Repeated subsetting of the same matrix using apply in R - r

Question

Motivation: I am currently trying to rethink my coding such as to exclude for-loops where possible. The below problem can easily be solved with conventional for-loops, but I was wondering if R offers a possibility to utilize the apply-family to make the problem easier.
Problem: I have a matrix, say X (n x k matrix) and two matrices of start and stop indices, called index.starts and index.stops, respectively. They are of size n x B and it holds that index.stops = index.starts + m for some integer m. Each pair index.starts[i,j] and index.stops[i,j] are needed to subset X as X[ (index.starts[i,j]:index.stops[i,j]),]. I.e., they should select all the rows of X in their index range.
Can I solve this problem using one of the apply functions?
Application: (Not necessarily important for understanding my problem.) In case you are interested, this is needed for a bootstrapping application with blocks in a time series application. The X represents the original sample. index.starts is sampled as replicate(repetitionNumber, sample.int((n-r), ceiling(n/r), replace=TRUE)) and index.stopsis obtained as index.stop = index.starts + m. What I want in the end is a collection of rows of X. In particular, I want to resample repetitionNumber times m blocks of length r from X.
Example:
#generate data
n<-100 #the size of your sample
B<-5 #the number of columns for index.starts and index.stops
#and equivalently the number of block bootstraps to sample
k<-2 #the number of variables in X
X<-matrix(rnorm(n*k), nrow=n, ncol = k)
#take a random sample of the indices 1:100 to get index.starts
r<-10 #this is the block length
#get a sample of the indices 1:(n-r), and get ceiling(n/r) of these
#(for n=100 and r=10, ceiling(n/r) = n/r = 10). Replicate this B times
index.starts<-replicate(B, sample.int((n-r), ceiling(n/r), replace=TRUE))
index.stops<-index.starts + r
#Now can I use apply-functions to extract the r subsequent rows that are
#paired in index.starts[i,j] and index.stops[i,j] for i = 1,2,...,10 = ceiling(n/r) and
#j=1,2,3,4,5=B ?

Answer 1

It's probably way more complicated than what you want/need, but here is a first approach. Just comment if that helps you in any way and I am happy to help.
My approach uses (multiple) *apply-functions. The first lapply "loops" over 1:B cases, where it first calculates the start and end points, which are combined into the take.rows (with subsetting numbers). Next, the inital matrix is subsetted by take.rows (and returned in a list). As a last step, the standard deviation is taken for each column of the subsetted matrizes (as a dummy function).
The code (with heavy commenting) looks like this:
# you can use lapply in parallel mode if you want to speed up code...
lapply(1:B, function(i){
starts <- sample.int((n-r), ceiling(n/r), replace=TRUE)
# [1] 64 22 84 26 40 7 66 12 25 15
ends <- starts + r
take.rows <- Map(":", starts, ends)
# [[1]]
# [1] 72 73 74 75 76 77 78 79 80 81 82
# ...
res <- lapply(take.rows, function(subs) X[subs, ])
# res is now a list of 10 with the ten subsets
# [[1]]
# [,1] [,2]
# [1,] 0.2658915 -0.18265235
# [2,] 1.7397478 0.66315385
# ...
# say you want to compute something (sd in this case) you can do the following
# but better you do the computing directly in the former "lapply(take.rows...)"
res2 <- t(sapply(res, function(tmp){
apply(tmp, 2, sd)
})) # simplify into a vector/data.frame
# [,1] [,2]
# [1,] 1.2345833 1.0927203
# [2,] 1.1838110 1.0767433
# [3,] 0.9808146 1.0522117
# ...
return(res2)
})
Does that point you in the right direction/gives you the answer?