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Question about curve fitting...

6 messages · S.O. Nyangoma, Dan Bolser, Bert Gunter +3 more

Original
S.O. Nyangoma
# 
I see that 
log(y)=log(k1)+k2*log(x)
use lm?

----- Original Message -----
From: Dan Bolser <dmb at mrc-dunn.cam.ac.uk>
Date: Wednesday, August 10, 2005 11:41 am
Subject: [R] Question about curve fitting...
Dan Bolser
#
On Wed, 10 Aug 2005, S.O. Nyangoma wrote:

            

      
      
      
    

        
Thats a nice solution in this instance, but in general how do I get R to
fit a particular function (formula) and return the parameters?

Cheers,
Dan.

            

      
      
      
    

        

      
      
    

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Bert Gunter
    

    

      
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As you appeared to have received no reply ...

1) Use nls() on the original equation

2) Transforming first and using linear fitting is **NOT** the same. The
error structures differ and therefore you get different results. The
greatest effect is on inferences -- i.e confidence intervals for the
parameters: the usual asymptotic intervals would be symmetric based on the
original scale, asymmetric based on the transformed. For reasonably
well-behaved data the point estimates shouldn't be too much different,
however. In any case, the linearization is a good way to find starting
values.


-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
"The business of the statistician is to catalyze the scientific learning
process."  - George E. P. Box

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Peter Dalgaard
    

    

      
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Dan Bolser <dmb at mrc-dunn.cam.ac.uk> writes:

            

      
      
      
    

        
nls() is your friend.

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Dimitris Rizopoulos
    

    

      
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I think here it's important to consider how the errors term come into 
the model. If "y = k1 * x^k2 * e" then in the log-scale you have a 
linear model; however if you assume that "y = k1 * x^k2 + e", the you 
want a nonlinear model (i.e., nls()). For instance,

x <- runif(500, 1, 3)
y <- 1 * x^2 + rnorm(500)
m <- nls(y ~ exp(k1 + k2 * log(x)), start = c("k1" = 1, "k2" = 2))
c(exp(coef(m)[1]), coef(m)[2])


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/336899
Fax: +32/16/337015
Web: http://www.med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Dan Bolser" <dmb at mrc-dunn.cam.ac.uk>
To: "S.O. Nyangoma" <S.O.Nyangoma at amc.uva.nl>
Cc: "R mailing list" <r-help at stat.math.ethz.ch>
Sent: Wednesday, August 10, 2005 4:53 PM
Subject: Re: [R] Question about curve fitting...

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Gabor Grothendieck
    

    

      
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Note that even if you decide that this distinction is applicable,
you may still wish to run a linear model prior to
nls to get the starting values.

        
On 8/10/05, Dimitris Rizopoulos <dimitris.rizopoulos at med.kuleuven.be> wrote: