Reshape matrix by taking two consecutive rows every two rows

Hello everyone,

I have seen quite some questions here on the forum that are about matrix reshaping. However, for my particular problem I did not find an answer. So, here we are:

I have got a matrix that looks like

[1 1 1 1;
 2 2 2 2;
 3 3 3 3;
 4 4 4 4;
 5 5 5 5;
 6 6 6 6;
 7 7 7 7;
 8 8 8 8]

Is it somehow possible to reshape it to

[1 1 1 1 3 3 3 3;
 2 2 2 2 4 4 4 4;
 5 5 5 5 7 7 7 7;
 6 6 6 6 8 8 8 8]

with the built-in functions like reshape, permutate, flip,...?

There would always be the way with for loops but I thought it might be neater (and possibly faster) to do it with the built-in functions.

Thanks for your help.

Answer 1

[Adam Danz]

x must be an n*m matrix where n is divisible of 4.

M is the reformatted matrix with size k*j where k=n/2 and j=m*2.

x = [1 1 1 1;
    2 2 2 2;
    3 3 3 3;
    4 4 4 4;
    5 5 5 5;
    6 6 6 6;
    7 7 7 7;
    8 8 8 8];

rowIdx =  [1 3 2 4]' + [0:4:size(x,1)-1];
M = reshape(x(rowIdx(:),:)', size(x,2)*2, [])'

M = 4×8

1 1 1 1 3 3 3 3 2 2 2 2 4 4 4 4 5 5 5 5 7 7 7 7 6 6 6 6 8 8 8 8

Another example,

x = [1 1 1 1;
    2 2 2 2;
    3 3 3 3;
    4 4 4 4;
    5 5 5 5;
    6 6 6 6;
    7 7 7 7;
    8 8 8 8
    9 9 9 9
    10 10 10 10
    11 11 11 11 
    12 12 12 12];

rowIdx =  [1 3 2 4]' + [0:4:size(x,1)-1];
M = reshape(x(rowIdx(:),:)', size(x,2)*2, [])'

M = 6×8

1 1 1 1 3 3 3 3 2 2 2 2 4 4 4 4 5 5 5 5 7 7 7 7 6 6 6 6 8 8 8 8 9 9 9 9 11 11 11 11 10 10 10 10 12 12 12 12

To get from M back to x,

k = reshape(M',size(M,2)/2,[]);
rowIdx =  [1 3 2 4]' + (0:4:size(k,2)-1);
xOrig = k(:,rowIdx(:))';

[Pascal Weller]

Thanks, that is exactly what I need. I knew there was a way to use reshape(). Just was not able to wrap my head around it. :D

[Pascal Weller]

Is there a way to do the vice versa transformation without a for loop but with reshape? So from M to x? I can only think of solutions involving for loops.

[Adam Danz]

See updated solution.

[Pascal Weller]

Amazing! Huge thanks! You make it look so easy...

Answer 2

[Taimoor Hasan Khan]

A = [1 1 1 1;
     2 2 2 2;
     3 3 3 3;
     4 4 4 4;
     5 5 5 5; 
     6 6 6 6;
     7 7 7 7;
     8 8 8 8];
 
A = cat( 2, [ A(1:2, :); A(5:6, :) ], [ A(3:4, :); A(7:8, :) ] )
            % vertical concat. #1     % vertical concat #2

Not sure how you could generalize the process without loops though, sorry...

[Pascal Weller]

In my case the matrix is some million rows long so I would definitely need to do the indexing with the help of a for loop. However, for smaller matrices your answer is a simple solution and much appreciated. I am sure it will be of good use for someone out there. So thanks for your contribution!