Understanding the behaviour of subsetting using 'which' - r

Question

I was trying to define a function generating all the prime numbers till n.
I came up with the following solution, which I compared to a solution readily available (given below for reference). Essentially there's just a line of difference in both the codes (indicated below)
sieve <- function(n){
sq.n <- sqrt(n)
vec <- 2:n
primes <- rep(0, times=(sq.n))
i <- 1
while (!(is.na(primes[i] < sq.n)) && (primes[i]) < (sq.n)) {
primes[i] <- vec[1]
vec <- vec[which(vec%%primes[i] != 0)] # This keeps all the numbers not divisible by
# the prime in question
i <- i + 1
}
return(c(primes[which(primes!=0)], vec))
}
Curious about efficiency, a google search yielded the following code -
getPrimeNumTilln <- function(n) {
a <- c(2:n)
l <- 2
r <- c()
while (l*l < n) {
r <- c(r,a[1])
a <- a[-(which(a %% l ==0))] # This removes all the numbers which are
# divisible by the prime in question
l <- a[1]
}
c(r,a)
}
Both solutions work okay. (The internet solution gives a wrong answer if n is the square of a prime, but that can be corrected easily)
And these are the microbenchmark results -
microbenchmark(sieve(100),getPrimeNumTilln(100),times=100)
Unit: microseconds
expr min lq mean median uq max neval
sieve(100) 142.107 153.106 165.85155 162.785 165.425 466.795 100
getPrimeNumTilln(100) 41.797 47.076 51.09312 49.276 51.036 126.269 100
I would like to understand the fair difference in the runtime of both the functions

Answer 1

The loop of the first function does 10 iterations for n = 100, the second function does 4.
sieve <- function(n){
sq.n <- sqrt(n)
vec <- 2:n
primes <- rep(0, times=(sq.n))
i <- 1
while (!(is.na(primes[i] < sq.n)) && (primes[i]) < (sq.n)) {
count <<- count + 1
primes[i] <- vec[1]
vec <- vec[which(vec%%primes[i] != 0)] # This keeps all the numbers not divisible by
# the prime in question
i <- i + 1
}
return(c(primes[which(primes!=0)], vec))
}
count <- 0
sieve(100)
#[1] 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
count
#[1] 10
getPrimeNumTilln <- function(n) {
a <- c(2:n)
l <- 2
r <- c()
while (l*l < n) {
count <<- count + 1
r <- c(r,a[1])
a <- a[-(which(a %% l ==0))] # This removes all the numbers which are
# divisible by the prime in question
l <- a[1]
}
c(r,a)
}
count <- 0
getPrimeNumTilln(100)
# [1] 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
count
#[1] 4