[R] converting to a language object

[ripley@stats.ox.ac.uk]

In this case a simler answer is that you want a formula, so use 
as.formula:

> eq <- "y ~ x1 + x2 +x2000"
> as.formula(eq)
y ~ x1 + x2 + x2000

once the quotation marks are in the right place.

It is much simpler to use lm(y ~ ., data=foo) !

A caveat: you may hit some limits on expression size if you try to put
2000 variables in a formula, and almost certainly you will hit
computational limits if you try to do a regression on it, as you need
n >> p = 2000, and the key computation is (I think) O(np^2).

On Fri, 7 Feb 2003, Marc Schwartz wrote:

> Arnab mukherji wrote:
> > Hi,
> > 
> > I have a query about the following:
> > 
> > CONTEXT: if we define:
> > 
> > eq <- y ~ x1 + x2
> > 
> > then is.language(eq) returns true. this works perfectly with commands
> > such as lm(eq,data=my.data)
> > 
> > THE PROBLEM: Now I have a big data set with about 2000 independent
> > variables. So I tried to automate this. I can pick off the names of
> > the variables and insert a plus in between them and get a string.
> > Thus I have
> > 
> > eq<- y ~ " x1 + x2 + ... +x2000" or eq<-"y ~ " x1 + x2 + ... +x2000"
> > 
> > from either case how can I typecast eq into a langugae object and get
> > the lm command to work ?
> > 
> > Does anyone have any suggestions?
> > 
> > thanks
> > 
> > Arnab.
> 
> 
> Arnab,
> 
> Once you have your full formula string constructed you can use:
> 
> eq <- eval(parse(text = string))
> 
> This will put the model formula into eq.
> 
> 
> For example try:
> 
> eq <- eval(parse(text = "y ~ x1 + x2"))
> 
> 
> See ?eval and ?parse for more information.
> 
> Hope that helps.
> 
> 
> Regards,
> 
> Marc Schwartz
> 
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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
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