Re: [R] Can't get function to run iteratively

From: Dimitris Rizopoulos <dimitris.rizopoulos_at_med.kuleuven.ac.be>
Date: Fri 08 Apr 2005 - 20:33:42 EST

R is a vectorized language, look at this:

fd <- function(b, gn, ..., eps=sqrt(.Machine$double.neg.eps)){

    n <- length(b)
    g0 <- gn(b, ...)
    out <- numeric(n)
    b. <- b + eps*max(abs(b), 1)
    c(gn(b., ...) - g0) / (b. - b)
}
##############
test <- function(x) x*x
fd(seq(0.1, 0.5, 0.1), test)
2*seq(0.1, 0.5, 0.1)

I hope it helps.

Best,
Dimitris



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

Address: Kapucijnenvoer 35, Leuven, Belgium

Tel: +32/16/336899
Fax: +32/16/337015
Web: http://www.med.kuleuven.ac.be/biostat/
     http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm


> Trying to do something very simple...
>
>> numerical.grad
> function(func,x, eps=1e-12) {
> # very simple (crude) numerical approximation
> f <-func(x)
> df <-1:length(x)
> for (i in 1:length(x)) {
> dx <- x
> dx[i] <- dx[i] +eps
> df[i] <- (func(dx)-f)/eps
> }
> df
> }
>> test<-function(x){x^2}
>> y<-seq(0.1,0.5,0.1)
>> numerical.grad(test,y)
> [1] 0.1999997 0.0000000 0.0000000 0.0000000 0.0000000
> Warning messages:
> 1: number of items to replace is not a multiple of replacement
> length
> 2: number of items to replace is not a multiple of replacement
> length
> 3: number of items to replace is not a multiple of replacement
> length
> 4: number of items to replace is not a multiple of replacement
> length
> 5: number of items to replace is not a multiple of replacement
> length
>>
>
>
> Any thoughts why this doesn't seem to be working ?
>
>
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R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Fri Apr 08 20:40:38 2005

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