3D array to 2D then back again

Question

I have the array

import numpy as np
a = np.array([[[11,12,13], [14,15,16]],
              [[21,22,23], [24,25,26]],
              [[31,32,33], [34,35,36]]])

# array([[[11, 12, 13],
#         [14, 15, 16]],

#        [[21, 22, 23],
#         [24, 25, 26]],

#        [[31, 32, 33],
#         [34, 35, 36]]])

#a.shape
#(3, 2, 3)

I need to reshape it to a 2D array with three columns:

step1 = a.transpose(1, 2, 0).reshape(-1, 3)
# array([[11, 21, 31],
#        [12, 22, 32],
#        [13, 23, 33],
#        [14, 24, 34],
#        [15, 25, 35],
#        [16, 26, 36]])

I then need to reshape it back to the original shape, but I cant figure out how?

step1.reshape(2, 3, 3).transpose(2, 0, 1)?

Nin17
– Nin17

2025-06-30 19:59:01 +00:00
Commented Jun 30 at 19:59 — Nin17
– Nin17, Commented Jun 30 at 19:59

ThomasIsCoding · Accepted Answer · 2025-07-01 13:46:46Z

3

You probably could try

step1 = a.reshape(3,-1).T
step2 = step1.T.reshape(a.shape)

print(f'step1 :\n {step1}\n')
print(f'step2 :\n {step2}\n')

and you will see

step1 :
 [[11 21 31]
 [12 22 32]
 [13 23 33]
 [14 24 34]
 [15 25 35]
 [16 26 36]]

step2 :
 [[[11 12 13]
  [14 15 16]]

 [[21 22 23]
  [24 25 26]]

 [[31 32 33]
  [34 35 36]]]

edited Jul 1 at 13:46

answered Jun 30 at 22:44

ThomasIsCoding

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2 Comments

hpaulj Jul 1 at 13:42

Is the explicit flatten needed? I think the reshape effectively does that before applying the new shape.

ThomasIsCoding Jul 1 at 13:47

@hpaulj aha, you are right, thanks for helping to improve the code :)

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