Hi Masao - A combination of the ismember and removing of zeros may work. If A is your first matrix (larger data set) and B is your second matrix (lookup dataset), then
[isInA,whereInA]=ismember(B(:,1),A(:,1))
will return the isInA indicating which elements of B are in A (1) and which elements aren't (0). The whereInA will indicate what those indices are in A. If all elements of B are in A then there is no problem and we could use the following to append the final column of A onto B:
B(isInA,4) = A(whereInA,5);
But if there is an element of B not in A then there is a zero in the same row of each of the isInA and whereInA matrices. Using that fact, and that zero indices on the left-hand side of the assignment will be ignored, we can do:
% remove the zeros from whereInA
whereInA(whereInA==0) = [];
% append the column of A onto B for those elements of B in A wrt time
B(isInA,4) = A(whereInA,5);
I added some dummy rows to your example of B (so that the first and third rows of this updated matrix would not match wrt time in A) and the result was as expected - the two rows that did match in A had their fourth column updated with the fifth column of A.
Geoff
