How can I efficiently (with no loops) compute the mean of each column in a matrix, taking different elements from each column?

I need to compute something similar to this:

M = [ 1  2  3  4  5; ...
      6  7  8  9 10; ...
     11 12 13 14 15];
idx = [ 1 2 3 2 1; ...
        2 3 3 3 3];
for col=1:size(M,2),
    v(1,col) = mean(M(idx(1,col):idx(2,col),col));
end;

How can I do this without the loop?

Answer 1

[Bruno Luong]

M = [ 1  2  3  4  5; ...
      6  7  8  9 10; ...
     11 12 13 14 15];
idx = [ 1 2 3 2 1; ...
        2 3 3 3 3];

m = size(M,1);
r = (1:size(M,1)).';
sum(M.*(r >= idx(1,:) & r <= idx(2,:)),1)./(diff(idx)+1)

Result

ans =

    3.5000    9.5000   13.0000   11.5000   10.0000

[Nieves Lopez]

Thank you very much!

Answer 2

[Bruno Luong]

M = [ 1  2  3  4  5; ...
      6  7  8  9 10; ...
     11 12 13 14 15];
idx = [ 1 2 3 2 1; ...
        2 3 3 3 3];
    
[m,n] = size(M);
mean(M(idx+(0:n-1)*m),1) % use "mean(M(bsxfun(@plus,idx,(0:n-1)*m)),1)" for R2016a or prior

Result

ans =

    3.5000    9.5000   13.0000   11.5000   10.0000

[Nieves Lopez]

Sorry, I meant from one value to the other, not only those values. For example, on the fifth column, it is a mean from the first to the third row. In this case, it is a coincidence that the mean of only the first and third row is equal to the mean of the first, second and third row, but it is this last operation the one I need.

[Bruno Luong]

I see. See other answer.