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    <p>Now you have clarified your question (again...) here is a solution (still using a for loop...). It includes "rule 7" - excluding runs of fewer than three elements; it also includes the second part of that rule - runs of fewer than three zeros don't count as zero. The new code looks like this:</p> <pre><code>A = [ 0 0 0 2 4 1 0 0 0 1 3 2; 0 0 0 5 1 1 1 1 0 0 0 1; 0 3 4 1 0 3 1 2 0 0 1 3; 0 0 0 0 1 3 4 5 0 0 0 0]; retVal = cell(1, size(A, 1)); for ri = 1:size(A,1) temp = [1 0 0 0 A(ri,:) 0 0 0 1]; % pad ends with 3 zeros + 1 % so that is always a "good run" isz = (temp == 0); % find zeros - pad "short runs of 0" with ones diffIsZ = diff(isz); f = find(diffIsZ == 1); l = find(diffIsZ == -1); shortRun = find((l-f)&lt;3); % these are the zeros that need eliminating for ii = 1:numel(shortRun) temp(f(shortRun(ii))+1:l(shortRun(ii))) = 1; end % now take the modified row: nz = (temp(4:end-3)~=0); dnz = diff(nz); % find first and last nonzero elements f = find(dnz==1); l = find(dnz==-1); middleValue = floor((f + l)/2); rule7 = find((l - f) &gt; 2); retVal{ri} = middleValue(rule7); end </code></pre> <p>You have to use a cell array for the return value since you don't know how many elements will be returned per row (per your updated requirement). </p> <p>The code above returns the following cell array:</p> <pre><code>{[5 11], [6], [7], [7]} </code></pre> <p>I appear still not to understand your "rule 7", because you say that "no columns in row 3 satisfy this condition". But it seems to me that once we eliminate the short runs of zeros, it does. Unless I misinterpret how you want to treat a run of non-zero numbers that goes right to the edge (I assume that's OK - which is why you return <code>11</code> as a valid column in row 1; so why wouldn't you return <code>7</code> for row 3??)</p>
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    1. COYour solution looks very good. However, that `fl` is undefined. Probably `nz`?
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      1. USLuis Mendo
    2. COSorry, I mean `dnz`
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    3. CO@LuisMendo Thanks for your answer, but still this is not working with multiple values in one row, I've tried A = [ 0 0 2 4 1 0 1 3 2 0; 0 3 5 5 1 0 0 2 2 0; 0 3 4 1 0 3 1 2 0 0; 0 0 0 0 1 0 0 0 0 0]; Also I've added two conditions in step 7 so if it is possible I'm asking if you could please check it?
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