Count how many times a number appears in a matrix and assign that increasing value to a new matrix

I have a 32 x 6 matrix consisting of repeats of 1-16 in each column in a specific order, e.g., :

    X = [6  1  1  8  16  1

        2  16  6  6  9  8

        7  9  10  13  1  16

        15  5  3  6  3  11

        9  3  13  1  7  12

        10  11  8  7  5  7

         etc etc];

So given this set up, each instance of 1 - 16 will appear 12 times, twice in each column. What I need to do is create another matrix of the same size, indexing for each number where those repeats are occurring. So for example, if a column looked like this:

        1
        3
        3
        2
        1

it would return:

        1
        1
        2 (the second time 3 is appearing)
        1
        2 (the second time 1 is appearing)

I was thinking about using some kind of count function for each number and then perhaps applying a multiplication to it? Would be fine if these had to be separate matrices for each of the 16 numbers if that makes it any simpler. Any suggestions?

Answer 1

[Andrei Bobrov]

out = ones(size(X));
for j1 = 1:size(X,2)
    [u,i1] = unique(X(:,j1));
    id = i1(histc(X(:,j1),u) > 1);
    out(id,j1) =  out(id,j1) + 1;
end

or

out = zeros(size(X));
for j1 = 1:size(X,2)
    b = unique(X(:,j1));
    for jj = 1:numel(b)
        t = abs(X(:,j1) - b(jj)) < eps(100);
        out(t,j1) = 1:nnz(t);
    end
end

[Lauren Shone]

Thanks, this produces a beautiful result on a column by column basis, but do you know if there is a way to keep the cumulation going across columns? i.e., once you move to the second column the indexes become 3 and 4, then for the third column 5 and 6 etc? Alternatively, do you think it would work to reshape X and try and apply a similar function to what you have suggested that works it way down the one column if going across doesn't work? Have tried a few variations but nothing coming up so far.

[Lauren Shone]

Although on second thought, given the small size of the matrix I can easily just do a more manual replace column by column, Thanks again!

Answer 2

[Matt J]

This relies on the assumption that each number appears exactly twice in each column (as in the posted example):

out = ones(size(X));
for ii = 1:size(X,2)
    [~,idx] = unique(X(:,ii),'last');
    out(idx,ii)=2;
end

Answer 3

[Azzi Abdelmalek]

A=[]
for k=1:6
  A=[A [randperm(16)' ;randperm(16)']]
end

% for the above example

idx=[];
for k=1:16
  [ii,~]=find(A==k)
  idx=[idx;reshape(ii,2,[])];
end