a=[ 0 0 0 0]
a=[a; 0 0 0 0]
a=[a; 0 0 0 0]
a=[a; 0 0 0 0]
a=[a; 0 0.10 0 0]
a=[a; 0 0.15 1.82 3.62]
a=[a; 0.02 0.24 3.58 7.20]
a=[a; 0.02 0.28 5.13 10.49]
a=[a; 0.03 0.37 6.63 13.76]
a=[a; 0.02 0.41 7.95 16.76]
[r c]=find(a);
loc=[r c];
[loca locb]=ismember(1:size(a,2),loc(:,2));
rows=r(locb)
May need some more extensive testing, but something like this. I suspect it will do something nasty if a column has only zero values.
Alternatively, you may find a loop based solution to be a little less obtuse, and possibly more robust
a=[ 0 0 0 0]
a=[a; 0 0 0 0]
a=[a; 0 0 0 0]
a=[a; 0 0 0 0]
a=[a; 0 0.10 0 0]
a=[a; 0 0.15 1.82 3.62]
a=[a; 0.02 0.24 3.58 7.20]
a=[a; 0.02 0.28 5.13 10.49]
a=[a; 0.03 0.37 6.63 13.76]
a=[a; 0.02 0.41 7.95 16.76]
[rows cols]=size(a); % Find size of matrix
rowLocs=zeros(1,cols); % Set up output vector
for col=1:cols % loop through each column of the matrix
row=1;
while row<=rows % loop through the rows in each column
if a(row,col)~=0 % if a non-zero is found, record row and stop the row loop
rowLocs(col)=row;
break;
end
row=row+1; % next row
end
end
rowLocs % Show locations of first non-zero element in each column
