I have the following matrix

Keeping the 20 row and 20 column fixed (so the 21st rows and columns because I started at 0)...how do I push each row and column back one spot?

I need to push the 0th row and column to the 19th row and column, the 1st row and column to the 0th row and column..so pretty much swapping out every row and column with the one before it while keeping the 20th row and column fixed.

$\begin{bmatrix}a_{0,0} &a_{0,1}&....&a_{0,20}\\.&.&&.\\.&.&&.\\.&.&&.\\a_{20,0} & a_{20,1}&....&a_{20,20}\end{bmatrix}$



Small example on 5x5 matrix:

pp = Table[p[i, j], {i, 5}, {j, 5}]

enter image description here

One way:

pp[[#, #]] &@Insert[Rest@Range[5], 1, -2]

enter image description here

Or another:

pp[[;; 4, ;; 4]] = RotateLeft[pp[[;; 4, ;; 4]], {1, 1}];
  • $\begingroup$ If you don't want to modify pp, just create new m = pp and do this on m. $\endgroup$ – Kuba Apr 16 '15 at 11:56
  • $\begingroup$ It's not working up to par with mine. Just making sure, if I have: p[0,1] p[0,2]................p[0,20] . . . . . . . . . p[20,1] p[20,2]..............p[20,20] Would my code be like: pp[[;; 20, ;; 20]] = RotateLeft[pp[[;; 20, ;; 20]], {0, 0}]; pp // MatrixForm? $\endgroup$ – John Yates Apr 16 '15 at 12:01
  • $\begingroup$ @JohnYates What do you mean by not working? Yes, something like that. p.s. MatrixForm is only to show result nicely. $\endgroup$ – Kuba Apr 16 '15 at 12:14
  • $\begingroup$ @kabu I added an image file with the respective problem $\endgroup$ – John Yates Apr 16 '15 at 12:30
  • $\begingroup$ @JohnYates You rotate by {0,0}, not {1,1}... $\endgroup$ – Kuba Apr 16 '15 at 12:32

I'm a bit late coming in here with this, but let there be a different approach anyway. It so happens, that the permutation you describe is simply Cycles[{Range@20}]:

(* {20, 1, 2, ..., 19, 21} *)

With mat = Table[p[i,j],{i,0,20},{j,0,20}] You can try this:

    MapAt[Permute[#, Cycles[{Range@20}]] &, 
        Transpose@MapAt[Permute[#, Cycles[{Range@20}]] &, mat, {;; 20}],
    {;; 20}]

This will return you the complete desired output without having to modify parts of the matrix step-by-step.

Note to self and question to audience: because I have to permute first rows, then columns, I needed to make transpositions. It would be great if there's a way to have MapAt act on an entire column, similar to how here it acted on rows.

  • 1
    $\begingroup$ You can MapAt e.g.: {All, 1}. mathematica.stackexchange.com/a/31173/5478 $\endgroup$ – Kuba Apr 16 '15 at 13:42
  • 1
    $\begingroup$ In this case this does not work, and probably never will, simply because of the way data is ordered in the list. Instead of operating with Permute on a list of elements constituting the first column of the matrix it operates with Permute on each element in the column separately and, naturally, returns an error. $\endgroup$ – LLlAMnYP Apr 16 '15 at 13:47

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