How to circular shift a matrix for every 1 to 6 elements until its end?
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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.
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More Answers (4)
Guillaume
on 15 Dec 2017
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))
Andrei Bobrov
on 15 Dec 2017
Edited: Andrei Bobrov
on 15 Dec 2017
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)));
Jos (10584)
on 15 Dec 2017
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)
arun
on 15 Dec 2017
0 votes
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