Margining two vectors on precisely defined intervals - r

Question

Given are two vectors:
vec_nums <- 1:20
vec_ltrs <- letters[1:10]
I would like to write a function that would merge them some each element from the second vectors appears on the precisely defined position within the first vector. For example, running:
vec_mrg <- funMergeVectsByPlace(x = vec_num, y = vec_ltrs, position = 3)
Should return vec_mrg of the following content:
[1] "a" "b" "1" "c" "d" "2" "f" "g" "3" "i" "j" "4" "l" "m" "5" ...
Desired characteristics:
The function places element from the vector passed via the y = on the position given in the position = counting from the left hand side. So position = 3 should be understood as *every third place" accounting for 3, 6, ...
The function should work on numeric string and factor vectors and return an ordered factor.
The function should work on factor, string and numeric vectors
In case of vector y being shorter than than the number of inserts in the x the function should return remaining part of x without any additions
Suggested structure
I would envisage for the function to be of this structure:
funMergeVectsByPlace <- function(x,y position = 3) {
# Convert
vec_a <- as.character(x)
vec_b <- as.character(y)
# Missing part
# Combine two vectors
# Create ordered factor
vec_fac <- factor(vec_mrg,
# levels =
# I want the levels to reflect the order of elements in the vec_merg
)
# Return
return(vec_fac)
}
Samples
Simplest
Concerning attempts, simplest approach:
vec_mrg <- c(vec_nums, vec_ltrs)
vec_mrg <- order(vec_mrg)
But this would not create the order
Loop
for (i in 1:length(vec_nums)) {
pos <- position
vec_nums[pos] <- vec_ltrs[i]
pos <- pos + pos
# i will be out of bounds and the way to move the other vector is missing
}

Answer 1

vec_mrg <- function(x,y,pos) {
res <- y
counter <- seq(floor(length(y)/(pos-1)))
for(i in counter) {
res <- append(res, x[i], seq(pos-1,by=pos, length.out=length(counter))[i])
}
res
}
vec_mrg(vec_nums, vec_ltrs, 3)
#[1] "a" "b" "1" "c" "d" "2" "e" "f" "3" "g" "h" "4" "i" "j"
#[15] "5"

Answer 2

A loop-free solution:
funMergeVectsByPlace <- function( x, y, position )
{
n <- min( length(y)%/%(position-1), length(x) )
A <- rbind( matrix(head(y,n*(position-1)),position-1), head(x,n) )
rest <- c( x[-(1:n)], y[-(1:(n*(position-1)))] )
c(c(A),rest)
}
Speed comparison with Lafortunes solution:
> library(microbenchmark)
> vec_nums <- 1:20
> vec_ltrs <- letters[1:10]
> microbenchmark(Lafortune = vec_mrg(vec_nums,vec_ltrs,3),
+ mra68 = funMergeVectsByPlace(vec_nums,vec_ltrs,3),
+ times .... [TRUNCATED]
Unit: microseconds
expr min lq mean median uq max neval
Lafortune 137.677 143.112 161.12006 146.734 153.980 2931.512 10000
mra68 77.443 81.067 92.13208 83.331 86.954 2718.204 10000
Larger vectors:
> vec_nums <- 1:2000
> vec_ltrs <- letters[rep(1:10,100)]
> microbenchmark(Lafortune = vec_mrg(vec_nums,vec_ltrs,3),
+ mra68 = funMergeVectsByPlace(vec_nums,vec_ltrs,3),
+ times .... [TRUNCATED]
Unit: milliseconds
expr min lq mean median uq max neval
Lafortune 32.993883 40.991796 63.758011 51.171020 90.122351 456.9748 1000
mra68 1.101865 1.489533 2.468496 1.751299 3.338881 230.0460 1000
> v1 <- vec_mrg(vec_nums,vec_ltrs,3)
> v2 <- funMergeVectsByPlace(vec_nums,vec_ltrs,3)
>
Notice that the vec_mrg function does not append the rest of the x vector to the result, but funMergeVectsByPlace does. Otherwise the results are the same:
> v1 <- vec_mrg(1:20,letters[1:10],3)
> v2 <- funMergeVectsByPlace(1:20,letters[1:10],3)
> v1
[1] "a" "b" "1" "c" "d" "2" "e" "f" "3" "g" "h" "4" "i" "j" "5"
> v2
[1] "a" "b" "1" "c" "d" "2" "e" "f" "3" "g" "h" "4" "i" "j" "5" "6" "7" "8" "9" "10" "11" "12" "13" "14" "15" "16" "17" "18" "19" "20"
> identical(v1,v2[1:length(v1)])
[1] TRUE
>
Neither vec_mrg nor funMergeVectsByPlace return factors. If one includes factor(...), both functions are getting slower, but funMergeVectsByPlace is still faster than vec_mrg.