[R] nls and factor

[Prof Brian Ripley]

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
>
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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
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