[Rd] Converting expression to a function
John C Nash
nashjc at uottawa.ca
Sun Mar 18 18:01:20 CET 2012
Previously, I've posted queries about this, and thanks to postings and messages in
response have recently had some success, to the extent that there is now a package called
nlmrt on the R-forge project https://r-forge.r-project.org/R/?group_id=395 for solving
nonlinear least squares problems that include small or zero residual problems via a
Marquardt method using a call that mirrors the nls() function. nls() specifically warns
against zero residual problems.
However, I would still like to be able to convert expressions with example vectors of
parameters to functions that optim() and related functions can use. The code below gets
"almost" there, but
1) Can the code be improved / cleaned up?
2) Can the eval() of the output of the Form2resfun be avoided?
3) Can the extraction of the parameter names be embedded in the function rather than put
separately?
Off-list responses are likely best at this stage, while the tedious details are sorted
out. I will post a summary in a couple of weeks of the results. Collaborations re: this
and the larger package welcome, as there is considerable testing and tuning to do, but
preliminary experience is encouraging.
John Nash
# --------- code block -----------
rm(list=ls()) # clear workspace
Form2resfun <- function(f, p ) {
cat("In Form2resfun\n")
xx <- all.vars(f)
fp <- match(names(p), xx) # Problem in matching the names of params
xx2 <- c(xx[fp], xx[-fp])
ff <- vector("list", length(xx2))
names(ff) <- xx2
sf<-as.character(f)
if ((length(sf)!=3) && (sf[1]!="~")) stop("Bad model formula expression")
lhs<-sf[2] # NOTE ORDER formula with ~ puts ~, lhs, rhs
rhs<-sf[3]
# And build the residual at the parameters
resexp<-paste(rhs,"-",lhs, collapse=" ")
fnexp<-paste("crossprod(",resexp,")", sep="")
ff[[length(ff) + 1]] <- parse(text=fnexp)
# want crossprod(resexp)
myfn<-as.function(ff, parent.frame())
}
# a test
y<-c(5.308, 7.24, 9.638, 12.866, 17.069, 23.192, 31.443,
38.558, 50.156, 62.948, 75.995, 91.972) # for testing
t<-1:length(y) # for testing
f<- y ~ b1/(1+b2*exp(-1*b3*t))
p<-c(b1=1, b2=1, b3=1)
b<-p
npar<-length(b)
for (i in 1:npar){
bbit<-paste(names(b)[[i]],"<-",b[[i]])
eval(parse(text=bbit))
}
tfn<-Form2resfun(f, b)
ans<-eval(tfn(t=t,y=y, b))
print(ans)
# --------- end code block -----------
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