[R] Function update problem

[ripley@stats.ox.ac.uk]

You can use substitute: something like (untested)

for(i in 1:100){
    form <- substitute(.~.+x[,i], list(i=i))
    model <- update(model, form)
    ## do something useful in here
}
and you do not need to update unchanged arguments!

However, why are you rewriting add1.default, when there is add1.lm?

On Thu, 13 Feb 2003, lun li wrote:

> Dear all,
> 
> I am trying an automatic model selection for a multiple linear regression 
> using function lm and update. But, I meet a problem when using update. The 
> problem is the function update can not update when variables as a vector(for 
> example,x is a matrix with 100 regression variables). The code is as below:
> 
>   > model<-lm(y~x1,singular.ok=T,na.action=na.omit)
>   > for(i in 1:100){
>   > model<-update(model,.~.+x[,i],singular.ok=T,na.action=na.omit)}
> 
> If the above code is represented as below, I can get the correct result. 
> However, I must use the loops.
> 
> >model<-lm(y~x1,singular.ok=T,na.action=na.omit)
> >model<-update(model,.~.+x[,1],singular.ok=T,na.action=na.omit)
> >model<-update(model,.~.+x[,2],singular.ok=T,na.action=na.omit)
>         ......
> >model<-update(model,.~.+x[,100],singular.ok=T,na.action=na.omit)

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
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