
– LMC, Commented Jun 26 at 2:31

Lucenaposition · Accepted Answer · 2025-06-26 02:04:06Z

2

You can do it like this:

import numpy as np
arr=np.array([[[ 0,  9,  0,  2], [ 6,  8,  0,  2]],
 [[ 0,  9,  1,  5], [ 4,  8,  1,  5]],
 [[ 1,  0,  1,  2], [ 6,  6,  1,  2]],
 [[ 0,  9,  1,  5], [ 5,  5,  1,  5]],
 [[ 0,  9,  1,  5], [ 6,  5,  1,  5]],
 [[ 1,  0,  1,  2], [ 4,  7,  1,  2]],  
 [[ 0,  9,  3,  4], [ 6,  4,  3,  4]]])
new_arr=list(arr)
new_arr.sort(key=lambda x:[x[0][0],x[1][0],x[0][1],x[1][1]])
new_arr=np.array(new_arr)
print(new_arr)

edited Jun 26 at 2:04

answered Jun 26 at 0:40

Lucenaposition

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Comments

Onyambu · Accepted Answer · 2025-06-27 02:10:38Z

1

Another method include:

import numpy as np

rule = [0,0],[1,0],[0,1],[1,1]
new_arr[np.lexsort(np.fliplr(new_arr[:,*np.array(rule).T]).T)]

You could also simply do:

new_arr[np.lexsort(np.fliplr(new_arr[:,[0,1,0,1],[0,0,1,1]]).T)]

answered Jun 27 at 2:10

Onyambu

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2 Comments

LMC Jun 27 at 12:58

So [0,0] is the primary key but it's passed last to lexsort. Still don't get why but guess it's a numpy internal stuff.

Onyambu Jun 27 at 14:57

According to documentation, if you are sorting by a and breaking the ties by b then you use lexsort((b,a)) and not lexsort((a, b)). ie the key starts from last to the firs. In this case, I first sort using 0,0,1,1 then 0,1,0,1

LMC · Accepted Answer · 2025-06-26 02:40:10Z

0

Using numpy.lexsort and specifying keys in reverse order and formatted as requested by OP.

import numpy as np

new_arr = np.array([[[ 0,  9,  0,  2], [ 6,  8,  0,  2]],
 [[ 0,  9,  1,  5], [ 4,  8,  1,  5]],
 [[ 1,  0,  1,  2], [ 6,  6,  1,  2]],
 [[ 0,  9,  1,  5], [ 5,  5,  1,  5]],
 [[ 0,  9,  1,  5], [ 6,  5,  1,  5]],
 [[ 1,  0,  1,  2], [ 4,  7,  1,  2]],  
 [[ 0,  9,  3,  4], [ 6,  4,  3,  4]]])

keys = (new_arr[:,1,1], new_arr[:, 0, 1], new_arr[:,1,0], new_arr[:,0,0])
sorted_indices = np.lexsort(keys)

arr = new_arr[sorted_indices]
print(arr)

[[[0 9 1 5] [4 8 1 5]]
 [[0 9 1 5] [5 5 1 5]]
 [[0 9 3 4] [6 4 3 4]]
 [[0 9 1 5] [6 5 1 5]]
 [[0 9 0 2] [6 8 0 2]]
 [[1 0 1 2] [4 7 1 2]]
 [[1 0 1 2] [6 6 1 2]]]

edited Jun 26 at 2:40

answered Jun 26 at 2:27

LMC

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