[R] Non-linear curve fitting (nls): starting point and quality of fit

[Greg Snow]

One thing to note is that there are more than one model that can be
called exponential.  Two of the common ones are:

y = exp( a + b*x + error )
y = exp( a + b*x ) + error

The common way to fit the first is to take the log of both sides and
just fit a linear model with log(y), I expect (but am not sure) that
that is what Excel does.  It is likely that the reason you get
different results is that you are fitting different models with the 2
programs.  You first need to decide which is the correct model (and it
may be different from the 2 already mentioned), then worry about
fitting that model and what goes with that.

On Mon, Jun 4, 2012 at 6:19 AM, Nerak <nerak.t at hotmail.com> wrote:
> Hi all,
>
> Like a lot of people I noticed that I get different results when I use nls
> in R compared to the exponential fit in excel. A bit annoying because often
> the R^2 is higher in excel but when I'm reading the different topics on this
> forum I kind of understand that using R is better than excel?
>
>  (I don't really understand how the difference occurs, but I understand that
> there is a different way in fitting, in excel a single value can make the
> difference, in R it looks at the whole function? I read this: "Fitting a
> function is an approximation, trying to find a minimum. Think of frozen
> mountain lake surrounded by mountains. Excel's Solver will report the
> highest tip of the snowflake on the lake, if it finds it. nls will find out
> that the lake is essentially flat compare to the surrounding and tell you
> this fact in unkind word." )
>
>
> I have several questions about nls:
>
> 1. The nls method doesn't give an R^2. But I want to determine the quality
> of the fit. To understand how to use nls I read "Technical note: Curve
> fitting with the R environment for Statistical Computing". In that document
> they suggested this to calculate R^2:
>
> RSS.p<-sum(residuals(fit)^2)
>  TSS<-sum((y-mean(y))^2)
>  r.squared<-1-(RSS.p/TSS)
>  LIST.rsq<-r.squared
>
> (with fit my results of the nls: formula y ~ exp.f(x, a, b) : y :
> a*exp(-b*x))
>
> While I was reading on the internet to find a possible reason why I get
> different results using R and excel, I also read lots of different things
> about the "R^2 problem" in nls.
>
> Is the method I'm using now ok, or should someone suggest to use something
> else?
>
> 2. Another question I have is like a lot of people about the singular
> gradient problem. I didn't know the best way to chose my starting values for
> my coefficients. when it was too low, I got this singular gradient error.
> Raising the value helped me to get rid of that error. Changing that value
> didn't change my coefficients nor R^2. I was wondering if that's ok, just to
> raise the starting value of one of my coefficients?
>
> The only things that change are the Achieved convergence tolerance and
> number of iterations to convergence. P values, residual standard error and
> the coefficients have always exactly the same results. What does the
> achieved convergence tolerance actually mean? What are its implications? (I
> suppose the time to calculate it changes)
>
> (the most useful information about nls and singular gradient error i found
> is this one (and that's why I started playing with changing the starting
> values):
> " if the estimate of the rank that results is less than the number of
> columns in the gradient (the number of nonlinear parameters), or less than
> the number of rows (the number of observations), nls stops.")
>
>
> I hope someone can help me with this questions. I would like to know what's
> happening and not just having to accept the results I get now :).
>
> Kind regards,
>
> Nerak
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/Non-linear-curve-fitting-nls-starting-point-and-quality-of-fit-tp4632295.html
> Sent from the R help mailing list archive at Nabble.com.
>
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-- 
Gregory (Greg) L. Snow Ph.D.
538280 at gmail.com