[R] nls and factor

[Andrew Robinson]

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
> 
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-- 
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