Re: [R] using ddply with segmented regression

2013-10-16 Thread Jess Inskip
Hi Paul, Thanks for starting this thread. I have been struggling with the same plyr problems. I have also had some trouble moving onto the next step - integrating the segmented output with ggplot2 or lattice. Have you had any luck in this? There is good documentation on adding lm into ggpl

Re: [R] using ddply with segmented regression

2013-10-14 Thread Prew, Paul
m...@comcast.net] Sent: Monday, October 14, 2013 5:30 PM To: Prew, Paul Cc: arun; R help Subject: Re: [R] using ddply with segmented regression On Oct 14, 2013, at 2:57 PM, Prew, Paul wrote: > Hello, the code provided by arun did the trick. Thank you very much, arun. > > However, I&

Re: [R] using ddply with segmented regression

2013-10-14 Thread arun
h Center  | Mail Stop ESC-F4412-A 655 Lone Oak Drive  |  Eagan, MN 55121-1560 -Original Message- From: arun [mailto:smartpink...@yahoo.com] Sent: Saturday, October 12, 2013 1:42 AM To: R help Cc: Prew, Paul Subject: Re: [R] using ddply with segmented regression Hi, Try: segmentf_df

Re: [R] using ddply with segmented regression

2013-10-14 Thread David Winsemius
> > Paul Prew | Statistician > 651-795-5942 | fax 651-204-7504 > Ecolab Research Center | Mail Stop ESC-F4412-A > 655 Lone Oak Drive | Eagan, MN 55121-1560 > > -Original Message- > From: arun [mailto:smartpink...@yahoo.com] > Sent: Saturday, Octobe

Re: [R] using ddply with segmented regression

2013-10-14 Thread Prew, Paul
y, October 12, 2013 1:42 AM To: R help Cc: Prew, Paul Subject: Re: [R] using ddply with segmented regression Hi, Try: segmentf_df <- function(df) { out.lm<-lm(deltaWgt~Cycle, data=df) segmented(out.lm,seg.Z=~Cycle, psi=(Cycle=NA),control=seg.control(stop.if.error=FALSE,n.boot=0)) } librar

Re: [R] using ddply with segmented regression

2013-10-11 Thread arun
Hi, Try: segmentf_df <- function(df) { out.lm<-lm(deltaWgt~Cycle, data=df) segmented(out.lm,seg.Z=~Cycle, psi=(Cycle=NA),control=seg.control(stop.if.error=FALSE,n.boot=0)) } library(plyr) library(segmented) dlply(df,.(Lot.Run),segmentf_df) $`J062431-1` Call: segmented.lm(obj = out.lm, seg.Z =