Question

I have a matrix like

frequencyarray =

    697    697    697    697    697    697      0      0    697
      0      0      0      0      0      0    770    770      0
      0      0      0      0      0      0      0      0      0
      0      0      0      0      0      0      0      0      0
   1209   1209   1209   1209   1209   1209   1209   1209   1209
      0      0      0      0      0      0      0      0      0
      0      0      0      0      0      0      0      0      0

I’d now like to eliminate all the zeros on a vector base from it and get this

frequencyarray2 =

    697    697    697    697    697    697    770    770    697
   1209   1209   1209   1209   1209   1209   1209   1209   1209

I only want to do this if exactly two non-zero numbers are in a vector. If a vector has more or less than two non-zero numbers, they should be replaced with zeros.

Therefore, if I have something like:

frequencyarray =

      0      0    697    697
      0      0    770    770
      0      0      0      0
      0      0      0      0
   1209   1209   1209   1209
      0      0      0      0
      0      0      0      0

the response should be:

frequencyarray2 =

      0      0      0      0

Using find, this doesn’t work, it just does everything in on single vector. I thought about using find and then reshape. But that does only work if exactly two numbers are non-zero per vector.

If possible, I’d like to avoid loops. The columns don’t have any relation between them.

Any idea is appreciated

Answer 1

Well… I’ve solved the problem. Even not in a way I’d desired, but it works… And it doesn’t have to be ultra-fast.

Here’s my solution, if someone is interested:

function [collapsed] = collapse(frequencies)
    fsize = size(frequencies,2);
    collapsed = zeros(2,fsize);
    for it=1:fsize
        col = frequencies(:,it);
        nonzero = frequencies(find(col != 0));
        if (size(nonzero,1) == 2)
            collapsed(:,it) = nonzero;
        endif
    endfor
endfunction