On 11/12/2020 6:50 p.m., Rolf Turner wrote:
I want to fit a model y = x/(x-a) where the value of a depends
on the level of a factor z. I cannot figure out an appropriate
syntax for nls(). The "parameter" a (to be estimated) should be a
vector of length equal to the number of levels of z.
I tried:
strt <- rep(3,length(levels(z))
names(strt=levels(z)
fit <- nls(y ~ x/(x - a[z]),start=strt,data=xxx)
but of course got an error:
Error in nls(y ~ x/(x - a[z]), start = strt, data = xxx) :
parameters without starting value in 'data': a
I keep thinking that there is something obvious that I should
be doing, but I can't work out what it is.
Does there *exist* an appropriate syntax for doing what I want
to do? Can anyone enlighten me? The data set "xxx" is given
in dput() form at the end of this message.
I don't know of anything easy here. I think you need to do some tricky
stuff with formulas to get what you want. For example,
pred <- function(x, z, ...) {
a <- unlist(list(...))
names(a) <- levels(xxx$z)
x/(x-a[z])
}
strt <- rep(3,length(levels(xxx$z)))
names(strt) <- levels(xxx$z)
fla <- y ~ pred(x, z)
fla[[3]] <- as.call(c(as.list(fla[[3]]), lapply(levels(xxx$z), as.name)))
fit <- nls(fla,start=strt,data=xxx)
That line starting fla[[3]] is the ugly part: it takes the simple
formula y ~ pred(x, z) and changes it to y ~ pred(x, z, a1, a2, a3, a4,
a5). As far as I know, nls() can't handle vector paramters, it only
deals with scalars, so this passes them each as a separate argument.
Maybe rlang or one of the other packages like that has code to make this
less obscure.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
