R help - May 2006 - Substitute and quotes

Hi folks!

 

I'm trying to update a model using:

 

form <- substitute(.~.-bad.var, list(bad.var=bad.var))

model.update<-update(model, form)

 

where bad.var is a character string like lag(X1, 10)

 

The problem is that substitute puts lag(X1, 10) in quotes - something I
don't want it to do.

It even does this if I use noquote(bad.var).  Any suggestions around
this?

 

Thanks!

 

John


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On Tue, 16 May 2006, Kerpel, John wrote:
> I'm trying to update a model using:
>
> form <- substitute(.~.-bad.var, list(bad.var=bad.var))
> model.update<-update(model, form)
>
> where bad.var is a character string like lag(X1, 10)
>
> The problem is that substitute puts lag(X1, 10) in quotes - something I
> don't want it to do.
This is just a symptom. The problem is that lag(X1,10) is a character 
string, and you want a parsed expression

The solution depends on why lag(X1,10) is a string.  If you have
   bad.var <- "lag(X1,10)"
in your code you can replace it by
   good.var <- quote(lag(X1,10))

If "lag(X1,10)" comes in as a string in input then you need to parse
it
    good.var <- parse(text=bad.var)


Also, the call to substitute() returns an unevaluated call rather than a 
formula, so you may need to eval() it before passing it to update().

 	-thomas