converting stata's by syntax to R

attach(dat)
dat<-dat[order(fam,wt),]

lis<-by(dat,fam,function(x) x[length(x$fam),])

do.call(rbind,lis)

I am struggling with migrating some stata code to R.  I have a data
frame containing, sometimes, repeat observations (rows) of the same
family.  I want to keep only one observation per family, selecting
that observation according to some other variable.  An example data
frame is:

# construct example data
fam <- c(1,2,3,3,4,4,4)
wt <- c(1,1,0.6,0.4,0.4,0.4,0.2)
keep <- c(1,1,1,0,1,0,0)
dat <- as.data.frame(cbind(fam,wt,keep))
dat

I want to keep the observation for which wt is a maximum, and where
this doesn't identify a unique observation, to keep just one anyway,
not caring which.  Those observations are indicated above by keep==1.
(Note, keep <- c(1,1,1,0,0,1,0) would be fine too, but not
c(1,1,1,0,0,0,1)).

The stata code I would use is
bys fam (wt): keep if _n==_N

This is my (long-winded) attempt in R:

# first keep those rows where wt=max_fam(wt)
maxwt <- by(dat,dat$fam,function(x) max(x[,2]))
maxwt <- sapply(maxwt,"[[",1)
maxwt.dat <- data.frame("maxwt"=maxwt,"fam"=as.integer(names(maxwt)))
dat <- merge(dat,maxwt.dat)
dat <- dat[dat$wt==dat$maxwt,]
dat

Now I am stuck - I want to keep either row with fam==4, and have tried
playing around with combinations of sample and apply or by, but with
no success.  I can only find an inefficient for-loop solution:
     
# identify those rows with >1 observation
more <- by(dat,dat$fam,function(x) dim(x)[1])
more <- sapply(more,"[[",1)
more.dat <- data.frame("more"=more,"fam"=as.integer(names(more)))
dat <- merge(dat,more.dat)

# sample from those for whom more>1
result<-dat[dat$more==1,]
for(f in unique(dat$fam[dat$more>1])) {
 rows <- rownames(dat[dat$fam==f,])
 result <- rbind(result,dat[sample(rows,1),])
}
result

I am sure that for something so simple in stata to be so complicated
in R must indicate ignorance of R on my part, but searches of help
files and RSiteSearch hasn't led to any better solution.

Any suggestions would be most helpful! Thanks, C.