# [R] nls does not accept start values

Petr PIKAL petr.pikal at precheza.cz
Mon Mar 2 10:48:33 CET 2009

```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

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

however

nls(fol[], 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 [] [], data.frame(x=x, y=y))
> >
> > # This works and gives me a result
> >
> > fit <- nls(form [] [], 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 [] [], data.frame(x=x, y=y), start=get(form
[]
> > []))
> >             ^^^^^^^^^^^^^^^^^^^
> > 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[], data.frame(x=x, y=y), start = form)
>
>
> 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