Shift rows by different amounts

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Ellie
Ellie am 31 Jul. 2015
Kommentiert: Ellie am 31 Jul. 2015
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

Akzeptierte Antwort

Cedric
Cedric am 31 Jul. 2015
Bearbeitet: Cedric am 31 Jul. 2015
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.
  2 Kommentare
Cedric
Cedric am 31 Jul. 2015
PS: Andrei's BSXFUN-based solution can beat the loop though.
Ellie
Ellie am 31 Jul. 2015
Thanks!

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Weitere Antworten (1)

Walter Roberson
Walter Roberson am 31 Jul. 2015
  1 Kommentar
Ellie
Ellie am 31 Jul. 2015
I have seen all of those and have attempted to modify them for my code but have been having issues.

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