[R] Fitting a curve to data

Greg Snow Greg.Snow at imail.org
Thu Jul 3 17:56:46 CEST 2008


Does something like this do what you want:

data(ethanol, package='lattice')
fit1 <- loess(NOx ~ E, data=ethanol, span=.6)

predfun <- function(x){
        tmp.df <- data.frame(E=x)
        predict(fit1, tmp.df)
}

integrate(predfun, 0.6, 1.2)


--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
(801) 408-8111



> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of Shirin Safa
> Sent: Tuesday, July 01, 2008 2:28 PM
> To: stephen sefick
> Cc: r-help at r-project.org
> Subject: Re: [R] Fitting a curve to data
>
> no this is not what I want.
> I was using "loess" function or smooth.spline.
> but For loess I don't know how would I be able to get the integral.
>
>
> On 7/1/08, stephen sefick <ssefick at gmail.com> wrote:
> >
> > ?lm
> > lm(x[,1]~x[,2])
> >
> >  On Tue, Jul 1, 2008 at 2:28 PM, Shirin Safa
> <sh.safa at gmail.com> wrote:
> >
> >> Hi
> >>
> >> I have a set of data like this:
> >>
> >>    *Time of Day* *Pct of Daily Volume* 9:45 7.50% 10 6.25% 10:15
> >> 4.45% 10:30 4.80% 10:45 4.45% 11:00 4.20% 11:15 2.50% 11:30 2.30%
> >> 11:45 2.25% 12:00 2.45% 12:15 2.60% 12:30 2.00% 12:45 2.05% 13:00
> >> 2.40% 13:15 1.90% 13:30 3.10% 13:45 2.90% 14:00 2.80% 14:15 2.50%
> >> 14:30 3.40% 14:45 4.40% 15:00 5.40% 15:15 4.00% 15:30 4.70% 15:45
> >> 6.20% 16:00 8.50%
> >>
> >> I want to fit a curve to the data and get the best
> function fitted,
> >> therefore I can calculate the integral of the function.
> >> How can I do that?
> >>
> >> Thanks,
> >> Shirin
> >>
> >>        [[alternative HTML version deleted]]
> >>
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> >>
> >
> >
> >
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>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
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>



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