[R] Fitting exponential curve to data points

Stephen Tucker brown_emu at yahoo.com
Tue Jul 24 11:37:24 CEST 2007


Well spoken. And since log transformations are nonlinear and 'compresses' the
data, it's not surprising to find that the fit doesn't look so nice while the
fit metrics tell you that a model does a good job.

--- ted.harding at nessie.mcc.ac.uk wrote:

> On 24-Jul-07 01:09:06, Andrew Clegg wrote:
> > Hi folks,
> > 
> > I've looked through the list archives and online resources, but I
> > haven't really found an answer to this -- it's pretty basic, but I'm
> > (very much) not a statistician, and I just want to check that my
> > solution is statistically sound.
> > 
> > Basically, I have a data file containing two columns of data, call it
> > data.tsv:
> > 
> > year  count
> > 1999  3
> > 2000  5
> > 2001  9
> > 2002  30
> > 2003  62
> > 2004  154
> > 2005  245
> > 2006  321
> > 
> > These look exponential to me, so what I want to do is plot these
> > points on a graph with linear axes, and add an exponential curve over
> > the top. I also want to give an R-squared for the fit.
> > 
> > The way I did it was like so:
> > 
> > 
> ># Read in the data, make a copy of it, and take logs
> > data = read.table("data.tsv", header=TRUE)
> > log.data = data
> > log.data$count = log(log.data$count)
> > 
> ># Fit a model to the logs of the data
> > model = lm(log.data$count ~ year, data = log.data)
> > 
> ># Plot the original data points on a graph
> > plot(data)
> > 
> ># Draw in the exponents of the model's output
> > lines(data$year, exp(fitted(model)))
> > 
> > 
> > Is this the right way to do it? log-ing the data and then exp-ing the
> > results seems like a bit of a long-winded way to achieve the desired
> > effect. Is the R-squared given by summary(model) a valid measurement
> > of the fit of the points to an exponential curve, and should I use
> > multiple R-squared or adjusted R-squared?
> > 
> > The R-squared I get from this method (0.98 multiple) seems a little
> > high going by the deviation of the last data point from the curve --
> > you'll see what I mean if you try it.
> 
> I just did. From the plot of log(count) against year, with the plot
> of the linear fit of log(count)~year superimposed, I see indications
> of a non-linear relationship.
> 
> The departures of the data from the fit follow a rather systematic
> pattern. Initially the data increase more slowly than the fit,
> and lie below it. Then they increase faster and corss over above it.
> Then the data increase less fast than the fit, and the final data
> point is below the fit.
> 
> There are not enough data to properly identify the non-linearity,
> but the overall appearance of the data plot suggests to me that
> you should be considering one of the "growth curve" models.
> 
> Many such models start of with an increasing rate of growth,
> which then slows down, and typically levels off to an asymptote.
> The apparent large discrepancy of your final data point could
> be compatible with this kind of behaviour.
> 
> At this point, knowledge of what kind of thing is represented
> by your "count" variable might be helpful. If, for instance,
> it is the count of the numbers of individuals of a species in
> an area, then independent knowledge of growth mechanisms may
> help to narrow down the kind of model you should be tring to fit.
> 
> As to your question about "Is this the right way to do it"
> (i.e. fitting an exponential curve by doing a linear fit of the
> logarithm), generally speaking the answer is "Yes". But of course
> you need to be confident that "exponential" is the right curve
> to be fitting in the first place. If it's the wrong type of
> curve to be considering, then it's not "the right way to do it"!
> 
> Hoping this help[s,
> Ted.
> 
> --------------------------------------------------------------------
> E-Mail: (Ted Harding) <ted.harding at nessie.mcc.ac.uk>
> Fax-to-email: +44 (0)870 094 0861
> Date: 24-Jul-07                                       Time: 10:08:33
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