dpb <email@example.com> wrote in message <firstname.lastname@example.org>... > On 1/1/2014 3:50 AM, Bruno Luong wrote: > > dpb <email@example.com> wrote in message <firstname.lastname@example.org>... > >> On 12/31/2013 2:16 AM, Bruno Luong wrote: > >> > B(ismember(A,B(:,1)),2:end) > >> > >> That returns the rows in B for which the indices are found in A, not > >> B, unfortunately, Bruno. It does get the first one correct as they're > >> both in position 2, but then it returns fourth row, not first where > >> the last element of A is found in first column of B... > > > > You are write, it learns me not to do things too fast. > > > >> > >> >> [~,lb]=ismember(A,B(:,1)); > >> >> B(lb(lb>0),2:end) > >> ans = > >> 0.1300 0.0010 0.8900 > >> 0.1200 0.0050 0.1000 > >> >> > >> > > > > Should use the first argument of ismember() rather than recomputing it. > > Also OP requires the original order, so sort() will fix the order: > > > > [t, iB] = ismember(A,B(:,1)); > > B(sort(iB(i)),2:end) > > Been puzzling over your last comment, Bruno...couldn't decide what you > meant by "recomputing it". I guess you're speaking of the logical test > to select nonzero elements instead of just a direct use of the logical > addressing vector. > > The problem I see with sort() is that I think in the general case the > results could be jumbled since the two matrices are given independently. > AFAICT there's no guarantee the magnitude order would be the same as > the position order in the B array of an arbitrary selection of values in > the A vector. Or, at least, that's my understanding of the OP's quest... > > -- sort is not working but indexing one is working fine. I checked with random element, my matrix size is 3822x14.