
Re: Finding endpoints for runs
Posted:
Jan 27, 2013 1:23 AM


Gautam Sethi <gautamsethi@gmail.com> wrote in message <cd7ccb2d8c834568a56522f22a9416bf@googlegroups.com>... > Dear all: > I'm trying to find the endpoints for runs of numbers based on a vector. Let > > >> Y = [1 3 4 6 7 8] > > Y = > > 1 3 4 6 7 8 > > I want an output vector Z of size(length(Y),2) where > > Z = > > 0 2 > 2 5 > 2 5 > 5 9 > 5 9 > 5 9 > > Here is the idea behind what I want. > > The first element of Y is 1, which is not followed by the next integer; thus, the integers that "surround" the integer 1 are 0 and 2, the first row of Z. > > The second element of Y is 3, which is followed by the next integer, 4. However there is a break after 4. Thus, the run of consecutive integers 3 and 4 is "surrounded" by the integers 2 and 5. Therefore, the next two rows of Z, corresponding to the two elements 3 and 4 of Y, are [2 5]. > > Likewise, the last three elements of Y are consecutive integers, "surrounded" by the integers 5 and 9. Hence the last three rows of Z are [5 9]. > > Thanks for your help!          Y = Y'; f = find([true;diff(Y)>1;true]); Z = zeros(size(Y,1),2); Z(1,:) = [Y(1)1,Y(f(2)1)+1]; Z(f(2:end1),:) = diff([Y(f(1:end1)),Y(f(2:end)1)],1,1); Z = cumsum(Z,1);
Roger Stafford

