Thank you Katharine. I am certain nprint is affecting my solution. Let me
know how I can send the data (~300Kb). The script I used it:
ST1 <- ST04
SM1 <- SM08
SR1 <- SRch2
ST <- ST1 [!is.na(SR1)]
SM <- SM1 [!is.na(SR1)]
SR <- SR1 [!is.na(SR1)]
q <- 0.90
p <- c("a"=-0.003, "b"=0.13, "c"=0.50, "E"=400)
SRf <- 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, SR, SRfcall)
{
res <- SR - do.call("SRfcall", c(list(ST = ST, SM = SM), as.list(p)))
abserr <- abs(res)
qnum <- quantile(abserr, probs=q, na.rm=TRUE)
res[abserr > qnum]=0
res
}
SRmodel<- nls.lm(par = p, fn = optim.f, SRfcall = SRf, ST = ST, SM = SM,
SR = SR, control = nls.lm.control(nprint=1))
summary(SRmodel)
SRpred <- do.call(SRf, c(list(ST = ST1, SM = SM1), SRmodel$par))
plot(SR1~SRpred)
a=c(0,2,4,6)
b=c(0,2,4,6)
lines(a,b,col=4)
Changing the nprint argument to 0 (or removing nprint) results in different
and completely wrong solutions. The same is true for this equivalent
simplified script from your example.
ST1 <- ST04
SM1 <- SM08
SR1 <- SRch2
ST <- ST1 [!is.na(SR1)]
SM <- SM1 [!is.na(SR1)]
SR <- SR1 [!is.na(SR1)]
q <- 0.9
p <- list(a=-0.003, b=0.13, c=0.50, E=400)
optim.f <- function(p, ST, SM, SR, q) {
res <- SR -
(p$a*SM^I(2)+p$b*SM+p$c)*exp(p$E*((1/(283.15-227.13))-(1/(ST+273.15-227.13))))
abserr <- abs(res)
qnum <- quantile(abserr, probs=q, na.rm=TRUE)
res[abserr > qnum] <- 0
res
}
SRmodel <- nls.lm(par = p, fn = optim.f, ST = ST, SM = SM, SR = SR, q = q,
control = nls.lm.control(nprint=1))
summary(SRmodel)
SRfun <- function(ST, SM, a, b, c, E)
{(a*SM^I(2)+b*SM+c)*exp(E*((1/(283.15-227.13))-(1/(ST+273.15-227.13))))}
SRpred <- do.call(SRfun, c(list(ST = ST1, SM = SM1), SRmodel$par))
plot(SR1~SRpred)
a=seq(0,6,1)
b=seq(0,6,1)
lines(a,b,col=4)
Here, however, I get the following additional error after summary(SRmodel):
Error in param/se : non-numeric argument to binary operator
although the solution is same as for the first script.
Note that a q of 95 is OK, a q of 90 is not, but a q of 50 is OK again...
To make your example reproducible you have to provide the data somehow;
I am pretty sure nprint doesn't effect the solution, but if it does this
would be a bug and I would appreciate a reproducible report.
The example in nls.lm is a little complicated in order to show how to use
an analytical expression for the gradient (possible to provide in the
argument jac); since you don't need this, note that your residual function
can be simplified into something along the lines of
p <- list(a=-0.003, b=0.13, c=0.50, E=400)
optim.f <- function(p, ST, SM, R, q) {
res <- R
-(p$a*SM^I(2)+p$b*SM+p$c)*exp(p$E*((1/(283.15-227.13))-(1/(ST+273.15-227.13))
abserr <- abs(res)
qnum <- quantile(abserr, probs=q, na.rm=T)
res[abserr > qnum] <- 0
res
}
Rmodel <- nls.lm(par = p, fn = optim.f, ST = ST, SM = SM, R = R, q = q)
On Thu, 8 May 2008, Fernando Moyano wrote:
> 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.
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