# [R] nls and factor

Manuel Gutierrez manuel_gutierrez_lopez at yahoo.es
Thu Apr 20 12:10:08 CEST 2006

```Thanks Andrew. I am now trying but without much
success. I don't now how to give start values for the
factor?.
Could you give me an example solution with my toy
example?

a<-as.factor(c(rep(1,50),rep(0,50)))
independ<-1:100
respo<-rep(NA,100)
respo[a==1]<-(independ[a==1]^2.3)+2
respo[a==0]<-(independ[a==0]^2.1)+3
library(nlme)
gnls(respo~independ^b+a,start=list(b=1.8))

Many thanks.
Manu

--- Andrew Robinson <A.Robinson at ms.unimelb.edu.au>
escribió:

> Manuel,
>
> I don't think that it works very easily.  Instead,
> try gnls() in the
> nlme package.
>
> Cheers
>
> Andrew
>
> On Thu, Apr 20, 2006 at 11:18:02AM +0200, Manuel
> Gutierrez wrote:
> > 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 at stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> http://www.R-project.org/posting-guide.html
>
> --
> Andrew Robinson
> Department of Mathematics and Statistics
> Tel: +61-3-8344-9763
> University of Melbourne, VIC 3010 Australia
> Fax: +61-3-8344-4599
> Email: a.robinson at ms.unimelb.edu.au
> http://www.ms.unimelb.edu.au
>

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