How do I create new rows from numbers repeating themselves in two or more rows of a matrix?

Take for example matrix a:

a=[1 2 3 4; 5 8 10 1; 2 3 11 12].

Numbers recurring in different rows should create a new row and be emitted from the row they were in before, giving us the matrix:

a=[1 0 0; 2 3 0; 4 0 0; 5 8 10; 11 12 0]

[Walter Roberson]

4 does not occur in any other row, so why is it not retained in the row with the 1?

Your duplicates, the 2 and the 3 in this example, do they always occur in pairs, or could they be single?

What should happen for numbers that have multiple duplicates?

Answer 1

[Guillaume]

This is not very easy, so why do you want to do that?

The following may do what you want:

a = num2cell(a, 2); %convert into a cell array of rows
height = numel(a);
result ={};
for row = 1:height;
   for otherrow = row+1:height;
      if isempty(a{row})
         break;
      end
      [isfound, where] = ismember(a{row}, a{otherrow});
      if any(isfound)
         result = [result; a{row}(isfound)]; %#ok<AGROW>
         a{row}(isfound) = []; %prevent extracted elements to be reprocessed
         a{otherrow}(where(isfound)) = []; %remove duplicate
      end
   end
   if ~isempty(a{row})
      result = [result; a{row}]; %#ok<AGROW> append non-duplicate
   end
end

Note that it puts non-recurring numbers after any recurring numbers in the same row.

[Natalie]

You need to run it again on your result in order to receive the desired matrix.

[Guillaume]

You need to run it again on your result in order to receive the desired matrix

I assume it's in response to Walter's question: What should happen for numbers that have multiple duplicates?.

My code above ignores the issue in that it only removes a number and its first duplicate from further processing. A third duplicate will thus be considered a unique value again.