[R] nls does not accept start values

[Petr PIKAL]

Thank you

It was simplified version of my problem. I want to elaborate a function 
which can take predefined list of formulas, some data and evaluate which 
formulas can fit the data. I was inspired by some article in Chemical 
engineering in which some guy used excel solver for such task. I was 
curious if I can do it in R too. I am not sure if nls is appropriate tool 
for such task but I had to start somewhere.

Here is a function which takes list of formulas and data and gives a 
result for each formula.

modely <- function(formula, data, ...){
ll <- length(formula)   #no of items in formula list
result2 <- vector("list", ll) #prepare results
result1 <- rep(NA, ll)
for(i in 1:ll) {
fit<-try(nls(formula[[i]], data))
if( class(fit)=="try-error") result1[i] <- NA  else result1[i] <- 
sum(resid(fit)^2)
if( class(fit)=="try-error") result2[[i]] <- NA  else result2[[i]] <- 
coef(fit)
}

ooo<-order(result1) #order results according to residual sum

#combine results into one list together with functions used

result <- mapply(c, "sq.resid" = result1, result2) 
names(result) <- as.character(formula)
# output
result[ooo]
}

# data
x <-1:10
y <-1/(.5-x)+rnorm(10)/100

# list of formulas
fol <- structure(list(a = y ~ 1/(a - x), b = y ~ a * x^2 + b * log(x), 
    c = y ~ x^a), .Names = c("a", "b", "c"))

modely(fol, data.frame(x=x, y=y)

does not use "correct" model because when using default start values it 
results in

> nls(fol[[1]], data.frame(x=x, y=y))
Error in numericDeriv(form[[3]], names(ind), env) : 
  Missing value or an infinity produced when evaluating the model

however

 nls(fol[[1]], data.frame(x=x, y=y), start=list(a=mean(y)))

gives correct result. Therefore I started think about how to add a 
"better" starting value for some fits as a second part of my formula list 
to define structure like>

list(a= formula1, start.formula1, b=formula2, start.formula2, ....)

I wonder If you can push me to better direction.

Thanks again
Best regards
Petr




Uwe Ligges <ligges at statistik.tu-dortmund.de> napsal dne 02.03.2009 
09:41:45:

> Petr PIKAL wrote:
> > Hi to all
> > 
> > OK as I did not get any response and I really need some insight I try 
> > again with different subject line
> > 
> > I have troubles with correct evaluating/structure of nls input
> > 
> > Here is an example
> > 
> > # data
> > x <-1:10
> > y <-1/(.5-x)+rnorm(10)/100
> > 
> > # formula list
> > form <- structure(list(a = list(quote(y ~ 1/(a - x)), 
"list(a=mean(y))")), 
> >  .Names = "a")
> > 
> > # This gives me an error due to not suitable default starting value
> > 
> > fit <- nls(form [[1]] [[1]], data.frame(x=x, y=y))
> > 
> > # This works and gives me a result
> > 
> > fit <- nls(form [[1]] [[1]], data.frame(x=x, y=y), 
start=list(a=mean(y)))
> > 
> > *** How to organise list "form" and call to nls to enable to use other 

> > then default starting values***.
> > 
> > I thought about something like
> > 
> > fit <- nls(form [[1]] [[1]], data.frame(x=x, y=y), start=get(form 
[[1]] 
> > [[2]]))
> >             ^^^^^^^^^^^^^^^^^^^ 
> > but this gives me an error so it is not correct syntax. (BTW I tried 
eval, 
> > assign, sustitute, evalq and maybe some other options but did not get 
it 
> > right.
> > 
> > I know I can put starting values interactively but what if I want them 

> > computed by some easy way which is specified by second part of a list, 

> > like in above example.
> 
> If you really want to orgnize it that way, why not simpler as in:
> 
> form <- list(y ~ 1/(a - x), a = mean(y))
> fit <- nls(form[[1]], data.frame(x=x, y=y), start = form[2])
> 
> 
> Uwe Ligges
> 
> 
> > If it matters
> > WXP,  R2.9.0 devel.
> > 
> > Regards
> > Petr
> > 
> > petr.pikal at precheza.cz
> > 
> > ______________________________________________
> > 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.