# [R] ... and 5 cents more.

Mon May 21 13:54:52 CEST 2007

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>
> I was solving similar problem some time ago.
> Here is my script.
> I had a data frame, containing a response and several other variables,
> which were assumed predictors.
> I was trying to choose the best linear approximation.
> This approach now seems to me useless, please, don't blame me for that.
> However, the script might be useful to you.
>
> <code>
> library(forward)
>
> # dfr is a data.frame, that contains everything.
> # The response variable is named med5x
> # The following lines construct linear models for all possibe formulas
> # of the form
> # med5x~T+a+height
> # med5x~a+height+RH
> # T, a, RH, etc are the names of possible predictors
>
> inputs<-names(dfr)[c(10:30,1)]  # dfr was a very large data frame,
> containing lot of variables.
> # here we have chosen only a subset of them.
>
> for(nc in 11:length(inputs)){ # the linear models were assumed to have at
> least 11 terms
> # now we are generating character vectors containing formulas.
>
>   formulas<-paste("med5x",sep="~",
>
> fwd.combn(inputs,nc,fun=function(x){paste(x,collapse="+")}))
>
> # and then, are trying to fit every
>
>   for(f in formulas){
>     lms<-lm(eval(parse(text=f)),data=dfr)
>
>
> cat(file="linear_models.txt",f,sum(residuals(lms)^2),"\n",sep="\t",append=TRUE)
>   }
> }
> </code>
>
> Hmm, looking back, I see that this is rather inefficient script.
> For example, the inner cycle can easily be replaced with the apply
> function.
>
>

lm(as.formula(f),data=dfr)
do.call("lm",list(formula=f,data=dfr))

also should work in the inner cycle.

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