[R] plotting the result of a nonlinear regression

[lauramorgana at bluewin.ch]

Thank you for the suggestion of using predict!
If someone is interested here is the loop I used:

seq(0,max(subset(dati, Fiume=="Laveggio")$Portata), length=100)->seqQ

for(i in 1:35)
{
  jpeg(paste("result",i,".jpg"))
  plot(subset(dati,Fiume=="Laveggio")$PTG.P~subset(dati,Fiume=="Laveggio")$Portata, 
       main=print(formList[[i]]), sub=print(summary(resultList[[i]])$sigma))
  lines(seqQ, predict(resultList[[i]], list(Portata = seqQ)))
  dev.off()
  }

----Messaggio originale----
Da: ssefick at gmail.com
Data: 10.02.2009 18.15
A: <lauramorgana at bluewin.ch>
Copia: <r-help at r-project.org>
Oggetto: Re: [R] plotting the result of a nonlinear regression

?predict

On Tue, Feb 10, 2009 at 11:06 AM, lauramorgana at bluewin.ch
<lauramorgana at bluewin.ch> wrote:
> Hello,
> to plot the result of a singular non linear regression (using nls) I usually use the function plotfit, for 
example:
>
> r.PTG.V<-nls(PTG.P~ fz1(Portata, a,b), data=dati, start=list(a=10, b=10), nls.control(maxiter=200), 
algorithm='port',
> trace=TRUE, na.action=na.omit, lower=list(a=0, b=10), upper=list(a=100, b=100))
> plotfit(r.PTG.V)
>
> I tried to use the function plotfit on the result of the following for loop but I got an error message:
> for (i in 1:length(formList))
> {
>    resultList[[i]] <- nls(formList[[i]], data=subset(dati, Fiume=="Laveggio"), start=startList7[[i]],
>    nls.control(maxiter=1000, warnOnly=TRUE), algorithm='port', na.action=na.omit,lower=lowerList7[[i]],
>    upper=upperList7[[i]])
> }
> plotfit(resultList[[1]]
>
> "Error in diff(as.numeric(y[ord])) :
>  (list) object cannot be coerced to type 'double'"
>
> Could somebody tell me what this error means? how can I fix this problem?
> Any suggestion appreciated!!
> Thanks,
> Laura
>
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>



-- 
Stephen Sefick

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