Shift rows by different amounts

I have a N^2 x N^2 matrix where N = 4, and I wish to shift rows 1, 5, 9, and 13 by 0. Rows 2, 6, 10, 14 by 4. Rows 3, 7, 11, 15 by 8. Finally rows 4, 8, 12, 16 by 12. Ideally it would be a general code, as I plan to apply it to more than just this example. I have tried many things but to no avail. Any help would be appreciated. Thanks

 采纳的回答

Cedric
Cedric 2015-7-31
编辑:Cedric 2015-7-31
N = 4 ;
A = repmat( 1:N^2, N^2, 1 ) ; % Dummy example.
for k = 2 : N
A(k:N:N^2,:) = circshift( A(k:N:N^2,:), (k-1) * N, 2 ) ;
end
With that, the the original A is:
A =
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
and the final:
A =
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12
9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8
5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12
9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8
5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12
9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8
5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12
9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8
5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4
I am not sure that the solutions that don't involve a FOR loop are more efficient ultimately, because the FOR loop has only N-1 iterations and no realloc or conversion to/from cell arrays. You'd have to profile every approach to be sure.

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PS: Andrei's BSXFUN-based solution can beat the loop though.

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