Conversion from a Numpy 3D array to a 2D array

Question

Having the following 3D array (9,9,9):

>>> np.arange(729).reshape((9,9,9))
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  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  75  76  77  78  79  80]]
      ...
       [[648 649 650 651 652 653 654 655 656]
        [657 658 659 660 661 662 663 664 665]
        [666 667 668 669 670 671 672 673 674]
        [675 676 677 678 679 680 681 682 683]
        [684 685 686 687 688 689 690 691 692]
        [693 694 695 696 697 698 699 700 701]
        [702 703 704 705 706 707 708 709 710]
        [711 712 713 714 715 716 717 718 719]
        [720 721 722 723 724 725 726 727 728]]])

How do I reshape it to look like this 2D array (27,27):

Divakar · Accepted Answer · 2017-03-08T00:45:59.853

You need to go 6D with one reshaping basically splitting each axes into two, then transpose to push back the even axes (2nd, 4th and 6th) to the end and a final reshape back to 2D -

a.reshape(-1,3,3,3,3,3).transpose(0,2,4,1,3,5).reshape(27,27)

Sample run -

In [28]: a = np.arange(729).reshape((9,9,9))

In [29]: out = a.reshape(-1,3,3,3,3,3).transpose(0,2,4,1,3,5).reshape(27,27)

In [30]: out[0]
Out[30]: 
array([  0,   1,   2,   9,  10,  11,  18,  19,  20,  81,  82,  83,  90,
        91,  92,  99, 100, 101, 162, 163, 164, 171, 172, 173, 180, 181, 182])

In [31]: out[1]
Out[31]: 
array([  3,   4,   5,  12,  13,  14,  21,  22,  23,  84,  85,  86,  93,
        94,  95, 102, 103, 104, 165, 166, 167, 174, 175, 176, 183, 184, 185])

In [32]: out[2]
Out[32]: 
array([  6,   7,   8,  15,  16,  17,  24,  25,  26,  87,  88,  89,  96,
        97,  98, 105, 106, 107, 168, 169, 170, 177, 178, 179, 186, 187, 188])

In [33]: out[3]
Out[33]: 
array([ 27,  28,  29,  36,  37,  38,  45,  46,  47, 108, 109, 110, 117,
       118, 119, 126, 127, 128, 189, 190, 191, 198, 199, 200, 207, 208, 209])

In [34]: out[-1]
Out[34]: 
array([546, 547, 548, 555, 556, 557, 564, 565, 566, 627, 628, 629, 636,
       637, 638, 645, 646, 647, 708, 709, 710, 717, 718, 719, 726, 727, 728])

Generic case solution

BSZ = [3,3] # Block size
p,q = BSZ
out = a.reshape(p,q,p,q,p,q).transpose(0,2,4,1,3,5).reshape(p**3,q**3)

I am amazed at how you can think in `6D` after seeing the self-similarity of the "sudoku" blocks! — Jan Christoph Terasa, Mar 07 '17 at 16:22
Thank you, that's exactly it! But what if I have a different "ratio", like instead of using 3 by 3, use a by b (cenverting a 3D array `(a*b,a*b,a*b)` to a 2D array `(a*a*a,b*b*b)`)? — vasco_t, Mar 07 '17 at 23:39
@vasco_t Check out the just added `Generic case solution`. So, that `BSZ = [a,b]`. — Divakar, Mar 08 '17 at 00:46

gzc · Answer 2 · 2017-03-07T16:45:57.993

Divide and Conquer is very applicable to such problem. It's easier to understand compared with numpy magic. And this can solve not only 27 * 27 problem, also all 3 power problems such as case of 81 * 81 . Show the code first.

import numpy as np

# np.arange(729).reshape((9,9,9)).flatten()
arr = np.arange(729)

result =  np.empty((27,27))

def assign(width, start, end, anchor_x, anchor_y):
    if width > 3:
        sub_width = width / 3
        for i in range(3):
            for j in range(3):
                assign(
                    sub_width,
                    start + (i * 3 + j) * sub_width ** 2,
                    start + (i * 3 + j) * sub_width ** 2 + sub_width ** 2,
                    anchor_x + i * sub_width,
                    anchor_y + j * sub_width)

    else:
        result[anchor_x:anchor_x+3, anchor_y:anchor_y+3] = arr[start:end].reshape(3,3)

assign(27, 0, 729, 0, 0)

print(result)

Explanation:

Every time you divide a big n*n matrix to 9 smaller (n/3) * (n/3) matrix, and solve them each recursively. Until width of matrix equals to 3, stop recursion and copy 9 numbers (arr[start:end]) to result[anchor_x:anchor_x+3, anchor_y:anchor_y+3]

Thank you! That also solves the problem. But what if I have a different "ratio", like instead of using 3 by 3, use a by b (cenverting a 3D array `(a*b,a*b,a*b)` to a 2D array `(a*a*a,b*b*b)`)? — vasco_t, Mar 07 '17 at 23:43

Conversion from a Numpy 3D array to a 2D array

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