[R] function in nls argument

Thu May 8 19:07:52 CEST 2008

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
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
> 

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