[R] plot.gam

[Peter Dalgaard BSA]

Thomas Lumley <thomas at biostat.washington.edu> writes:

> Suppose we fit
> 
> 	model<-lm(y~sin(x)+ns(z,3))
> 
> 
> Given 'model' how can we extract x & z?  We can't work them out from
> model.frame(model), which only contains transformed versions.  Even
> assuming we can guess what the name of the variable is (not trivial --
> which is the variable in  ns(z,df)? In ns(df=df,z)?), we don't know where
> to find it.
> 
> It seems that we would have to assume that model$call$data contains the
> variables if evaluated in the calling frame. This means that plot.gam
> wouldn't necessarily work if called from inside a function, or if the
> original data set had been changed since the model was fitted, or if the
> default na.action had been changed or various other problems.
> 
> Ideas?

The obvious one: Let the user do the thinking:

d<-data.frame(x=0, z=seq(-3, 3, .1))
evalq(plot(z, predict(model, new=d), type='l'), d)

or even (I think it works. Inefficiently, though.)

curve(predict(model, new=data.frame(x=0, z=x)))


-- 
   O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)             FAX: (+45) 35327907
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._