From: dpb on 19 Apr 2010 09:17 isha ram wrote: > dpb <none(a)non.net> wrote in message > <hqafij$k8o$1(a)news.eternal-september.org>... >> isha ram wrote: >> > ImageAnalyst <imageanalyst(a)mailinator.com> wrote in message > >> <465323f6-adf8-467e-ad50-e88c49d08792(a)j5g2000vbl.googlegroups.com>... >> >> Wouldn't the first local min be the 1 at the first element? And >> >> wouldn't the second local min be the 1 at position 9? I don't follow >> >> your example. Can you try again? >> > > Q=[1 0.9 0.8 0.7 0.6 0.5 0.6 0.4 0.3 0.2 0.6 ....] >> > first local minimum=0.5 >> > second local minimum=0.2 >> > thanks >> >> >> Q=[1 0.9 0.8 0.7 0.6 0.5 0.6 0.4 0.3 0.2 0.6]; >> >> Q(diff(Q)>0) >> ans = >> 0.5000 0.2000 >> >> >> >> You're welcome... :) >> > thanks The above works for noiseless data--you may need some other nuance(s) if data aren't...I really was expecting a "but..." :) -- |