On Aug 26, 2010, at 1:48 PM, David Winsemius wrote:
On Aug 26, 2010, at 1:35 PM, Marlin Keith Cox wrote:
I need the parameters estimated for a non-linear equation, an
example of the
data is below.
# rm(list=ls()) I really wish people would add comments to
destructive pieces of code.
Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8)
Level<-c( 100, 110, 90, 95, 87, 60, 65, 61, 55, 57, 40,
41, 50,
47,
44, 44, 42, 38, 40, 37, 37, 35, 40, 34, 32, 20, 22,
25, 27,
29)
plot(Time,Level,pch=16)
You did not say what sort of "non-linear equation" would best suit,
nor did you offer any background regarding the domain of study.
There must be many ways to do this. After looking at the data, a
first pass looks like this:
> lm(log(Level) ~Time )
An effort to chide me offlist for not actually saying this was a
"linear model" has fallen on unreceptive ears, since it does estimate
a non-linear _equation_ which was what was requested. However, this
chiding has lead me to expand my repertoire of worked solutions to
include an nls() solution to a Weibull equation:
> fm1 = nls(Level~ exp(-A*Time^CC) , start = list(CC=.2, A = .1),
control=list(maxiter=200, minFactor=.00001), alg = 'plinear')
> fm1
Nonlinear regression model
model: Level ~ exp(-A * Time^CC)
data: parent.frame()
CC A .lin
0.7045 0.3250 96.5518
residual sum-of-squares: 680.6
Number of iterations to convergence: 5
Achieved convergence tolerance: 2.338e-06
> plot(Level~Time)
> points(unique(Time), predict(fm1,
data.frame(Time=unique(Time) ) ) , col="red", pch=4)
Maybe a Weibull model would be more appropriate.
Yes it appears to be.
--
David Winsemius, MD
West Hartford, CT
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