Re: [R] nls(): Levenberg-Marquardt, Gauss-Newton, plinear - PI curve fitting

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Tue 21 Jun 2005 - 21:12:44 EST

On 6/21/05, Gabor Grothendieck <ggrothendieck@gmail.com> wrote:
> On 6/21/05, Christfried Kunath <mailpuls@gmx.net> wrote:
> > Hello,
> >
> > i have a problem with the function nls().
> >
> > This are my data in "k":
> > V1 V2
> > [1,] 0 0.367
> > [2,] 85 0.296
> > [3,] 122 0.260
> > [4,] 192 0.244
> > [5,] 275 0.175
> > [6,] 421 0.140
> > [7,] 603 0.093
> > [8,] 831 0.068
> > [9,] 1140 0.043
> >
> > With the nls()-function i want to fit following formula whereas a,b, and c
> > are variables: y~1/(a*x^2+b*x+c)
> >
> > With the standardalgorithm "Newton-Gauss" the fitted curve contain an peak
> > near the second x,y-point.
> > This peak is not correct for my purpose. The fitted curve should descend
> > from the maximum y to the minimum y given in my data.
> >
> > The algorithm "plinear" give me following error:
> >
> >
> > phi function(x,y) {
> > k.nls<-nls(y~1/(a*(x^2)+b*x+c),start=c(a=0.0005,b=0.02,c=1.5),alg="plinear")
> > coef(k.nls)
> > }
> >
> > phi(k[,1],k[,2])
> >
> > Error in qr.solve(QR.B, cc) : singular matrix `a' in solve
> >
> >
> > I have found in the mailinglist
> > "https://stat.ethz.ch/pipermail/r-help/2001-July/012196.html" that is if t
> > he data are artificial. But the data are from my measurment.
> >
> > The commercial software "Origin V.6.1" solved this problem with the
> > Levenberg-Marquardt algorithm how i want.
> > The reference results are: a = 9.6899E-6, b = 0.00689, c = 2.72982
> >
> > What are the right way or algorithm for me to solve this problem and what
> > means this error with alg="plinear"?
> >
> > Thanks in advance.
>
> This is not a direct answer to your question but log(y) looks nearly linear
> in x when plotting them together and log(y) ~ a + b*x or
> y ~ a*exp(b*x) will always be monotonic. Also, this model uses only 2
> rather than 3 parameters.
>

One other comment. If you do want to use your model try fitting 1/y first to get your starting value since that model has a unique solution:

> res1 <- nls(1/y ~ a*x^2 + b*x + c, start = list(a=0,b=0,c=0))
> res2 <- nls(y ~ 1/(a*x^2 + b*x + c), start = as.list(coef(res1)))
> res2

Nonlinear regression model
  model: y ~ 1/(a * x^2 + b * x + c)
   data: parent.frame()

           a b c
9.690187e-06 6.885577e-03 2.729825e+00
 residual sum-of-squares: 0.000547369



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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Jun 21 21:17:29 2005

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