Numpy array [? 5 5] to five arrays with [? 5]

Question

I have a [? 5 5] array: (?=3 in this case)

[[[ 0  1  2  3  4]
 [ 5  6  7  8  9]
 [10 11 12 13 14]
 [15 16 17 18 19]
 [20 21 22 23 24]]
[[25 26 27 28 29]
 [30 31 32 33 34]
 [35 36 37 38 39]
 [40 41 42 43 44]
 [45 46 47 48 49]]
[[50 51 52 53 54]
 [55 56 57 58 59]
 [60 61 62 63 64]
 [65 66 67 68 69]
 [70 71 72 73 74]]]

I want to have 5 (the # of row) separate array have (? 5) like this:

[array([[ 0,  1,  2,  3,  4],
   [25, 26, 27, 28, 29],
   [50, 51, 52, 53, 54]]), 
 array([[ 5,  6,  7,  8,  9],
   [30, 31, 32, 33, 34],
   [55, 56, 57, 58, 59]]), 
 array([[10, 11, 12, 13, 14],
   [35, 36, 37, 38, 39],
   [60, 61, 62, 63, 64]]), 
 array([[15, 16, 17, 18, 19],
   [40, 41, 42, 43, 44],
   [65, 66, 67, 68, 69]]), 
 array([[20, 21, 22, 23, 24],
   [45, 46, 47, 48, 49],
   [70, 71, 72, 73, 74]])]

Is this a simple preferably one/two numpy operation way to do this?

I'd propose renaming the question into Numpy array [3 5 5] to [5 3 5] or ...to 5x [3 5] to make it more useful to others. Apparently the word transpose is not obvious to everybody looking at multidimensional data for the first time, so this is where the value in keeping this question would be. Otherwise I'm sure this would be a duplicate to other existing transpose questions. — roadrunner66
– roadrunner66, Commented Apr 11, 2016 at 17:48
@roadrunner66 I use ?, since it could be anything. Do you think changing it to 5 is better for others? Thanks! — Sung Kim
– Sung Kim, Commented Apr 11, 2016 at 20:13
Yeah, I'm not sure, I just wanted to find away that others with a similar problem (who don't know the word transpose) will find your question. Maybe reorder dimensions of numpy array ? — roadrunner66
– roadrunner66, Commented Apr 11, 2016 at 22:10

roadrunner66 · Accepted Answer · 2016-04-11 05:45:16Z

2

Yes, there is : numpy.transpose. You get to chose any sequence in the axes of your 3D structure.

import numpy as np

aaa=np.array([[[ 0 , 1,  2,  3,  4],
 [ 5,  6,  7,  8,  9],
 [10, 11, 12, 13, 14],
 [15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24]],
[[25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34],
 [35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44],
 [45, 46, 47, 48, 49]],
[[50, 51, 52, 53, 54],
 [55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64],
 [65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]])

bb= np.transpose(aaa,axes=[1,0,2])
print bb

output:

[[[ 0  1  2  3  4]
  [25 26 27 28 29]
  [50 51 52 53 54]]

 [[ 5  6  7  8  9]
  [30 31 32 33 34]
  [55 56 57 58 59]]

 [[10 11 12 13 14]
  [35 36 37 38 39]
  [60 61 62 63 64]]

 [[15 16 17 18 19]
  [40 41 42 43 44]
  [65 66 67 68 69]]

 [[20 21 22 23 24]
  [45 46 47 48 49]
  [70 71 72 73 74]]].

To access the subarrays, just use indexing like so:

 c= b[0]
 print c

Output:

 [[ 0  1  2  3  4]
 [25 26 27 28 29]
 [50 51 52 53 54]]

edited Apr 11, 2016 at 5:45

answered Apr 11, 2016 at 5:26

roadrunner66

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3 Comments

Sung Kim Over a year ago

Thanks! Then, should I apply split to get five arrays? Can I do it in one operation?

roadrunner66 Over a year ago

No split needed. To access the five subarrays, just use indexing : print b[0] print b[1] etc. or reassign to whatever you like : c1=b[0] etc.

Imanol Luengo Over a year ago

@SungKim Additionally, you could just write a, b, c, d, e = np.transpose(aaa, axes=(1, 0, 2)).

Imanol Luengo · Accepted Answer · 2016-04-11 08:31:25Z

2

Just to propose an alternative solution, np.swapaxes is another alternative to np.transpose when only a pair of axes is involved.

a, b, c, d, e = arr.swapaxes(0, 1) # swap axes 0, 1

swapaxes will always return a view of the array, with the same result as the np.transpose proposed by @roadrunner66. It is usually just sightly faster than transpose and heavily used inside numpy's code to put the important dimension up-front.

edited Apr 11, 2016 at 8:31

answered Apr 11, 2016 at 8:19

Imanol Luengo

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