The semantic gymnastics below aren't necessary (nor are necessarily they a
bad idea). This is basically FAQ 7.21. So, e.g.
z <- c("x1","x2")
mod <- lm(y ~ get(z[1]) + get(z[2]))
Bert Gunter
Genentech
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Charilaos Skiadas
Sent: Sunday, December 30, 2007 10:23 AM
To: Daniel O'Shea
Cc: r-help@r-project.org
Subject: Re: [R] refering to variable names in lm where the variable name
isin another variable
On Dec 30, 2007, at 12:49 PM, Daniel O'Shea wrote:
> I am trying to refer to a variable name in a lm regression where
> the variable name is in another variable, but it does seem to
> work. Here is an example:
>
> y<-rnorm(10)
> dat<-data.frame(x1=rnorm(10),x2=rnorm(10),x3=rnorm(10))
> nam<-c('x1','x2','x3')
> library(gtools)
> com<-combinations(3,2,1:3)
> mod<-lm(y~nam[com[1,1]],data=dat)
>
> #error in model frame....:variable lengths differ().
>
> I also get the error if i just refer to variable x1 as nam[1] in
> the lm. any suggestions. I am trying to set up a for loop that
> will perform an all subsets regression and calculate the AIC for each.
There's probably a number of ways to go about it. The problem is that
nam[1] is a string vector, while you want the underlying "symbol". In
your case, I think the simplest solution would be:
frm <- formula(paste("y~",nam[1]))
mod<-lm(frm,data=dat)
This would also work:
frm <- bquote(y~.(var1), list(var1=as.name(nam[1])))
mod<-lm(frm,data=dat)
This however does have a possible sideeffect, as in the following:
frm <- bquote(y~.(var1), list(var1=as.name(nam[2])))
mod2 <- update(mod)
mod2 is now the model for x2, even though we didn't give it an
explicit new formula (The old formula had kept the difference in frm.
You can probably avoid this by:
mod<-lm(bquote(y~.(var1), list(var1=as.name(nam[1]))),data=dat)
Though again the name of the formula is not very pretty. The best
one, from the point of view of getting the correct call in lm,
probably is:
eval(bquote(lm(y~.(var1), data=dat), list(var1=as.name(nam[1]))))
> Dan
Hope this helps,
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
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