Simple. One line.
A = randi([-5,5],[10,7])
A =
0 -2 -5 -1 -3 1 0
2 1 -5 5 5 5 -2
3 0 4 -5 -2 5 -2
-3 5 -5 1 3 5 0
5 4 -4 -3 4 -5 -1
4 1 -4 -1 -1 -3 -5
2 4 -1 0 -2 5 1
0 1 3 0 3 5 -5
3 4 -1 1 -1 1 2
-2 5 -4 -1 1 -3 -1
[~,col] = max(A > 0,[],2)
Why does it work? A > 0 returns an array that is true (1) only when a number is positive.
Max returns the location of the first such element (among equal maxes) that it finds. And since the A>0 matrix is a boolean one, it is composed of only 0 and 1 elements.
Will this fail if there are no positive elements at all in a row? Well, yes. In that case, it will return 1 for those rows.
But that is true of almost any of the schemes suggested by others. If you want a scheme that will return perhaps a NaN for those rows where there were no positive elements, you could patch it by one more test. I'll regenerate a new matrix with only a few columns, but many negative elements, to insure at least one row will have all non-positive elements.
A = randi([-10,5],[10,3])
A =
-4 2 -8
-1 -3 -4
-6 5 -2
-5 -10 4
4 -1 2
-4 4 0
-2 -4 -4
-9 -9 -1
-7 3 2
3 3 -7
[~,col] = max(A > 0,[],2);
col(~any(A>0,2)) = NaN
col =
2
NaN
2
3
1
2
NaN
NaN
2
1
As you can see, in this final result, rows {2,7,8} had no positive elements at all, so a NaN was generated for those rows.
Or, perhaps you don't want to see a NaN generated. What we might do here is to append an extra column, that is ALWAYS positive. This is a common trick one can employ.
[~,col] = max([A,ones(size(A,1),1)] > 0,[],2)
So for a 3 column array, it returns the number 4 where there were NO positive elements.
In the end, it is important that you decide what to do when NO elements are positive in a row, as you should expect to see a bug in your code at some point.
