Thanks! that worked. I also tested with the below method although your solution is faster and done in fewer steps.
inter <- intercept(segmented.mod) inter.m <- as.matrix(inter$x) inter.row <- nrow(inter.m) answer <- inter.m[inter.row,1] PD On Tue, Jul 5, 2016 at 2:28 PM, <ruipbarra...@sapo.pt> wrote: > Hello, > > Try > > dimnames(inter$x)[[1]] > > You could have seen this by inspecting 'inter': > > str(inter) > > Hope this helps, > > Rui Barradas > > > Citando Narendra Modi <bjpmodi2...@gmail.com>: > > I am able to perform regression on a dataset as below: > > plot(x,y) > lin.mod <- lm(y~x) > m <- mean(x) > m > > segmented.mod <- segmented(lin.mod, seg.Z = ~x, psi= m) > > plot(segmented.mod, add=T) > sl <- slope(segmented.mod) > inter <- intercept(segmented.mod) > > summary(segmented.mod) # Show Summary > sl # show all the slopes > inter # show all the intercepts > > > In my dataset, the above method correctly identifies the breakpoints and > hence I get two intercepts. > > inter > > $x > Est. > intercept1 -3.269 > intercept2 -19.980 > > What I am interested is the "intercept2" value. How can I obtain this? > > The method needs to be dynamic as in if the next dataset has 3 intercepts, > I would like to get "intercept3 value. > > PD > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.htmland provide commented, minimal, > self-contained, reproducible code. > > > ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
