create a randomised vector from matrix

I'm relatively new to Matlab so this might be a very simple problem to solve: I have a 37 x 8 matrix (ascending order numbers, 1:296) I need to create a 296 x 1 vector in which each row will take a number from the matrix rows, with the exception that it cannot be the same row as the previous 8 rows.

Any help would be much appreciated!

Answer 1

[Image Analyst]

Try this:

newVector = yourMatrix(:);

This will produce a column vector. No element of newVector will have come from the same row as any of the previous 8 elements of newVector did.

[Darya Frank]

Hi Image Analyst, Thanks for your answer! I actually meant that newVector should be of random order (rather than getting the column order into a vector), as I need 10 different lists that would also satisfy the no-repeat within 8 rows rule. Any insight? Thanks!

[Image Analyst]

I don't think it's possible to be totally random. You can use randperm(numel(yourMatrix)) to get a list of linear indexes in a random order, and then just keep trying until you get the condition where no element is in the same row as any other element within 8 elements of it, but I don't think you'll ever succeed. I believe you'll have to impose some kind of order onto it, which is what I did. Good luck trying tough. Let us know if you succeed, and post an actual example where it worked.

If your matrix were square you could arrange it in a Latin Square pattern https://en.wikipedia.org/wiki/Latin_square but I don't think there's any Latin rectangle pattern.

[Darya Frank]

The way I was thinking was by some sort of elimination - so first create a random sequence of 8 rows and then for each new row randomly select a number from a row that hasn't appeared in the last 8 places - does that make sense?

Answer 2

[Azzi Abdelmalek]

A=reshape(1:37*8,37,8)  % Example
[n,m]=size(A)
idx=cell2mat(arrayfun(@(x) randperm(n)', 1:m,'un',0)')
idy=zeros(size(idx));
for k=1:n
  ii=find(ismember(idx,k));
  idy(ii)=randperm(m)';
end
ixy=sub2ind([n m],idx,idy);
out=A(ixy)

[Image Analyst]

I don't think this works. I added this check after the code:

% Let's see if any of the linear indexes is in the same row as any of the prior 8 indices
for k = 9:length(ixy)
  % Get row of kth index
  [rowk, colk] = find(A == ixy(k));
  % Check prior 8 elements
  for k2 = k-8 : k-1
    % Get row of k2'th index.
    % Find out the row in A where this number appears.
    [rowk2, colk2] = find(A == ixy(k2));
    % See if this row matches the  row where ixy(k) lies.
    if rowk2 == rowk
      fprintf('Index %d and %d are both in row %d.\n', ixy(k), ixy(k2), rowk);
    end
  end
end

and it found lots of cases where an index was in the same row of the original A matrix as one of the prior 8 indexes.

[Azzi Abdelmalek]

You are right