On Thu, 20 Apr 2006, Manuel Gutierrez wrote: > Is it possible to include a factor in an nls formula?
Yes. What do you intend by it? If you mean what it would mean for a lm formula, you need A[a] and starting values for A. There's an example on p.219 of MASS4. > I've searched the help pages without any luck so I > guess it is not feasible. > I've given it a few attempts without luck getting the > message: > + not meaningful for factors in: > Ops.factor(independ^EE, a) > > This is a toy example, my realworld case is much more > complicated (and can not be solved linearizing an > using lm) > a<-as.factor(c(rep(1,50),rep(0,50))) > independ<-rnorm(100) > respo<-rep(NA,100) > respo[a==1]<-(independ[a==1]^2.3)+2 > respo[a==0]<-(independ[a==0]^2.1)+3 > nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) > > Any pointers welcomed > Many Thanks, > Manu > > ______________________________________________ > [email protected] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
