Use of x`apply` to speed up loops - r

Question

This is a more focussed question based on another question I have open at Vectorize/Speed up Code with Nested For Loops
Basically, I want to speed up the execution of this code. I was thinking of using one of the apply family of functions. The apply function would have to use/perform the following:
Input: loop over regions 1 to 10; vectors sed and borewidth with preallocated dimensions filled with NAs
Process: fill data in each of sed and borewidth in the manner implemented in the inner for loop
Output: sed and borewidth vectors
Assumptions (h/t Simon Urbanek): the begin, finish points of each row are contiguous, sequential and for each region, begin at 0.
Code is as below:
for (region in 1:10) {
# subset standRef and sample by region code
standRef.region <- standRef[which(standRef$region == region),]
sample.region <- sample[which(sample$region == region),]
for (i in 1:nrow(sample.region))
{
# create a dataframe - locations - that includes:
# 1) those indices of standRef.region in which the value of the location column is greater than the value of the ith row of the begin column of sample.region
# 2) those indices of standRef.region in which the value of the location column is less than the value of the ith row of the finish column of sample.region
locations <- standRef.region[which((standRef.region$location > sample.region$begin[i]) & (standRef.region$location < sample.region$finish[i])),]
sed[end_tracker:(end_tracker + nrow(locations))] <- sample.region$sed[i]
borewidth[end_tracker:(end_tracker + nrow(locations))] <- sample.region$borewidth[i]
# update end_tracker to the number of locations rows for this iteration
end_tracker <- end_tracker + nrow(locations)
}
cat("Finished region", region,"\n")
}
Sample Data for borewidth andsed. Edit: corrected formatting error in dput
structure(list(region = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
begin = c(0L, 2253252L, 7091077L, 9120205L, 0L, 135094L,
941813L, 5901391L, 6061324L), finish = c(2253252L, 7091077L,
9120205L, 17463033L, 135094L, 941813L, 5901391L, 6061324L,
7092402L), sed = c(3.31830840984048, 1.38014704208403, 6.13049140975458,
2.10349875097134, 0.48170587509345, 0.13058713509175, 9.13509713513509,
6.13047153058701, 3.81734081501503), borewidth = c(3L, 5L,
2L, 1L, 1L, 1L, 2L, 4L, 4L)), .Names = c("region", "begin",
"finish", "sed", "borewidth"), class = "data.frame", row.names = c(NA,
-9L))
TIA.

Answer 1

With some extra assumptions based on the data you posted (incl. the other question), this is one way you could do it:
index <- unlist(lapply (unique(standRef$region), function(reg) {
reg.filter <- which(standRef$region == reg)
samp.filter <- which(sample$region == reg)
samp.filter[cut(standRef$location[reg.filter],c(0L,sample$finish[samp.filter]),labels=F)]
}))
sed <- sample$sed[index]
borewidth <- sample$borewidth[index]
The extra assumption is that your samples are contiguous, sequential (all your examples were) and start at 0. This allows us to use cut() on the $finish instead of treating each interval separately. One difference is that you code left gaps at the breaks, but I'm assuming that was not intentional.