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I've been trying to do something like a numpy.array_split(), but to split it like this instead:

enter image description here So It would return an array (for example let's call it output[] ) with n 2D subarrays inside of it.

For example (for n = 3):

  • output[0] would return the (yellow) subarray with columns a1, a4, a7, a10,
  • output[1] would return the (red) subarray with columns a2, a5, a8,
  • output[2] would return the (blue) subarray with columns a3, a6, a9.
def split(arr, n):
    output= [[] for _ in range(n)]
    for v, help in zip(arr, cycle(out)):
        help.append(v)
    return output

I don't know how to combine rows into one 2D array, so I have many 1D arrays instead of one 2D.

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  • I don't know how to combine rows into one 2D array, so I have many 1D arrays instead of one 2D. Commented Aug 29, 2022 at 12:09
  • edit your question and post your code (properly formatted as code) Commented Aug 29, 2022 at 12:09

3 Answers 3

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2

Not sure if a native solution exists but you can use:

# get groups
group = np.arange(a.shape[1])%n
# groups sorting order
order = np.argsort(group)
# get counts of each group (in order as the output is sorted)
_, idx = np.unique(group, return_counts=True)
# split the reindexed array
out = np.split(a[:, order], np.cumsum(idx[:-1]), axis=1)

Output:

[array([[ 0,  3,  6,  9],
        [10, 13, 16, 19],
        [20, 23, 26, 29],
        [30, 33, 36, 39],
        [40, 43, 46, 49]]),
 array([[ 1,  4,  7],
        [11, 14, 17],
        [21, 24, 27],
        [31, 34, 37],
        [41, 44, 47]]),
 array([[ 2,  5,  8],
        [12, 15, 18],
        [22, 25, 28],
        [32, 35, 38],
        [42, 45, 48]])]

Used input:

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]])
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Comments

1

You can use slicing [start:end:step] in numpy.array. (understanding-slicing)

n = 3
m = inp.shape[1] // n
output = [inp[:, i::3] for i in range(m)]

Example

import numpy as np

inp = np.arange(50).reshape(5, 10)

n = 3
m = inp.shape[1] // n
output = [inp[:, i::3] for i in range(m)]
print(output)

# [
#     array([[ 0,  3,  6,  9],     [0::3] -> columns : 0, 3, 6, 9
#            [10, 13, 16, 19],
#            [20, 23, 26, 29],
#            [30, 33, 36, 39],
#            [40, 43, 46, 49]]), 
 
#     array([[ 1,  4,  7],         [1::3] -> columns : 1, 4, 7
#            [11, 14, 17],
#            [21, 24, 27],
#            [31, 34, 37],
#            [41, 44, 47]]), 
    
#     array([[ 2,  5,  8],         [2::3] -> columns : 2, 5, 8
#            [12, 15, 18],
#            [22, 25, 28],
#            [32, 35, 38],
#            [42, 45, 48]])
# ]

Input array:

#column: 0,  1,  2,  3,  4,  5,  6,  7,  8,  9

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]])
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Comments

1

Another possible solution, based on map:

n = 3
_, ncols = m.shape
out = list(map(lambda x: m[:, np.arange(0, ncols)[x::n]], np.arange(0,n)))
out

Output:

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