[R] Q: selecting a name when it is known as a string

[D. R. Evans]

D. R. Evans said the following at 09/04/2007 04:14 PM :
> I am 100% certain that there is an easy way to do this, but after

I have reconsidered this and now believe it to be essentially impossible
(or at the very least remarkably difficult) although I don't understand why
it is so :-(

At least, I spent another two hours trying variations on the suggestions I
received, but still nothing worked properly.

It sure seems like it _ought_ to be easy, because of the following argument:

If I type an expression such as "A <- <something>" then R is perfectly
capable of parsing the <something> and executing it and assigning the
result to A. So it seems to follow that it ought to be able to parse a
string that contains exactly the same sequence of characters (after all,
why should the R parsing engine care whether the input string comes from
the terminal or from a variable?) and therefore it should be possible to
assign "<something>" to a variable and then have R parse that variable
precisely as if it had been typed.

That was my logic as to why this ought to be easy, anyway. (And there was
the subsidiary argument that this is easy in the other languages I use, but
R is sufficiently different that I'm not certain that that argument carries
much force.)

It does seem that there are several ways to make the

  lo <- loess(percent ~ ncms * ds, d, control=loess.control(trace.hat =
> 'approximate'))

command work OK if the right hand side is in a character variable, but I
haven't been able to find a way to make

  grid <- data.frame(expand.grid(ds=MINVAL:MAXVAL, ncms=MINCMS:MAXCMS))

work.

I always end up with a parse error or a complaint that "'newdata' does not
contain the variables needed" when I perform the next task:

  plo <- predict(lo, grid).

So I guess I have to stick with half a dozen compound "if" statements, all
of which do essentially the same thing :-(