# Re: [R] function in nls argument

From: Katharine Mullen <kate_at_few.vu.nl>
Date: Thu, 08 May 2008 21:24:22 +0200 (CEST)

To make your example reproducible you have to provide the data somehow; I am pretty sure nprint doesn't effect the solution, but if it does this would be a bug and I would appreciate a reproducible report.

The example in nls.lm is a little complicated in order to show how to use an analytical expression for the gradient (possible to provide in the argument jac); since you don't need this, note that your residual function can be simplified into something along the lines of

p <- list(a=-0.003, b=0.13, c=0.50, E=400)

optim.f <- function(p, ST, SM, R, q) {
res <- R -(p\$a*SM^I(2)+p\$b*SM+p\$c)*exp(p\$E*((1/(283.15-227.13))-(1/(ST+273.15-227.13))  abserr <- abs(res)
qnum <- quantile(abserr, probs=q, na.rm=T)  res[abserr > qnum] <- 0
res
}

Rmodel <- nls.lm(par = p, fn = optim.f, ST = ST, SM = SM, R = R, q = q)

On Thu, 8 May 2008, Fernando Moyano wrote:

> I solved the problem arising from using certain quantile values simply
> by printing the iterations with the argument nprint. This gave me
> correct estimates. I have no idea why.
>
> ----- Original Message ----
> From: elnano <nanomoyano_at_yahoo.com>
> To: r-help_at_r-project.org
> Sent: Thursday, May 8, 2008 5:43:31 PM
> Subject: Re: [R] function in nls argument
>
>
> I've basically solved the problem using the nls.lm function from the
> minpack.lm (thanks Katharine) with some modifications for ignoring residuals
> above a given percentile. This is to avoid the strong influence of points
> which push my modeled vs. measured values away from the 1:1 line.
> I based it on the example given for nls.lm. Here it is:
>
> R # soil respiration data
> ST <- ST [!is.na(R)] # soil temeprature data. Had to remove na to make
> nls.lm work
> SM <- SM [!is.na(R)] # soil moisture data
> R <- R [!is.na(R)]
> q <- 0.95 # quantile
>
> p <- c("a"=-0.003, "b"=0.13, "c"=0.50, E=400) # model parameters with
> estimated values
>
> Rf <- function(ST, SM, a, b, c, E)
> {
> expr <-
> expression((a*SM^I(2)+b*SM+c)*exp(E*((1/(283.15-227.13))-(1/(ST+273.15-227.13)))))
> eval(expr)
> }
>
> optim.f <- function(p, ST, SM, R, Rfcall)
> {
> res <- R - do.call("Rfcall", c(list(ST = ST, SM = SM), as.list(p))) #
> the nls.lm example divides this by sqrt(R), I don't know why. I removed
> that.
> abserr <- abs(res)
> qnum <- quantile(abserr, probs=q, na.rm=T) # calculate the "q"
> quantile of the absolute errors
> res[abserr > qnum]=0 # convert
> residuals above qnum to 0
> res
> }
>
> Rmodel<- nls.lm(par = p, fn = optim.f, Rfcall = Rf, ST = ST, SM = SM, R
> = R)
>
> summary(Rmodel)
>
> The only error I still get is when using lower q values. A q of around 0.95
> or less (depending on the dataset) gives me completely wrong parameter
> estimates resulting in negative predicted values. Maybe someone has a
> suggestion here.
>
> Fernando
>
>
>
> Katharine Mullen wrote:
> >
> > The error message means that the gradient (first derivative of residual
> > vector with respect to the parameter vector) is not possible to work with;
> > calling the function qr on the gradient multiplied by the square root of
> > the weight vector .swts (in your case all 1's) fails.
> >
> > If you want concrete advice it would be helpful to provide the commented,
> > minimal, self-contained, reproducible code that the posting guide
> > requests. what are the values of ST04, SM08b, ch2no, and tower?
> >
> > General comments: If your goal is to minimize sum( (observed -
> > predicted)^2), the function you give nls to minimize (optim.fun in your
> > case) should return the vector (observed - predicted), not the scalar sum(
> > (observed - predicted)^2). You may want to see the nls.lm function in
> > package minpack.lm for another way to deal with the problem.
> >
> > On Wed, 7 May 2008, Fernando Moyano wrote:
> >
> >> Greetings R users, maybe there is someone who can help
> >> me with this problem:
> >>
> >> I define a function "optim.fun" and want as output the
> >> sum of squared errors between predicted and measured
> >> values, as follows:
> >>
> >> optim.fun <- function (ST04, SM08b, ch2no, a, b, d, E)
> >> {
> >> predR <-
> >> (a*SM08b^I(2)+b*SM08b+d)*exp(E*((1/(283.15-227.13))-(1/(ST04+273.15-227.13))))
> >> abserr <- abs(ch2no-predR)
> >> qnum <- quantile(abserr, probs=0.95, na.rm=T)
> >>
> >> is.na(abserr) <- (abserr > qnum)
> >> errsq <- sum(abserr^2, na.rm=T)
> >> errsq
> >> }
> >>
> >> Then I want to optimize parameters a,b,d and E as to
> >> minimize the function output with the following:
> >>
> >> optim.model<-nls(nulldat ~ optim.fun(ST04, SM08b,
> >> ch2no, a, b, d, E), data=tower,
> >> start=list(a=-0.003,b=0.13,d=0.50, E=400), na.action =
> >> na.exclude )
> >>
> >> were nulldat=0
> >> At this point I get the following error message:
> >>
> >> Error in qr.default(.swts * attr(rhs, "gradient")) :
> >> NA/NaN/Inf in foreign function call (arg 1)
> >> Warning messages:
> >> 1: In if (na.rm) x <- x[!is.na(x)] else if
> >> (any(is.na(x))) stop("missing values and NaN's not
> >> allowed if 'na.rm' is FALSE") ... :
> >> the condition has length > 1 and only the first
> >> element will be used
> >> (this warning is repeated 12 times)
> >>
> >> Question: what does the error mean? What am I doing
> >> wrong?
> >> Thanks a bunch.
> >> Nano
> >> Jen, Germany
> >> Max Planck for Biogeochemistry
> >>
> >>
> >>
> >>
> >>
> >> ____________________________________________________________________________________
> >>
> >> [[elided Yahoo spam]]
> >>
> >> ______________________________________________
> >> R-help_at_r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> > ______________________________________________
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> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
>
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