[R] getting variable length numerical gradient
Randall R Schulz
rschulz at sonic.net
Mon Sep 26 15:53:50 CEST 2005
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.)
On Sunday 25 September 2005 04:07, Dimitris Rizopoulos wrote:
> 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
More information about the R-help
mailing list