Dear Catalin,
You can look at ?nls.
Alternatively, you could also consider a linear model as follows, where "d"
is your data:
# plot your data
with(d, plot(cls, proc, las = 1))
# linear model
fit <- lm(proc ~ I(1/cls) + I((1/cls)^2), data = d)
summary(fit)
# plotting
with(d, plot(cls, proc, las = 1))
grid <- seq(min(d$cls), max(d$cls), length = 1000)
points(grid, predict(fit, data.frame(cls = grid)), type = "l", col = 2)
HTH,
Jorge.-
On Wed, Apr 10, 2013 at 7:01 PM, catalin roibu <catalinro...@gmail.com>wrote:
> Hello all!
>
> I have a problem with a double exponential equation.
> this are my data's>
> structure(list(proc = c(1870.52067384719, 766.789388745793,
> 358.701545859122,
> 237.113777545511, 43.2726259059654, 148.985133316262, 92.6242882655781,
> 88.4521557193262, 56.6404686159112, 27.0374477259404, 34.3347291080268,
> 18.3226992991316, 15.2196612445747, 5.31600719692165, 16.7015717397302,
> 16.3923389973684, 24.2702542054496, 21.247247993673, 18.3070717608672,
> 2.8811892177331, 3.18018869564679, 8.74204132937479, 7.11596966047229
> ), cls = c(0.25, 0.5, 0.75, 1, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5,
> 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10)), .Names = c("proc",
> "cls"), row.names = c("0.25", "0.5", "0.75", "1", "11", "1.5",
> "2", "2.5", "3", "3.5", "4", "4.5", "5", "5.5", "6", "6.5", "7",
> "7.5", "8", "8.5", "9", "9.5", "10"), class = "data.frame")
> I want to compute a double exponential equation like this:
> proc=a*exp(b*class)+c*exp(d*class)
> or
> proc=a*exp(b*class)+c*class
> or a power, logarithmic equation.
> Is there a possibility to calculate R squared for each model?
>
> Thank you!
>
> --
> ---
> Catalin-Constantin ROIBU
> Lecturer PhD, Forestry engineer
> Forestry Faculty of Suceava
> Str. Universitatii no. 13, Suceava, 720229, Romania
> office phone +4 0230 52 29 78, ext. 531
> mobile phone +4 0745 53 18 01
> +4 0766 71 76 58
> FAX: +4 0230 52 16 64
> silvic.usv.ro
>
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>
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>
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