Is it possible to include a factor in an nls formula? 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 ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
