Question about curve fitting...
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...
On Wed, 10 Aug 2005, S.O. Nyangoma wrote:
I see that log(y)=log(k1)+k2*log(x) use lm?
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.
----- 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...
Meta: This question is somewhat long and has two parts, I would be very happyfor someone just to nudge me in the right direction with the manual / tutorial, as I am somewhat lost... 1) How do I fit a curve of the form "y = k1 * x^k2" ? I want to estimate values of k1 and k2 given the x/y data I have, and I can't work out how to get R to calculate and return their estimates. 2) Given the value of k1 and k2 for population A, how can I test if population B has significantly different values of k1 and k2? Sorry for the basic question. I think I just need to read up on a few functions. I have about 50 xy pairs in total if that makes a difference. Dan.
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