bind(1, x1, x2))
> >
> > ## The true Hessian
> > solve(summary(fm)$cov.unscaled)
> >
> >
> > I hope it helps.
> >
> > Best,
> > Dimitris
> >
> >
> > Dimitris Rizopoulos
> > Ph.D. Student
> > Biostatistical Centre
>
Dimitris,
On Monday 26 September 2005 07:16, Dimitris Rizopoulos wrote:
> 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, ...
>
> I hope it is more clear now.
t; <[EMAIL PROTECTED]>
To: "R Help"
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 w
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.
> 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/(0)16/336899
> Fax: +32/(0)16
From: "Antonio, Fabio Di Narzo" <[EMAIL PROTECTED]>
To:
Sent: Sunday, September 25, 2005 11:37 AM
Subject: [R] getting variable length numerical gradient
> Hi all.
> I have a numerical function f(x), with x being a vector of generic
> size (say k=4), and I wanna take the nu
Hi all.
I have a numerical function f(x), with x being a vector of generic
size (say k=4), and I wanna take the numerically computed gradient,
using deriv or numericDeriv (or something else).
My difficulties here are that in deriv and numericDeric the function
is passed as an expression, and one h
