[R] [EXT] Re: nls() syntax

[Andrew Robinson]

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: apro using 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-bounces using r-project.org> on behalf of Rolf Turner <r.turner using auckland.ac.nz>
Sent: Saturday, December 12, 2020 1:39:11 PM
To: Gabor Grothendieck <ggrothendieck using gmail.com>
Cc: r-help using r-project.org <r-help using r-project.org>
Subject: [EXT] Re: [R] nls() syntax

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On Fri, 11 Dec 2020 19:20:26 -0500
Gabor Grothendieck <ggrothendieck using 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.turner using 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))

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