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

Gabor Grothendieck ggrothendieck at gmail.com
Tue Jun 21 12:57:19 CEST 2005


On 6/21/05, Christfried Kunath <mailpuls at 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.




More information about the R-help mailing list