circshift columns of array by different shift size

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Sidafa
Sidafa 2015-4-10
评论： omar A.alghafoor 2020-10-12
I'm trying to do the following with arrayfun and circshift
s_dfp = magic(4);
s_hh1p = circshift(s_dfp(:,1),[1 -1]);
s_hh2p = circshift(s_dfp(:,2),[1 -2]);
s_hh3p = circshift(s_dfp(:,3),[1 -3]);
s_hh4p = circshift(s_dfp(:,4),[1 -4]);
HH = [s_hh1p s_hh2p s_hh3p s_hh4p];
s_dfp =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
HH =
4 14 15 1
16 2 3 13
5 11 10 8
9 7 6 12
Each column is shifted by its column number. I would like to do this for arbitrary size.
Thanks in advance.
  1 个评论
Image Analyst
Image Analyst 2015-4-10
This first column is shifted down 1 because the first element, 16, is now in row 2 instead of row 1. However if you were to circularly shift column 2 down, the top element, 2, would be in row 3 instead of row 1. But you have it in row 2. And in the third column, then 3 shifted down 3 rows would be in row 4, not row 2. Finally if you were to circularly shift column 4 down 4, the 13 would land in the same spot (row 1). So, other than the first column being shifted down 1, I can't see what you're doing. In fact, columns 1, 2, 3, and 4 are all shifted down by 1 row, NOT by their column number.

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采纳的回答

Jon
Jon 2015-4-10
I think this does what you want.
HH=cell2mat(arrayfun(@(k) circshift(d(:,k),k),1:size(d,2),'uni',0))
HH =
4 7 10 13
16 14 6 8
5 2 15 12
9 11 3 1)
You may need to change the sign of the second k in the circshift command to get the elements to move in the direction you want.

更多回答（2 个）

Image Analyst
Image Analyst 2015-4-10
I don't see that what you did matches what you said. But anyway, try this and see if it does what you want:
s_dfp = magic(4)
outputMatrix = s_dfp; % Initialize.
[rows, columns] = size(s_dfp);
for col = 1 : columns
% Extract just this column only.
thisColumn = s_dfp(:, col);
% Shift it down "col" rows.
shiftedColumn = circshift(thisColumn, -col)
% Put it back in
outputMatrix(:, col) = shiftedColumn;
end
% Print to command window
outputMatrix
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Image Analyst
Image Analyst 2020-10-12
You'd need to keep track of how much you shifted each column and shift it in the opposite direction. Or just keep your original image (don't delete it!).

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Star Strider
Star Strider 2015-4-10
One possibility is to vectorise it and use an anonymous function:
csm = @(Mtx,Col) circshift(Mtx(:,Col),[1 1-Col]); % Arbitrary Matrix
callcsm = @(Mtx) csm(Mtx, 1:size(Mtx,2));
s_dfp = magic(4);
HH = callcsm(s_dfp);
The ‘csm’ (circularly shift matrix) function does the shifting, and the ‘callcsm’ function requires only the matrix name to produce the vectorised result.

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