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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