# [R] How to define specially nested functions

Jerome Asselin jerome.asselin.stat at gmail.com
Fri Apr 29 06:25:14 CEST 2011

```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)[]
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)
 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

```