[R] using update( ) in a function?

[Thomas Lumley]

On Mon, 21 Apr 2003, Roger Peng wrote:

> I believe this is bug PR#1861.  Not sure what the workaround is.

No, the code as given doesn't get as far as that

> On Mon, 21 Apr 2003, Mahmood ARAI wrote:
>
> > Hello,
> >
> > using the function "update()" in the following function
> > does not work.
> >
> > # a function for running a model:  x1 ~ x2
> > # by updating the original model: fm1 <- lm(y ~ x1 + x2,data=df)
> > # update(fm1, x1 ~ . -x1, data=df) works at the command line
> > # but not in the function below.
> >
> >
> > R>   foo <- function(fm,x,df)
> > +     {
> > +     update(fm, x ~ . -x,data=df)
> > +     }
> > R>
> > R>
> > R>  df1 <- data.frame(y= 1:10, x1=(1:10)^2, x2=sqrt(1:10)^3)
> > R>  fm1 <- lm(y ~ x1+x2, data=df1)
> > R>
> > R>  foo(fm1, x1, df1)
> > Error in eval(expr, envir, enclos) : Object "x" not found

The main problem here is that the argument `x' is only going to have the
*value* of x1, where you need the symbol x1.  On top of that,
	update(fm, x ~ . -x,data=df)
uses the symbol `x' rather than its value, so even if its value were
correct it wouldn't work.

This works
> foo<-function(model,term,data){
+ ff<-substitute(term~.-term)
+ update(model,ff,data=data)
+ }
> data(trees)
> a<-lm(Girth~Height+Volume,data=trees)
> foo(a,Height,data=trees)

Call:
lm(formula = Height ~ Volume, data = data)

Coefficients:
(Intercept)       Volume
    69.0034       0.2319

but is not recommended style, as you are defeating the call-by-value
semantics of R.  A better version is
> bar<-function(model,term,data){
+ ff<-substitute(x~.-x,list(x=as.name(term)))
+  update(model,ff,data=data)
+ }
> bar(a,"Height",data=trees)

Call:
lm(formula = Height ~ Volume, data = data)

Coefficients:
(Intercept)       Volume
    69.0034       0.2319

and you could have versions that allowed `term' to be a quoted name or
expression rather than a string.

	-thomas