Shift rows by different amounts

Question

Ellie 2015-7-31

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/231858-shift-rows-by-different-amounts

评论： Ellie 2015-7-31

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

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Answer 1

Cedric 2015-7-31

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https://ww2.mathworks.cn/matlabcentral/answers/231858-shift-rows-by-different-amounts#answer_187791

编辑：Cedric 2015-7-31

在 MATLAB Online 中打开

 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 =
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16

and the final:

 A =
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
  14    15    16     1     2     3     4     5     6     7     8     9    10    11    12
  10    11    12    13    14    15    16     1     2     3     4     5     6     7     8
   6     7     8     9    10    11    12    13    14    15    16     1     2     3     4
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
  14    15    16     1     2     3     4     5     6     7     8     9    10    11    12
  10    11    12    13    14    15    16     1     2     3     4     5     6     7     8
   6     7     8     9    10    11    12    13    14    15    16     1     2     3     4
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
  14    15    16     1     2     3     4     5     6     7     8     9    10    11    12
  10    11    12    13    14    15    16     1     2     3     4     5     6     7     8
   6     7     8     9    10    11    12    13    14    15    16     1     2     3     4
   2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
  14    15    16     1     2     3     4     5     6     7     8     9    10    11    12
  10    11    12    13    14    15    16     1     2     3     4     5     6     7     8
   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.