Rob - is the matrix sorted on column 17? And can you quantify what you mean by really, really slow - does it take seconds, minutes, hours (!) to find the result? Are there that many rows in A?
Your final two lines of code
C(1,2) = mean(A(idxa(1):idxa(2)-1));
C(end,2) = mean(A(idxa(end):end));
ignores the 16th column of A. Is this intentional?
Since you want to average the elements of the 16th row of A that correspond to the unique values of the 17th row, then you could try something like the following using arrayfun which applies a function to each element of an input matrix
% add another element to idea which will be the number of rows of A plus one
% we use this last element to indicate where we should "stop" in calculating the
% average/mean for the final unique number
idxa = [idxa;size(A,1)+1];
% now use arrayfun to calculate the average for each set of unique values (from the
% 17th column) that are in the 16th column
D = arrayfun(@(x)mean(A(idxa(x):idxa(x+1)-1,16)),1:length(idxa)-1)';
The above is very similar to what you coded for the first and last means. We average all data in the 16th column of A for the rows idxa(x):idxa(x+1)-1. Our input to arrayfun is the anonymous function
@(x)mean(A(idxa(x):idxa(x+1)-1,16))
whose input parameter x is based on the second input to arrayfun
1:length(idxa)-1
which is just a list of indices from 1 to the end of idxa less one (since the last element of idxa is dummy that we added). So if A has 15 rows and idxa is
idxa =
1
3
4
5
7
8
11
15
16 <---this is the value we added
Then the means are calculated given the pair of row indices (1,2), (3,3), (4,4), (5,6), (7,7), (8,10), (11,14), and (15,15).
