circshift columns of array by different shift size

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.

[Image Analyst]

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.

Answer 1

[Jon]

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.

[JCHebert]

Brilliant! Thanks.

Answer 2

[Image Analyst]

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

[omar A.alghafoor]

How get invers this your code ?

[Image Analyst]

I don't understand the question. Yes, it was my code. But I don't know how to invers (inverse/invert) it. How do you invert code???

Please explain what you need in more detail.

[omar A.alghafoor]

How get return shifted Column this your code , to return original image

[Image Analyst]

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!).

[omar A.alghafoor]

thank you

Answer 3

[Star Strider]

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.