This is not the place to ask a question: do read the FAQ.
?nls shows that nls does have a 'subset' argument, and it does work.
Compare
> ?nls
> x <- 1:10
> y <- 2*x + 3 # perfect fit
> yeps <- y + rnorm(length(y), sd = 0.01) # added noise
> nls(yeps ~ a + b*x, start = list(a = 0.12345, b = 0.54321),
+ trace = TRUE)
1360.779 : 0.12345 0.54321
0.000778241 : 3.002946 1.999563
0.000778241 : 3.002946 1.999563
Nonlinear regression model
model: yeps ~ a + b * x
data: parent.frame()
a b
3.002946 1.999563
residual sum-of-squares: 0.000778241
> nls(yeps ~ a + b*x, start = list(a = 0.12345, b = 0.54321),
+ trace = TRUE, subset=1:9)
1056.760 : 0.12345 0.54321
0.0007055261 : 3.000838 2.000138
0.0007055261 : 3.000838 2.000138
Nonlinear regression model
model: yeps ~ a + b * x
data: parent.frame()
a b
3.000838 2.000138
residual sum-of-squares: 0.0007055261
I find your comments baffling: please provide a reproducible example of
your claims.
On Wed, 11 Oct 2006, [EMAIL PROTECTED] wrote:
> Full_Name: Tadashi Kadowaki
> Version: 2.4.0
> OS: Redhat Linux 9
> Submission from: (NULL) (58.12.166.67)
>
>
> Doesn't nls function support subset? It seems not to work.
> And, there are no information in the online help.
> Has it sunk into oblivion?
>
> ______________________________________________
> R-devel@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
