R help - Apr 2010 - Curve Fitting

I am having troubles in fitting functions of the form

y~a*x^b+c
 to data, for example

x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)

I tried for example with nls, which did only work with really good initial
guessed values.
Any suggestion, what I should use?
Thanks a lot
Thomas


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Thomas,

I think the issue of having reasonable starting values is inherent in all 
nonlinear optimization problems (unless they have some additional 
properties like convexity, for example).  Using a different algorithm may 
or may not help.  In fact, a vast majority of existing algorithms 
guarantees only local convergence provided that the starting point is 
within a certain distance from the solution.

If you are interested in this particular model, the following procedure 
seems to give reasonable starting point.

1.  Take a guess at c.  This is relatively easy since c is the value of y 
when a=0.  For your data a crude guess could be c=8.
2.  For a fixed value of c your model can be linearized by the following 
transformations
        y1 <- log(y-8)
        x1 <- log(x)
3.  Do
        summary(lm(y1~x1))
     and obtain values for intecept and slope
4.  Your starting values for nls are (a=exp(Intercept), b=slope, c=8) 

Hope this helps,

Andy

__________________________________
Andy Jaworski
518-1-01
Process Laboratory
3M Corporate Research Laboratory
-----
E-mail: apjaworski@mmm.com
Tel:  (651) 733-6092
Fax:  (651) 736-3122



From:
Thomas Bschorr <bschorr@phys.ethz.ch>
To:
r-help@r-project.org
Date:
04/30/2010 03:33 PM
Subject:
[R] Curve Fitting
Sent by:
<r-help-bounces@r-project.org>



I am having troubles in fitting functions of the form

y~a*x^b+c
 to data, for example

x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)

I tried for example with nls, which did only work with really good initial 
guessed values.
Any suggestion, what I should use?
Thanks a lot
Thomas


                 [[alternative HTML version deleted]]

______________________________________________
R-help@r-project.org mailing list
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]]

You can use nls2 to try many starting values. It works just like nls but:

- if you give it a two row data frame as the start value it will
create a grid between the upper and lower values of each parameter and
then run an optimization starting at each such point on the grid
returning the best

- it has all the algorithms that nls has (because it calls nls) plus
an additional algorithm called "brute-force" which simply evaluates
the formula at the start value

Thus you can use it in two ways:

1. perform a brute force run and take the best to get a good starting
value and using that starting value run nls, or
2. run an nls from each starting value

Obviously the second approach takes more run time but has some
advantages in certain cases.  In many cases either approach will work.
> library(nls2)
> # 1. First approach
> fo <- y ~ a * x ^ b + c
> start <- data.frame(a = c(-10, 10), b = c(-10, 10), c = c(-10, 10))
> fm.brute <- nls2(fo, start = start, alg = "brute")
> fm <- nls2(fo, start = fm.brute)
> fm
Nonlinear regression model
  model:  y ~ a * x^b + c
   data:  <environment>
     a      b      c
3.1865 0.5025 7.0900
 residual sum-of-squares: 5.677e-05

Number of iterations to convergence: 7
Achieved convergence tolerance: 4.437e-06
> # 2. Second approach
> fm2 <- nls2(fo, start = start, control = nls.control(warnOnly = TRUE))
There were 32 warnings (use warnings() to see them)
> fm2
Nonlinear regression model
  model:  y ~ a * x^b + c
   data:  <environment>
     a      b      c
3.1865 0.5025 7.0900
 residual sum-of-squares: 5.677e-05

Number of iterations to convergence: 9
Achieved convergence tolerance: 1.149e-07



On Fri, Apr 30, 2010 at 4:30 PM, Thomas Bschorr <bschorr at phys.ethz.ch>
wrote:
> I am having troubles in fitting functions of the form
>
> y~a*x^b+c
> ?to data, for example
>
> x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
> y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)
>
> I tried for example with nls, which did only work with really good initial
guessed values.
> Any suggestion, what I should use?
> Thanks a lot
> Thomas
>
>
> ? ? ? ?[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> 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.
>