Re: [R] How to define specially nested functions

From: Jerome Asselin <jerome.asselin.stat_at_gmail.com>
Date: Fri, 29 Apr 2011 00:25:14 -0400

On Thu, 2011-04-28 at 23:08 -0400, Chee Chen wrote:
> Dear All,
> I would like to define a function: f(x,y,z) with three arguments x,y,z, such that: given values for x,y, f(x,y,z) is still a function of z and that I am still allowed to find the root in terms of z when x,y are given.
> For example: f(x,y,z) = x+y + (x^2-z), given x=1,y=3, f(1,3,z)= 1+3+1-z is a function of z, and then I can use R to find the root z=5.
>
> Thank you.
> -Chee

Interesting exercise.

I've got this function, which I think it's doing what you're asking.

f <- function(x,y,z)
{

	fcall <- match.call()
	fargs <- NULL
	if(fcall$x == "x")
		fargs <- c(fargs, "x")
	if(fcall$y == "y")
		fargs <- c(fargs, "y")
	if(fcall$z == "z")
		fargs <- c(fargs, "z")
	
	ffunargs <- as.list(fargs)
	names(ffunargs) <- fargs
	
	argslist <- list(fcall)
	ffun <- append(argslist, substitute( x+y + (x^2-z) ), after=0)[[1]]
	as.function(append(ffunargs, ffun))

}

This yields.

> f(3, 2, z)
function (z = "z")
3 + 2 + (3^2 - z)
<environment: 0x132fdb8>
> f(3, 2, z)(3)
[1] 11

I haven't figured out how to get rid of the default argument value shown here as 'z = "z"'. That doesn't prevent it to work, but it's less pretty. If you find a better way, let me know.

HTH,
Jerome



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