getting variable length numerical gradient

Randall,

thanks for your comments; however, you have to take into account what 
is the purpose of the function here! The goal is to approximate 
*partial* derivatives numerically, using in fact the definition of the 
partial derivatives. If you recall this definition I hope that you can 
see why I change the ith element of the x vector and not the whole 
one. You could also test your approach with the original one in the 
logistic regression example and see the difference.

I hope it is more clear now.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Randall R Schulz" <rschulz at sonic.net>
To: "R Help" <R-Help at stat.math.ethz.ch>
Sent: Monday, September 26, 2005 3:53 PM
Subject: Re: [R] getting variable length numerical gradient


Dimitris,

I'm new to R programming, and I'm trying to learn the proper way to do
certain things. E.g., I had a piece of code with explicit iteration to
apply some computations to a vector. It was pretty slow. I found a way
to utilize R's built-in vectorization and it was sped up considerably.

So I want to ask about the code you supplied. Please see below.

(By the way, this message is best viewed using a mono-spaced font.)

maybe you can find the following function useful (any comments are
greatly appreciated):

fd <- function(x, f, scalar = TRUE, ..., eps =
sqrt(.Machine$double.neg.eps)){
    f <- match.fun(f)
    out <- if(scalar){
        ...
    } else{
        n <- length(x)
        res <- array(0, c(n, n))
        f0 <- f(x, ...)
        ex <- pmax(abs(x), 1)
        for(i in 1:n){

This (following) statement will create a copy of the entire "x" vector
on each iteration. It doesn't look like that's what you would want to
do:

            x. <- x

The computation described by this statement could be vectorized 
outside
the loop:

            x.[i] <- x[i] + eps * ex[i]

            res[, i] <- c(f(x., ...) - f0) / (x.[i] - x[i])
        }
        res
    }
    out
}

Offhand, I cannot tell for sure if the last line of that loop is
vectorizable, but I have a hunch it is.

So at a minimum, it seems this fragment of your code:

for(i in 1:n){
x. <- x
x.[i] <- x[i] + eps * ex[i]
res[, i] <- c(f(x., ...) - f0) / (x.[i] - x[i])
}

Could be more efficiently and succinctly replaced with this:

        x. <- x + eps * ex
        for (in in 1:n)
            res[, i] <- c(f(x., ...) - f0) / (x.[i] - x[i])


Could your someone else with R programming experience comment?


Thanks.


Randall Schulz

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