How to define specially nested functions

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

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.

If solving the equation for z with given x and y is the
main purpose, then try the following

  f <- function(x,y,z) x+y + (x^2-z)
  uniroot(f, c(0, 10), x=1, y=3)$root
  [1] 5

Hope this helps.

Petr Savicky.

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