[R] nls() syntax
Rolf Turner
r@turner @end|ng |rom @uck|@nd@@c@nz
Sat Dec 12 00:50:15 CET 2020
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.
cheers,
Rolf Turner
--
Honorary Research Fellow
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276
Data set "xxx":
structure(list(x = c(30, 40, 50, 60, 70, 80, 90, 100, 110, 120,
130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250,
30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160,
170, 180, 190, 200, 210, 220, 230, 240, 250, 30, 40, 50, 60,
70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190,
200, 210, 220, 230, 240, 250, 30, 40, 50, 60, 70, 80, 90, 100,
110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230,
240, 250, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140,
150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250), y = c(1.27,
1.16, 1.19, 1.15, 1.09, 1.07, 1.07, 1.05, 1.07, 1.03, 1.05, 1.07,
1.06, 1.03, 1.05, 1.04, 1.03, 1.03, 1.03, 1.02, 1.02, 1.01, 1.01,
1.21, 1.15, 1.1, 1.1, 1.06, 1.06, 1.05, 1.03, 1.07, 1.04, 1.04,
1.02, 1.04, 1.02, 1.04, 1.03, 1.01, 1.03, 1.01, 1, 1.02, 1.03,
1.02, 1.42, 1.27, 1.23, 1.14, 1.17, 1.08, 1.11, 1.06, 1.07, 1.08,
1.06, 1.07, 1.04, 1.03, 1.07, 1.04, 1.03, 1.03, 1.03, 1.04, 1.03,
1.03, 1.04, 1.85, 1.41, 1.35, 1.21, 1.22, 1.15, 1.14, 1.07, 1.1,
1.09, 1.1, 1.09, 1.08, 1.08, 1.09, 1.09, 1.07, 1.06, 1.03, 1.08,
1.05, 1.02, 1.05, 1.99, 1.6, 1.44, 1.4, 1.24, 1.3, 1.21, 1.23,
1.18, 1.18, 1.12, 1.15, 1.09, 1.07, 1.13, 1.1, 1.05, 1.13, 1.09,
1.03, 1.11, 1.07, 1.05), z = structure(c(1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), .Label = c("p1",
"p2", "p3", "p4", "p5"), class = "factor")), class = "data.frame", row.names = c(NA,
-115L))
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