Replicate certain values in vector determined by other vector - r

Question

I have a vector of values (say 1:10), and want to repeat certain values in it 2 or more times, determined by another vector (say c(3,4,6,8)). In this example, the result would be c(1,2,3,3,4,4,5,6,6,7,8,8,9,10) when repeating 2 times.
This should work for an arbitrary length range vector (like 200:600), with a second vector which is contained by the first. Is there a handy way to achieve this?

Answer 1

Akrun's is a more compact method, but this also will work
# get rep vector
reps <- rep(1L, 10L)
reps[c(3,4,6,8)] <- 2L
rep(1:10, reps)
[1] 1 2 3 3 4 4 5 6 6 7 8 8 9 10
The insight here is that rep will take an integer vector in the second argument the same length as the first argument that indicates the number of repetitions for each element of the first argument.
Note that this solution relies on the assumption that c(3,4,6,8) is the index or position of the elements that are to be repeated. Under this scenario, then d-b's comment has a one-liner
rep(x, (seq_along(x) %in% c(3,4,6,8)) + 1)
If instead, c(3,4,6,8) indicates the values that are to be repeated, then docendo-discimus's super-compact code,
rep(x, (x %in% c(3,4,6,8)) * (n-1) +1)
where n may be adjusted to change the number of repetitions. If you need to call this a couple times, this could be rolled up into a function like
myReps <- function(x, y, n) rep(x, (x %in% y) * (n-1) +1)
and called as
myReps(1:10, c(3,4,6,8), 2)
in the current scenario.

Answer 2

We can try
i1 <- v1 %in% v2
sort(c(v1[!i1], rep(v1[i1], each = 2)))
#[1] 1 2 3 3 4 4 5 6 6 7 8 8 9 10
Update
For the arbitrary vector,
f1 <- function(vec1, vec2, n){
i1 <- vec1 %in% vec2
vec3 <- seq_along(vec1)
c(vec1[!i1], rep(vec1[i1], each = n))[order(c(vec3[!i1],
rep(vec3[i1], each=n)))]
}
set.seed(24)
v1N <- sample(10)
v2 <- c(3,4,6,8)
v1N
#[1] 3 10 6 4 7 5 2 9 8 1
f1(v1N, v2, 2)
#[1] 3 3 10 6 6 4 4 7 5 2 9 8 8 1
f1(v1N, v2, 3)
#[1] 3 3 3 10 6 6 6 4 4 4 7 5 2 9 8 8 8 1

Answer 3

Here's another approach using sapply
#DATA
x = 1:10
r = c(3,4,6,8)
n = 2 #Two repetitions of selected values
#Assuming 'r' is the index of values in x to be repeated
unlist(sapply(seq_along(x), function(i) if(i %in% r){rep(x[i], n)}else{rep(x[i],1)}))
#[1] 1 2 3 3 4 4 5 6 6 7 8 8 9 10
#Assuming 'r' is the values in 'x' to be repeated
unlist(sapply(x, function(i) if(i %in% r){rep(i, n)}else{rep(i, 1)}))
#[1] 1 2 3 3 4 4 5 6 6 7 8 8 9 10
Haven't tested these thoroughly but could be possible alternatives. Note that the order of the output will be considerably different with this approach.
sort(c(x, rep(x[x %in% r], n-1))) #assuming 'r' is values
#[1] 1 2 3 3 4 4 5 6 6 7 8 8 9 10
sort(c(x, rep(x[r], n-1))) #assuming 'r' is index
#[1] 1 2 3 3 4 4 5 6 6 7 8 8 9 10

Answer 4

I suggest this solution just to emphasize the cool usage of append function in base R:
ff <- function(vec, v, n) {
for(i in seq_along(v)) vec <- append(vec, rep(v[i], n-1), after = which(vec==v[i]))
vec
}
Examples:
set.seed(1)
ff(vec = sample(10), v = c(3,4,6,8), n = 2)
#[1] 3 3 4 4 5 7 2 8 8 9 6 6 10 1
ff(vec = sample(10), v = c(2,5,9), n = 4)
#[1] 3 2 2 2 2 6 10 5 5 5 5 7 8 4 1 9 9 9 9