[R] optim error - repost

[Katharine Mullen]

try:
vol <- rep(c(0.03, 0.5, 2, 4, 8, 16, 32), 3)
time <- rep(c(2,4,8),each=7)
p.mated <- c(0.47, 0.48, 0.43, 0.43, 0.26, 0.23, NA, 0.68, 0.62, 0.64,
0.58, 0.53, 0.47, 0.24, 0.8, 0.79, 0.71, 0.56, 0.74, 0.8, 0.47)

eury <- data.frame(vol=vol, time=time, p.mated=p.mated)
eury <- na.omit(eury)

p0 <- c(f=0.87, b=0.1, d=10)
eury.fit <- function(p, time, vol, p.mated)
{
  f <- p[1]
  b <- p[2]
  d <- p[3]
  xx <- f*(1-exp(-b*time)-(b/(b-d/vol))*(exp(-d/vol*time)-exp(-b*time)))
  ## minimize sum of square differences
  sum((p.mated - xx)^2)
}

eury.opt <- optim(par=p0, fn=eury.fit, method = "BFGS", hessian = TRUE,
                  time=eury$time, vol=eury$vol,
                  p.mated=eury$p.mated, control=list(trace=1))
##  done with nls
eury.newfit1 <- nls(p.mated ~ f * ( 1 -
exp(-b*time)-(b/(b-d/vol))*(exp(-d/vol*time)-exp(-b*time))),
                    data=eury, start=list(f=.87, b=0.1, d=10),
                    trace=TRUE)

coef(eury.newfit1)

eury.opt$par

On Sun, 1 Jun 2008 keunhchoi at gmail.com wrote:

> Here is a clean version. I did this with nls and  it works (see below), but
> I need to  do it with optim. Keun-Hyung
>
> # optim
> vol<-rep(c(0.03, 0.5, 2, 4, 8, 16, 32), 3)
> time<-rep(c(2,4,8),each=7)
> p.mated<-c(0.47, 0.48, 0.43, 0.43, 0.26, 0.23, NA, 0.68, 0.62, 0.64, 0.58,
> 0.53, 0.47,
> 0.24, 0.8, 0.79, 0.71, 0.56, 0.74, 0.8, 0.47)
> eury<-data.frame(vol=vol, time=time, p.mated=p.mated)
> eury<-na.omit(eury); eury
>
> p0<- c(f=0.87, b=0.1, d=10)
> eury.fit <- function (f, time, vol)
> {
>             f<-p[1]; b<-p[2]; d<-p[3]
>             p.mated = p[1] * ( (1 -
> exp(-p[2]*time))-(p[2]/(p[2]-(p[3]/vol)))
>                            * (exp(-p[3]/vol*time)-exp(-p[2]*time)))
> }
> eury.opt<- optim(p0, fn=eury.fit, NULL, method = "BFGS", hessian = TRUE)
>
> # I received the following error message.
> Error in fn(par, ...) : argument "time" is missing, with no default
>
> ##  done with nls - this  works
> eury.newfit1 <- nls(p.mated ~ f * ( 1 -
> exp(-b*time)-(b/(b-d/vol))*(exp(-d/vol*time)-exp(-b*time))),
>                                             data=eury, start=list(f=.87,
> b=0.1, d=10))
> v <- log10(range(eury$vol))
> y <- expand.grid(vol=10^seq(min(v), max(v), length=100), time=c(2,4,8))
> y$pred.mate.new <- predict(eury.newfit1,y)
>
> plot (eury$vol, eury$p.mated, type="n", log="x",  xlab="Volume L", ylab =
> "Fraction Mating")
> for (i in c(2, 4, 8))
> {
>          points(eury$vol[eury$time==i], eury$p.mated[eury$time==i], pch=16,
> col=(i/2)+1)
>          lines(y$vol[y$time==i], y$pred.mate.new[y$time==i], lwd=3, col=i)
> }
> legend(.005,.2, c(" 2h","4h","8h"), col=c(2,4,8), lwd=3)
>
> 	[[alternative HTML version deleted]]
>
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