How to circular shift a matrix for every 1 to 6 elements until its end?

I have a matrix 12528x246, I would like to the circular shift of the 2nd dimension(246). The shift should be done for every consecutive 6 elements from 1 position till the end of the matrix starting from 1 to 246. For e.g, 1:6(perform circular shift for 1 position) then repeat the steps for 7:12,......up to 241:246. Is it possible to do such an operation using circular shift? I tried my best but couldn't find the correct logic.

[KL]

For e.g, 1:6(perform circular shift for 1 position) then repeat the steps for 7:12

repeat how? by increasing the shift value by 2,3 and so on?

Answer 1

[KL]

I'm only guessing. If you want to do circshifts for blocks of columns, use mat2cell first, do the shifting and convert it back using cell2mat. Small example,

A = repmat((1:18),10,1); %dummy data with 18 columns
sz = size(A);
splits = 6;
B = mat2cell(A,sz(1),repmat(splits,1,sz(2)/splits));
B_shifted = cell2mat(cellfun(@(x) circshift(x,1,2),B,'uni',0));

A =
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18
     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18

B_shifted =  
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17
     6     1     2     3     4     5    12     7     8     9    10    11    18    13    14    15    16    17

Answer 2

[Guillaume]

If I understood correctly, all you have to do is reshape your original array into rows or columns of 6 elements, perform the circular shift, and reshape back. Because you want to operate on columns and matlab is row based, you'll have to transpose back and forth, so:

shift = 1;  %or -1 maybe, you haven't specified the direction
shiftedmatrix = reshape(circshift(reshape(yourmatrix.', 6, []), shift, 1).', size(yourmatrix))

Answer 3

[Andrei Bobrov]

A - your array

k = 6;
[m,n] = size(A);
out = A1(sub2ind([m,n],repelem(hankel([(2:m)';1],1:n/k),1,k),repelem(1:n,m,1)));

Answer 4

[Jos (10584)]

Clever indexing will do the trick:

A = repmat((1:18),10,1); % dummy data with 18 columns
S = 6

ix = 0:size(A,2)-1
ix0 = S * floor(ix/S) 
ix1 = rem(ix,S)
ix1shift = rem(ix1+S-1,S)
ix2 = ix0 + ix1shift + 1
B = A(:,ix2)

% all this can be done in one step, of course, which makes it an "easy-to-read" one-liner :D
B2 = A(:,S * floor((0:size(A,2)-1)/S) + rem(rem(0:size(A,2)-1,S)+S-1,S)+1)

Answer 5

[arun]

Thank you all for your ideas and contributions, but it seems cell2mat concept is easier than other ideas