It's a scaling problem:
If you do this:
datum <- datum[order(datum$X),]
with(datum, plot(Y~X))
with(datum, lines(X, 3400*exp(-1867*X)))
you'll see that your initial guess is just so far gone that the nls()
optimizer can't handle it.
If you try a more reasonable initial guess it works fine:
fit <- nls(Y ~ a*exp(-k * X), datum, start=c(a=3400, k=1.867))
Michael
On Wed, Nov 30, 2011 at 12:14 PM, chuck.01 <charliethebrow...@gmail.com> wrote:
> Hello,
> I have data like the following:
>
> datum <- structure(list(Y = c(415.5, 3847.83333325, 1942.833333325,
> 1215.22222233333,
> 950.142857325, 2399.5833335, 804.75, 579.5, 841.708333325, 494.053571425
> ), X = c(1.081818182, 0.492727273, 0.756363636, 0.896363636,
> 1.518181818, 0.499166667, 1.354545455, 1.61, 1.706363636, 1.063636364
> )), .Names = c("Y", "X"), row.names = c(NA, -10L), class = "data.frame")
>
>
> with(datum, plot(Y~X))
>
> As you can see there is a non-linear association between X and Y, and I
> would like to fit an appropriate model. I was thinking an exponential decay
> model might work well.
>
> I tried the following (a and k starting values are based off of a lm() fit),
> but get an error.
>
> fit <- nls(Y ~ a*exp(-k * X), datum, start=c(a=3400, k=1867))
>
> Error in nlsModel(formula, mf, start, wts) :
> singular gradient matrix at initial parameter estimates
>
> I have never attempted to fit a non-linear model before, and thus the model
> may be inappropriately specified, or it is also possible that I have no idea
> what I am doing.
>
> Would someone please offer some advice.
>
> Thanks.
> Chuck
>
>
> --
> View this message in context:
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> Sent from the R help mailing list archive at Nabble.com.
>
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