Not really understand what you intended to do. But based on your code, you could get an array by accumulating the vector you get within each loop. First, you’d better not overwrite your loop variable I. Second in your interp1 you don’t have an exact value for the original x. I guess you assume the default 1:length(I) unless this is not your intention. Let’s use newI for the interp1 output:
newI = interp1(I, xq);
D = [D newI(:)];
Before for I = R, you need also add D =[];
It’ll be better if you reallocate D first with the supposed size and assign newI to its column. E.g.
D = nan(5, size(R, 2));
for iC = 1:size(R, 2)
I = R(:, iC);
... % your interp code
D(:, iC) = newI(:);
end
You can test them. The second solution should be faster.
