[R] using eval() with pre-built expression inside function

[Roger D. Peng]

If you look at the beginning of lm(), you'll see that match.call() is 
used and name of the function (in this case "f1") is replaced with 
"model.frame".  Does something like this work?

f1 <- function(formula, data) {
	mf <- match.call(expand.dots = FALSE)
	mf[[1]] <- as.name("model.frame")
	eval(mf, parent.frame())
}

-roger

Heather Turner wrote:
> I'm trying to evaluate a pre-built expression using eval(), e.g.
> 
> dataset <- data.frame(y = runif(30, 50,100), x = gl(5, 6))
> 
> # one like this
> mf <- expression(model.frame(y~x))
> eval(mf, dataset, parent.frame())
> 
> # rather than this
> eval(expression(model.frame(y~x)), dataset, parent.frame())
> 
> In the example above there is no problem, the problem comes when I try to do a similar thing within a function, e.g.
> 
> f1 <- function(formula, data) {
>     mt <- terms(formula)
>     mf <- as.expression(as.call(c(as.name("model.frame"), formula = mt)))
>     eval(mf, data, parent.frame())
> }
> 
> 
>>f1(formula = y ~ x, data = dataset)
> 
> Error in eval(expr, envir, enclos) : Object "y" not found
> 
> I can get round this by building a call to eval using paste, e.g.
> 
> f2 <- function(formula, data) {
>     mt <- terms(formula)
>     mf <- as.expression(as.call(c(as.name("model.frame"), formula = mt)))
>     direct <- parse(text = paste("eval(expression(", mf,
>                "), data, parent.frame())"))
>     print(direct)
>     eval(direct)
> }
> 
> 
>>f2(formula = y ~ x, data = dataset)
> 
> expression(eval(expression(model.frame(formula = y ~ x)), data, 
>     parent.frame()))
>           y x
> 1  92.23087 1
> 2  63.43658 1
> 3  55.24448 1
> 4  72.75650 1
> 5  67.58781 1
> ...
> 
> but this seems rather convoluted. Can anyone explain why f1 doesn't work (when f2 does) and/or suggest a neater way of dealing with this?
> 
> Thanks
> 
> Heather
> 
> Mrs H Turner
> Research Assistant
> Dept. of Statistics
> University of Warwick
> 
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-- 
Roger D. Peng
http://www.biostat.jhsph.edu/~rpeng/