Forming an array with elements in specific positions from multiple arrays in Python

Question

I have an array A. I want to take the first, second,...,ninth elements of each A[0],A[1],A[2] to form a new array B. I present the current and expected outputs.

import numpy as np

A=np.array([np.array([[1, 2, 3],
              [4, 5, 6 ],
              [7, 8, 9]]),
       np.array([[[10, 11, 12],
               [13, 14, 15 ],
               [16, 17, 18]]]),
       np.array([[[19, 20, 21],
               [22, 23, 24],
               [25, 26, 27]]])], dtype=object)

for t in range(0,len(A)):
    B=A[0][t][0]
    print([B])

The current output is

[1]
[4]
[7]

The expected output is

array([[1,10,19],
        [2,11,20],
        [3,12,21],
        [4,13,22],
        [5,14,23],
        [6,15,24],
        [7,16,25],
        [8,17,26],
        [9,18,27]])

If A is a numpy array then it is just A.reshape(A.shape[0], -1).T — paime
– paime, Commented Jul 1, 2022 at 11:11
your first submatrix is misshapen. it just has 2 dimensions. the other two have 3 dimensions. you should try really hard to avoid numpy arrays of dtype object — Christoph Rackwitz
– Christoph Rackwitz, Commented Jul 1, 2022 at 18:38

Cardstdani · Accepted Answer · 2022-06-30 13:38:04Z

3

You can traverse the array, append all values as columns and transpose the resulting matrix:

import numpy as np

A=np.array([np.array([[1, 2, 3],
              [4, 5, 6 ],
              [7, 8, 9]]),
       np.array([[[10, 11, 12],
               [13, 14, 15 ],
               [16, 17, 18]]]),
       np.array([[[19, 20, 21],
               [22, 23, 24],
               [25, 26, 27]]])], dtype=object)
               
out = np.array([A[i].flatten() for i in range(len(A))]).transpose()
#out = np.array([i.flatten() for i in A]).transpose() #Second option
print(out)

Output:

[[ 1 10 19]
 [ 2 11 20]
 [ 3 12 21]
 [ 4 13 22]
 [ 5 14 23]
 [ 6 15 24]
 [ 7 16 25]
 [ 8 17 26]
 [ 9 18 27]]

edited Jun 30, 2022 at 13:38

answered Jun 30, 2022 at 13:32

Cardstdani

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Salvatore Daniele Bianco · Accepted Answer · 2022-06-30 13:37:07Z

1

B = np.array([a.ravel() for a in A]).T

#array([[ 1, 10, 19],
#       [ 2, 11, 20],
#       [ 3, 12, 21],
#       [ 4, 13, 22],
#       [ 5, 14, 23],
#       [ 6, 15, 24],
#       [ 7, 16, 25],
#       [ 8, 17, 26],
#       [ 9, 18, 27]])

answered Jun 30, 2022 at 13:37

Salvatore Daniele Bianco

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Comments

Christoph Rackwitz · Accepted Answer · 2022-07-01 18:45:24Z

Your array is broken.

If you fix it, you can deal with it a lot easier. .squeeze() shakes useless dimensions out of a numpy array.

numpy lets you transpose dimensions.

ravel flattens an array. That's a valid approach but I've chosen to do the flattening using reshape

a1 = np.array([[1, 2, 3],
               [4, 5, 6],
               [7, 8, 9]])
a2 = np.array([[[10, 11, 12],
                [13, 14, 15],
                [16, 17, 18]]])
a3 = np.array([[[19, 20, 21],
                [22, 23, 24],
                [25, 26, 27]]])

A = [a1, a2, a3]
print([a.shape for a in A])
# [(3, 3), (1, 3, 3), (1, 3, 3)]

A = [a.squeeze() for a in A]
print([a.shape for a in A])
# [(3, 3), (3, 3), (3, 3)]

A = np.array(A)
print(repr(A))
# array([[[ 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]]])

A.transpose((1, 2, 0)) # explicit form of A.transpose()
# array([[[ 1, 10, 19],
#         [ 2, 11, 20],
#         [ 3, 12, 21]],
#
#        [[ 4, 13, 22],
#         [ 5, 14, 23],
#         [ 6, 15, 24]],
#
#        [[ 7, 16, 25],
#         [ 8, 17, 26],
#         [ 9, 18, 27]]])

A.transpose((1, 2, 0)).reshape((-1, 3))
# array([[ 1, 10, 19],
#        [ 2, 11, 20],
#        [ 3, 12, 21],
#        [ 4, 13, 22],
#        [ 5, 14, 23],
#        [ 6, 15, 24],
#        [ 7, 16, 25],
#        [ 8, 17, 26],
#        [ 9, 18, 27]])

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