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 · 2017-03-08 00:45:59Z

4

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)

edited Mar 8, 2017 at 0:45

answered Mar 7, 2017 at 16:17

Divakar

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3 Comments

Jan Christoph Terasa Over a year ago

I am amazed at how you can think in 6D after seeing the self-similarity of the "sudoku" blocks!

vasco_t Over a year ago

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))?

Divakar Over a year ago

@vasco_t Check out the just added Generic case solution. So, that BSZ = [a,b].

gzc · Accepted Answer · 2017-03-07 16:45:57Z

1

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]

edited Mar 7, 2017 at 16:45

answered Mar 7, 2017 at 16:26

gzc

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1 Comment

vasco_t Over a year ago

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))?

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