Hi
Prof Brian Ripley <rip...@stats.ox.ac.uk> napsal dne 02.03.2009 10:24:52:
> It is not very clear what you are trying to do here, and
>
> > form <- structure(list(a = list(quote(y ~ 1/(a - x)),
"list(a=mean(y))")),
> > .Names = "a")
>
> is using a historic anomaly (see the help page).
>
> I am gussing you want to give nls an object containing a formula and
> an expression for the starting value. It seems you are re-inventing
You are correct as usually.
> self-starting nls models: see ?selfStart and MASS$ ca p. 216.
> One way to use them in your example is
>
> mod <- selfStart(~ 1/(a - x), function(mCall, data, LHS) {
> structure(mean(eval(LHS, data)), names="a")
> }, "a")
>
> nls(y ~ mod(x, a))
>
> But if you want to follow ypur route, youer starting values would be
> better to be a list that you evaluate in an appropriate context
> (which y is this supposed to be?). nls() knows where it will find
> variables, but it is not so easy for you to replicate its logic
> without access to its evaluation frames.
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
I tried to establish such structure to get more appropriate starting
values
list(a= list(formula1, start.formula1), b=list(formula2, start.formula2),
....)
But did not manage yet to get correct syntax for let say mean of response
values. I try to look more closely what I can achieve with selfStart
Thank you again
Best regards
Petr
>
> On Mon, 2 Mar 2009, 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 it matters
> > WXP, R2.9.0 devel.
> >
> > Regards
> > Petr
> >
> > petr.pi...@precheza.cz
>
>
> --
> Brian D. Ripley, rip...@stats.ox.ac.uk
> Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
> University of Oxford, Tel: +44 1865 272861 (self)
> 1 South Parks Road, +44 1865 272866 (PA)
> Oxford OX1 3TG, UK Fax: +44 1865 272595
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