Reshape 3D array to 2D array Python

Question

If I have a 3D array of ([4,3,3]) like this:

[[0,1,2]   [[9,10,11 ]    [[18,19,20]   [[27,28,29] 
 [3,4,5]    [12,13,14]     [21,22,23]    [30,31,32]
 [6,7,8]] , [15,16,17]] ,  [24,25,26]] , [33,34,35]]

How would I convert it to a 2D array of ([6,6]) like this so that the 1st half of arrays are at the top half of the 160x160 and the 2nd half are at the bottom:

[[0,1,2,9,10,11]
[3,4,5,12,13,14]
[6,7,8,15,16,17]
[18,19,20,27,28,29] 
[21,22,23,30,31,32]
[24,25,26,33,34,35]]

My array creation:

qDCTReversed = np.zeros((400,8,8), dtype=np.int)

And I need a (160,160) array.

The problem is that the input array do not seem shaped 4, 3,3 — Dani Mesejo
– Dani Mesejo, Commented Feb 8, 2019 at 12:44
You're data structure is not clear, neither in your comment or in your post. Therefore I'm downvoting it. — Nathan
– Nathan, Commented Feb 8, 2019 at 12:59

markuscosinus · Accepted Answer · 2019-02-08 13:37:14Z

2

A very fast one line solution using no for-loops is this:

# initialization
qDCTReversed = np.arange(4*3*3).reshape((4,3,3)) 

# calculation
qDCTReversed = qDCTReversed.reshape((2,2,3,3)).transpose((0,2,1,3)).reshape((6,6))

or for the (400,8,8) array:

qDCTReversed.reshape((20,20,8,8)).transpose((0,2,1,3)).reshape((160,160))

Speed comparison:

Mstaino's answer: 0.393 ms

yatu's answer: 0.138 ms

This answer: 0.016 ms

edited Feb 8, 2019 at 13:37

answered Feb 8, 2019 at 13:19

markuscosinus

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2 Comments

yatu Over a year ago

The problem with this answer is that you are hardcoding everything. The reshapes will have to be modified according to the shape of qDCTReversed

markuscosinus Over a year ago

The numbers can easiliy be replaced by calculations from qDCTReversed.shape. The reason I hardcoded it is, so that it's easier to read and understand. I can edit it if wished or needed.

Nathan · Accepted Answer · 2019-02-08 12:56:40Z

1

You can do this by looping over the list as such:

a = [[[ 0, 1, 2], [ 9,10,11]],
     [[ 3, 4, 5], [12,13,14]],
     [[ 6, 7, 8], [15,16,17]],
     [[18,19,20], [27,28,29]],
     [[21,22,23], [30,31,32]],
     [[24,25,26], [33,34,35]]]

b = [[i for j in k for i in j ] for k in a]
print(b)

outputs:

[ 0,  1,  2,  9, 10, 11]
[ 3,  4,  5, 12, 13, 14]
[ 6,  7,  8, 15, 16, 17]
[18, 19, 20, 27, 28, 29]
[21, 22, 23, 30, 31, 32]
[24, 25, 26, 33, 34, 35]

edited Feb 8, 2019 at 12:56

answered Feb 8, 2019 at 12:45

Nathan

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3 Comments

Nils Werner Over a year ago

OP is talking about numpy arrays, not lists

Harry Reid Over a year ago

a[0] is not "[[0,1,2], [9,10,11 ]]" it is "[[0,1,2],[3,4,5],[6,7,8],[9,10,11 ],[12,13,14]]"

Nathan Over a year ago

@HarryReid That wasn't clear from the original post, let me fix that

Tarifazo · Accepted Answer · 2019-02-08 13:03:12Z

1

The reshape you ask can be done with:

x = np.arange(36).reshape((4,3,3))
np.vstack(np.hstack(x[2*i:2+2*i]) for i in range(x.shape[0]//2))
>>array([[ 0,  1,  2,  9, 10, 11],
   [ 3,  4,  5, 12, 13, 14],
   [ 6,  7,  8, 15, 16, 17],
   [18, 19, 20, 27, 28, 29],
   [21, 22, 23, 30, 31, 32],
   [24, 25, 26, 33, 34, 35]])

answered Feb 8, 2019 at 13:03

Tarifazo

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2 Comments

Harry Reid Over a year ago

Thanks I appreciate it. Is there a way I can scale it to an array of (400,8,8)? When I try it outputs an array of (1600x16)

Tarifazo Over a year ago

Yes, replace the 2's by 20's: np.vstack(np.hstack(x[20*i:20+20*i]) for i in range(x.shape[0]//20)).shape.

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