Python NumPy Convert Columns to Rows

Question

Python 2.7.10 and NumPy. I have a matrix like this:

[[[ 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]]]

Note: The real matrix will have real data, and not consecutive numbers.

I need to rotate, flip, or something (I have tried them all) so as to end up with this:

[[[ 2 5 8 11]
[ 1 4 7 10]
[ 0 3 6 9]

[[14 17 20 23]
[13 16 19 22]
[12 15 18 21]

[[26 29 32 35]
[25 28 31 34]
[24 27 30 33]

[[38 41 44 47]
[37 40 43 46]
[36 39 42 45]]]

Basically, I need the entire columns of the matrix to become the rows.

Thanks.

Divakar · Accepted Answer · 2015-08-09 05:35:57Z

3

Flip the positions of columns with [:,:,::-1] and use np.transpose to swap rows with columns -

In [25]: A
Out[25]: 
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]]])

In [26]: A[:,:,::-1].transpose(0,2,1)
Out[26]: 
array([[[ 2,  5,  8, 11],
        [ 1,  4,  7, 10],
        [ 0,  3,  6,  9]],

       [[14, 17, 20, 23],
        [13, 16, 19, 22],
        [12, 15, 18, 21]],

       [[26, 29, 32, 35],
        [25, 28, 31, 34],
        [24, 27, 30, 33]]])

answered Aug 9, 2015 at 5:35

Divakar

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5 Comments

JayJay123 Over a year ago

Thank you. I appreciate it. I did try both slicing and transpose, but just not the two together.

gbronner Over a year ago

I tested this vs the other approach (fliplr(a).transpose(0,2,1), and the f(f(f(a).T).T).transpose(0,2,1), and found that this took 5.3; the fliplr took 8.3, and the last one took 13.6 seconds using the timeit module.

Divakar Over a year ago

@gbronner Nice! Well I guess it would be interesting to test out A[:,:,::-1].swapaxes(1,2) too!

gbronner Over a year ago

@Divakar swapaxes seemed to be indistinguishable from transpose(0,2,1).

Divakar Over a year ago

@gbronner Yes, functionally they are same!

gbronner · Accepted Answer · 2015-08-10 14:25:05Z

2

Here's a simpler way to do it:

a=numpy.arange(48).reshape((4,4,3)
numpy.fliplr(a.swapaxes(1,2))
#or you could do
numpy.fliplr(a.transpose(0,2,1))

From what I can tell, flipud flips the last dimension, while fliplr flips the second to last dimension. In three dimensions, the last dimension is Z, while the second to last dimension is Y. Hence transposing the data, and flipping the Y dimension works.

Enjoy.

edited Aug 10, 2015 at 14:25

answered Aug 10, 2015 at 2:15

gbronner

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Comments

NJM · Accepted Answer · 2015-08-09 03:21:33Z

1

For each 2d subarray in your super-array you can apply the numpy function:

np.rot90() http://docs.scipy.org/doc/numpy/reference/generated/numpy.rot90.html

so:

import numpy as np

array= 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]]])

desired_output = np.array([np.rot90(sub_array) for sub_array in array])

answered Aug 9, 2015 at 3:21

NJM

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Comments

gbronner · Accepted Answer · 2015-08-10 01:55:09Z

0

transpose and flipud are what you are looking for; the swapaxes can also function as transpose Note that transpose has a version that operates on multiple dimensions.

There may be a simpler expression for this, but this has the advantage of not using elaborate indexing. Example, done in Python 2.7.3 with numpy

f=numpy.flipud
a=numpy.arange(48).reshape((4,4,3))
result=f(f(f(a).T).T).transpose(0,2,1)

In [2]: a=numpy.arange(48).reshape((4,4,3))
Out[3]:
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]]])
In [5]: f(f(f(a).T).T).transpose(0,2,1)
Out[5]:
array([[[ 2,  5,  8, 11],
    [ 1,  4,  7, 10],
    [ 0,  3,  6,  9]],

   [[14, 17, 20, 23],
    [13, 16, 19, 22],
    [12, 15, 18, 21]],

   [[26, 29, 32, 35],
    [25, 28, 31, 34],
    [24, 27, 30, 33]],

   [[38, 41, 44, 47],
    [37, 40, 43, 46],
    [36, 39, 42, 45]]])

.

edited Aug 10, 2015 at 1:55

answered Aug 9, 2015 at 3:24

gbronner

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3 Comments

JayJay123 Over a year ago

Did not work for me. However, I do appreciate the response.

wwii Over a year ago

This does not solve the problem, it transposes the whole array.

gbronner Over a year ago

Actually, it does work. It is just very ugly. I came up with a simpler version that is clearer. To be perfectly honest, the flipud/fliplr statements seem like they could be replaced with a more generic flip by axis, which would be more obvious as one goes from 2 to 3 dimensions.

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