Hi, Jerome and Phil,
Thank you for your solutions and I have studied carefully your codes but I have
further questions (since I guess the simple lines of codes may not do the real
job I am going to describe to you. Please forgive me for my shallowness!)
I guess I over-simplified my question, basically I need such a function as the
integrand for estimation of the expectation by Monte Carlo methods.
Please allow me to state the problem in more details:
I have to define a function for Monte Carlo computation of conditional
expectation and solve for the argument for which the expectation equals a
pre-specified value. Say, the integrand function is f(x,y,z), where x, z are
deterministic, y probabilistic and follows a distribution F.
I will have to feed x=x0 to f, then I sample from F for y and evaluate
f(x0,y,z), and use Monte Carlo method to get the expectation, which gives a
function of z; now that the expectation is a function of z only, say, E(z);
finally to solve for z such that E(z) = 0.5, for example.
The function f itself is very complicated and has high dimensional vectors as
arguments except z, which is a real number.
I am new in R but unexpectedly encountered this symbolic incapability of R as I
almost finished programming all major computations in R. I have been skillful
in Matlab and Mathematica (and it is very easy to do this in them) but as I am
now in statistics I would like to continue in R unless it really is not able to
do it (in that case I will have to recode in Mathematica).
Any of your further help is much appreciated!
Best regards,
-Chee
From: Jerome Asselin
Sent: Friday, April 29, 2011 12:25 AM
To: Chee Chen
Cc: R -Help
Subject: Re: [R] How to define specially nested functions
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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