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Both methods work for me but I think the second solution is a brilliant use of bsxfun. Thanks for your]]> Feb 4, 2013 11:24:08 AM Feb 4, 2013 11:24:08 AM naresh.pai@gmail.com 0 http://mathforum.org/kb/thread.jspa?messageID=8230615&tstart=0#8230615
k = b > 0;
c = b(k);
l = diff([0 c]);
m = bsxfun(@le,(1:max(l))',l);
r = zeros(size(m));
r(m) =]]> Feb 2, 2013 4:41:06 AM Feb 2, 2013 4:41:06 AM b.luong@fogale.findmycountry 1 http://mathforum.org/kb/thread.jspa?messageID=8216114&tstart=0#8216114 c = b(k);
l = diff([0 c]);
s = c(end);
c(end) = [];
i = zeros(1,s);
i(c+1) = 1;
i = 1+cumsum(i);
j =]]> Feb 1, 2013 9:04:08 AM Feb 1, 2013 9:04:08 AM b.luong@fogale.findmycountry 2 http://mathforum.org/kb/thread.jspa?messageID=8214394&tstart=0#8214394 > "Jos (10584)" wrote in message]]> Feb 1, 2013 4:51:08 AM Feb 1, 2013 4:51:08 AM #10584@fileexchange.com 0 http://mathforum.org/kb/thread.jspa?messageID=8203626&tstart=0#8203626 > "Naresh Pai" wrote in message]]> Jan 31, 2013 10:36:08 AM Jan 31, 2013 10:36:08 AM naresh.pai@gmail.com 4 http://mathforum.org/kb/thread.jspa?messageID=8200709&tstart=0#8200709 > I have a vector with indices, for]]> Jan 31, 2013 5:20:08 AM Jan 31, 2013 5:20:08 AM #10584@fileexchange.com 5 http://mathforum.org/kb/thread.jspa?messageID=8193807&tstart=0#8193807
b = [1 4 7]

Using these indices, I am trying to develop a matrix that looks like this
a =[1 0]]> Jan 30, 2013 4:38:08 PM Jan 30, 2013 4:38:08 PM naresh.pai@gmail.com 6