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Antonio, Fabio Di Narzo

2005-Sep-25 09:37 UTC

[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 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 have to pass the list of variables
involved as a char vector... So, it's a pure R programming question.


Have a nice sunday,
Antonio, Fabio Di Narzo.

Dimitris Rizopoulos

2005-Sep-25 11:07 UTC

head link

[R] getting variable length numerical gradient

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){
        if(length(f0 <- f(x, ...)) != length(x))
            stop("'f' must be vectorized")
        x. <- x + eps * pmax(abs(x), 1)
        c(f(x., ...) - f0) / (x. - x)
    } else{
        n <- length(x)
        res <- array(0, c(n, n))
        f0 <- f(x, ...)
        ex <- pmax(abs(x), 1)
        for(i in 1:n){
            x. <- x
            x.[i] <- x[i] + eps * ex[i]
            res[, i] <- c(f(x., ...) - f0) / (x.[i] - x[i])
        }
        res
    }
    out
}


## Examples

x <- seq(-3.3, 3.3, 0.1)
all.equal(fd(x, pnorm, mean = 0.5), dnorm(x, mean = 0.5))


# Approximate the Hessian matrix for a logistic regression

# the score vector function
gn <- function(b, y, X){
    p <- as.vector(plogis(X %*% b))
    -colSums(X * (y - p))
}

# We simulate some data and fit the logistic regression
n <- 800
x1 <- runif(n,-3, 3); x2 <- runif(n, -3, 3)
pr <- plogis(0.8 + 0.4 * x1 - 0.3 * x2)
y <- rbinom(n, 1, pr)
fm <- glm(y ~ x1 + x2, binomial)

## The Hessian using forward difference approximation
fd(fm$coef, gn, scalar = FALSE, y = y, X = cbind(1, x1, x2))

## The true Hessian
solve(summary(fm)$cov.unscaled)


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)16/337015
Web: http://www.med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Antonio, Fabio Di Narzo" <antonio.fabio at gmail.com>
To: <R-help at stat.math.ethz.ch>
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 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 have to pass the list of 
> variables
> involved as a char vector... So, it's a pure R programming question.
>
>
> Have a nice sunday,
> Antonio, Fabio Di Narzo.
>
> ______________________________________________
> R-help at 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
> 

Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

Antonio, Fabio Di Narzo

2005-Sep-27 13:27 UTC

head link

[R] getting variable length numerical gradient

2005/9/26, Dimitris Rizopoulos <dimitris.rizopoulos at
med.kuleuven.be>:> AFAIK the deriv() is for symbolic derivatives, so I don't know if it
> will work in your case.
numericDeriv surely computes numerical gradient. The problem here is
its interface, which requires the vector of symbols involved in the
expression to differenziate.

>
> Best,
> Dimitris
>
>
> ----- Original Message -----
> From: "Antonio, Fabio Di Narzo" <antonio.fabio at
gmail.com>
> To: "Dimitris Rizopoulos" <dimitris.rizopoulos at
med.kuleuven.be>
> Cc: <R-help at stat.math.ethz.ch>
> Sent: Monday, September 26, 2005 10:58 AM
> Subject: Re: getting variable length numerical gradient
>
>
> Tnx very much Dimitris,
> your code does what I need. I've just adapted it to my needs (e.g., I
> don't deal with scalar functions), and so solved my problem.
>
> Given this, is there a way to use the deriv function in the base
> package, within this context (variable length vector of indipendent
> variables)?
>
> Best,
> Antonio, Fabio Di Narzo.
>
> On 9/25/05, Dimitris Rizopoulos <dimitris.rizopoulos at
med.kuleuven.be>
> wrote:
> > maybe you can find the following function useful (any comments are
> > greatly appreciated):
> >
> > fd <- function(x, f, scalar = TRUE, ..., eps > >
sqrt(.Machine$double.neg.eps)){
> >     f <- match.fun(f)
> >     out <- if(scalar){
> >         if(length(f0 <- f(x, ...)) != length(x))
> >             stop("'f' must be vectorized")
> >         x. <- x + eps * pmax(abs(x), 1)
> >         c(f(x., ...) - f0) / (x. - x)
> >     } else{
> >         n <- length(x)
> >         res <- array(0, c(n, n))
> >         f0 <- f(x, ...)
> >         ex <- pmax(abs(x), 1)
> >         for(i in 1:n){
> >             x. <- x
> >             x.[i] <- x[i] + eps * ex[i]
> >             res[, i] <- c(f(x., ...) - f0) / (x.[i] - x[i])
> >         }
> >         res
> >     }
> >     out
> > }
> >
> >
> > ## Examples
> >
> > x <- seq(-3.3, 3.3, 0.1)
> > all.equal(fd(x, pnorm, mean = 0.5), dnorm(x, mean = 0.5))
> >
> >
> > # Approximate the Hessian matrix for a logistic regression
> >
> > # the score vector function
> > gn <- function(b, y, X){
> >     p <- as.vector(plogis(X %*% b))
> >     -colSums(X * (y - p))
> > }
> >
> > # We simulate some data and fit the logistic regression
> > n <- 800
> > x1 <- runif(n,-3, 3); x2 <- runif(n, -3, 3)
> > pr <- plogis(0.8 + 0.4 * x1 - 0.3 * x2)
> > y <- rbinom(n, 1, pr)
> > fm <- glm(y ~ x1 + x2, binomial)
> >
> > ## The Hessian using forward difference approximation
> > fd(fm$coef, gn, scalar = FALSE, y = y, X = cbind(1, x1, x2))
> >
> > ## The true Hessian
> > solve(summary(fm)$cov.unscaled)
> >
> >
> > 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)16/337015
> > Web: http://www.med.kuleuven.be/biostat/
> >      http://www.student.kuleuven.be/~m0390867/dimitris.htm
> >
> >
> > ----- Original Message -----
> > From: "Antonio, Fabio Di Narzo" <antonio.fabio at
gmail.com>
> > To: <R-help at stat.math.ethz.ch>
> > 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 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 have to pass the list of
> > > variables
> > > involved as a char vector... So, it's a pure R programming
> > > question.
> > >
> > >
> > > Have a nice sunday,
> > > Antonio, Fabio Di Narzo.
> > >
> > > ______________________________________________
> > > R-help at 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
> > >
> >
> >
> > Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
> >
> >
>
>
> Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
>
>

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