Create a sequence from vectors with start and end positions

Question

Given two separate vectors of equal length: f.start and f.end, I would like to construct a sequence (by 1), going from f.start[1]:f.end[1] to f.start[2]:f.end[2], ..., to f.start[n]:f.end[n].

Here is an example with just 6 rows.

   f.start  f.end
[1,]   45739 122538
[2,]  125469 202268
[3,]  203563 280362
[4,]  281657 358456
[5,]  359751 436550
[6,]  437845 514644

Crudely, a loop can do it, but is extremely slow for larger datasets (rows>2000).

f.start<-c(45739,125469,203563,281657,359751,437845)
f.end<-c(122538,202268,280362,358456,436550,514644)
f.ind<-f.start[1]:f.end[1]
for (i in 2:length(f.start))
{
 f.ind.temp<-f.start[i]:f.end[i]
 f.ind<-c(f.ind,f.ind.temp)
}

I suspect this can be done with apply(), but I have not worked out how to include two separate arguments in apply, and would appreciate some guidance.

Answer 1

You can try using mapply or Map, which iterates simultaneously on your two vectors. You need to provide the function as first argument:

vec1 = c(1,33,50)
vec2 = c(10,34,56)

unlist(Map(':',vec1, vec2))
# [1]  1  2  3  4  5  6  7  8  9 10 33 34 50 51 52 53 54 55 56

Just replace vec1 and vec2 by f.start and f.end provided all(f.start<=f.end)

Answer 2

Your loop is going to be slow as you are growing the vector f.ind. You will also get an increase in speed if you pre-allocate the length of the output vector.

# Some data (of length 3000)
set.seed(1)
f.start <- sample(1:10000, 3000)
f.end <- f.start + sample(1:200, 3000, TRUE)

# Functions
op <- function(L=1) {
        f.ind <- vector("list", L)
            for (i in 1:length(f.start)) {
                f.ind[[i]] <- f.start[i]:f.end[i]
             }
        unlist(f.ind)
        }

op2 <- function() unlist(lapply(seq(f.start), function(x) f.start[x]:f.end[x]))
col <- function() unlist(mapply(':',f.start, f.end))

# check output
all.equal(op(), op2())
all.equal(op(), col())

A few benchmarks

library(microbenchmark)

# Look at the effect of pre-allocating
microbenchmark(op(L=1), op(L=1000), op(L=3000), times=500)
#Unit: milliseconds
#         expr       min        lq     mean    median        uq       max neval cld
#    op(L = 1) 46.760416 48.741080 52.29038 49.636864 50.661506 113.08303   500   c
# op(L = 1000) 41.644123 43.965891 46.20380 44.633016 45.739895  94.88560   500  b 
# op(L = 3000)  7.629882  8.098691 10.10698  8.338387  9.963558  60.74152   500 a  

# Compare methods - the loop actually performs okay
# I left the original loop out
microbenchmark(op(L=3000), op2(), col(), times=500)
#Unit: milliseconds
#        expr      min       lq     mean   median        uq      max neval cld
# op(L = 3000) 7.778643 8.123136 10.119464 8.367720 11.402463 62.35632   500   b
#        op2() 6.461926 6.762977  8.619154 6.995233 10.028825 57.55236   500  a 
#        col() 6.656154 6.910272  8.735241 7.137500  9.935935 58.37279   500  a 

So a loop should perform okay speed wise, but of course the Colonel's code is a lot cleaner. The *apply functions here wont really give much speed up in the calculation but they do offer tidier code and remove the need for pre-allocation.