[R] Predict polynomial problem

Charles C. Berry cberry at tajo.ucsd.edu
Tue Jan 19 02:36:28 CET 2010


On Mon, 18 Jan 2010, Barry Rowlingson wrote:

> I have a function that fits polynomial models for the orders in n:
>
> lmn <- function(d,n){
>  models=list()
>  for(i in n){
>    models[[i]]=lm(y~poly(x,i),data=d)
>  }
>  return(models)
> }
>
> My data is:
>
> > d=data.frame(x=1:10,y=runif(10))
>
> So first just do it for a cubic:
>
> > mmn = lmn(d,3)
> > predict(mmn[[3]])
>        1         2         3         4         5         6         7         8
> 0.6228353 0.5752811 0.5319524 0.4957381 0.4695269 0.4562077 0.4586691 0.4798001
>        9        10
> 0.5224893 0.5896255
>
> and lets extrapolate a bit:
>
> > predict(mmn[[3]],newdata=data.frame(x=c(9,10,11)))
>        1         2         3
> 0.5224893 0.5896255 0.6840976
>
> now let's to it for cubic to quintic:
>
> > mmn = lmn(d,3:5)
>
> check the cubic:
>
> > predict(mmn[[3]])
>        1         2         3         4         5         6         7         8
> 0.6228353 0.5752811 0.5319524 0.4957381 0.4695269 0.4562077 0.4586691 0.4798001
>        9        10
> 0.5224893 0.5896255
>
> - thats the same as last time. Extrapolate?
>
> > predict(mmn[[3]],newdata=data.frame(x=c(9,10,11)))
> Error: variable 'poly(x, i)' was fitted with type "nmatrix.3" but type
> "nmatrix.5" was supplied
> In addition: Warning message:
> In Z/rep(sqrt(norm2[-1L]), each = length(x)) :
>  longer object length is not a multiple of shorter object length
>
> it falls over. I can't see the difference between the objects,
> summary() looks the same. Is something wrapped up in an environment
> somewhere, or some lazy evaluation thing, or have I just done
> something stupid?
>

Its the environment thing.

I think you want something like this:

 	models[[i]]=lm( bquote( y ~ poly(x,.(i)) ), data=d)

Use
 	terms( mmn[[3]] )

both with and without this change and


 	ls( env = environment( formula( mmn[[3]] ) ) )
 	get("i",env=environment(formula(mmn[[3]])))
 	sapply(mmn,function(x) environment( formula( x ) ) )


to see what gives.

HTH,

Chuck

> Here's a complete example you can paste in - R --vanilla < this.R
> gives the error above - R 2.10.1 on Ubuntu, and also on R 2.8.1 I had
> lying around on a Windows box:
>
> d = data.frame(x=1:10,y=runif(10))
>
> lmn <- function(d,n){
>  models=list()
>  for(i in n){
>    models[[i]]=lm(y~poly(x,i),data=d)
>  }
>  return(models)
> }
>
> mmn = lmn(d,3)
> predict(mmn[[3]])
> predict(mmn[[3]],newdata=data.frame(x=c(9,10,11)))
>
> mmn2 = lmn(d,3:5)
> predict(mmn2[[3]])
> predict(mmn2[[3]],newdata=data.frame(x=c(9,10,11)))
>
>
> Barry
>
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>

Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901



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