Hi Soren,
Thanks for the reply. The main reason I want to compute the derivative is to
add it to the derivative table in R so that I can use it later in my
optimization. And, my complicated function doesn’t have an analytical form of
the derivative as it involves distribution function of a bivariate normal
distribution.
Thanks,
Debangan
--
Debangan Dey,
PhD Student in Biostatistics,
Johns Hopkins Bloomberg School of Public Health,
Baltimore, MD
From: Søren Højsgaard <sor...@math.aau.dk>
Date: Wednesday, April 14, 2021 at 6:56 PM
To: Debangan Dey <dd...@jhu.edu>, r-help@r-project.org <r-help@r-project.org>
Cc: Mikkel Meyer Andersen <m...@math.aau.dk>
Subject: Re: [R] NLSR package newDeriv function
External Email - Use Caution
One approach is to compute things exact using the caracas package; see
below.
Best regards
Søren
----
> library(caracas)
> f <- function(x,y){x+y}
> def_sym(x,y,z)
> f <- x+y^2+y*z^2
> f
[caracas]: 2 2
x + y + y⋅z
>
> d1 <- der(f, c(x,y,z))
> d2 <- der2(f, c(x,y,z))
>
> d1
[caracas]: ⎡ 2 ⎤
⎣1 2⋅y + z 2⋅y⋅z⎦
> d2
[caracas]: ⎡0 0 0 ⎤
⎢ ⎥
⎢0 2 2⋅z⎥
⎢ ⎥
⎣0 2⋅z 2⋅y⎦
>
> tex(d1)
[1] "\\left[\\begin{matrix}1 & 2 y + z^{2} & 2 y
z\\end{matrix}\\right]"
> tex(d2)
[1] "\\left[\\begin{matrix}0 & 0 & 0\\\\0 & 2 & 2 z\\\\0 & 2 z & 2
y\\end{matrix}\\right]"
>
> as_expr(d1)
expression(cbind(1, 2*y + z^2, 2*y*z))
> as_expr(d2)
expression(rbind(cbind(0, 0, 0), cbind(0, 2, 2*z), cbind(0, 2*z, 2*y)))
>
On Wed, 2021-04-14 at 19:35 +0000, Debangan Dey wrote:
> Hi,
>
> I am trying to solve a non-linear least square which has a function
> from R^3 -> R. Is it possible to define gradient using newDeriv for a
> 3-variate scalar function?
>
> I am trying to use the genD function in numDeriv package to define
> numerical gradient and treat them as a function. So far, I have
> failed to do it for simple function as follows �
>
> f <- function(x,y){x+y}
> grad.f <- function(x,y){z <- c(x,y)
> f2 <- function(z){f(z[1],z[2])}
> gd <- genD(f2,z)$D[,1:length(z)]
> return(gd)
> }
> newDeriv(f(x,y), grad.f(x,y))
>
> This derivative definition is not working as the following error pops
> up �
>
> fd <- fnDeriv(~ f(x,y), c("x",�y�))
> fd(c(1,2))
> Error in .grad[, "x"] <- grad.f(x) :
> number of items to replace is not a multiple of replacement length
>
>
> I know the above example would work if I just give expressions
> instead of using grad function in numDeriv but I am trying to use
> numDeriv::grad here to see if my toy example can be translated to the
> bigger function I would later use it on.
>
> Thanks,
> Debanga
>
> --
> Debangan Dey,
> PhD Student in Biostatistics,
> Johns Hopkins Bloomberg School of Public Health,
> Baltimore, MD
>
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>
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