Hi Rolf, It might also be worth experimenting to see if
y ~ 1 / ( 1 - a[z]/x ) yields the same result. It might be cleaner if x appears only once in the expression. Cheers, Andrew -- Andrew Robinson Director, CEBRA; Professor of Biosecurity School of BioSciences University of Melbourne, VIC 3010 Australia Tel: (+61) 0403 138 955 Email: a...@unimelb.edu.au Website: http://cebra.unimelb.edu.au/ I acknowledge the traditional owners of the land I inhabit and pay my respects to their elders. ________________________________ From: R-help <r-help-boun...@r-project.org> on behalf of Rolf Turner <r.tur...@auckland.ac.nz> Sent: Saturday, December 12, 2020 1:39:11 PM To: Gabor Grothendieck <ggrothendi...@gmail.com> Cc: r-help@r-project.org <r-help@r-project.org> Subject: [EXT] Re: [R] nls() syntax UoM notice: External email. Be cautious of links, attachments, or impersonation attempts On Fri, 11 Dec 2020 19:20:26 -0500 Gabor Grothendieck <ggrothendi...@gmail.com> wrote: > The start= argument should be as follows: > > nls(y ~ x/(x - a[z]),start=list(a = strt),data=xxx) Nya-ha! I, uh, clearly wasn't thinking clearly! (Feel a bit *duh* now!) Thanks Gabor. cheers, Rolf > > On Fri, Dec 11, 2020 at 6:51 PM Rolf Turner <r.tur...@auckland.ac.nz> > 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. > > > > 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)) ______________________________________________ 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. [[alternative HTML version deleted]] ______________________________________________ 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.
