12

I have a vector in R

vec <- c("x", "y", "z")

I want to generate a new vector with elements, in order, based off the elements of vec. I want the result to be:

"x" "x.1" "y" "y.1" "z" "z.1"

Now, I know that I could do the following:

vec <- c(vec, paste0(vec,".1"))

But this yields

"x" "y" "z" "x.1" "y.1" "z.1"

... which is not what I want.

Improve this question
3
  • 1
    stackoverflow.com/questions/2953093/… Commented Apr 7, 2025 at 6:46
  • @Edward while that question is older, this one has gained more attention, and received more answers. I will close that one in favor of this. See: meta.stackexchange.com/a/147651 p.s. It doesn't seem you've voted to close this, but someone did. Commented Apr 7, 2025 at 16:02
  • 1
    @M-- that post is related, but not necessarily a duplicate, which is why I didn’t vote to close. Commented Apr 8, 2025 at 0:09

9 Answers 9

Reset to default
17
Answer recommended by R Language Collective

You can use make.unique + rep

make.unique(rep(vec, each = 2))
[1] "x"   "x.1" "y"   "y.1" "z"   "z.1"
Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

10

Here is another base R approach using `vapply`:

1. \(x) c(x, paste0(x, ".1")) this anonymous function takes one element x and returns a vector c("x", "x.1")

2. vapply(vec, ..., character(2)) applies the function to each element of vec. With character(2) we indicate to expect exactly two characters. The result is a matrix.

3. Using as.vector(...) flattens the matrix column-wise

as.vector(vapply(vec, \(x) c(x, paste0(x, ".1")), character(2)))

output:

[1] "x"   "x.1" "y"   "y.1" "z"   "z.1"
Improve this answer

2 Comments

@Friede That is at your discretion whether you consider it useful or not. Just comment. You are not in a position to give the OP advice on what they should do. So stop preventing users from participating here and creating a productive environment. At least it brings vapply back to mind and shows how to turn a matrix into a vector. So I think it's great—otherwise, I wouldn’t have posted it!
We have make.unique: "Makes the elements of a character vector unique by appending sequence numbers to duplicates." With rep it is concise, easy to understand, complete. It's already suggested. With vapply(), as.vector(), c(), paste0() we add functions which are not needed or redundand for the given case. "At least it brings vapply back to mind and shows how to turn a matrix into a vector." seems to confirm my impression. Sometimes answers are written to answer questions and and sometimes they are written to write one. I strongly recommend to use make.unique() + rep().
9

You could also use rep + paste

vec <- c("x", "y", "z")
paste0(rep(vec, each = 2), c("", ".1"))

[1] "x"   "x.1" "y"   "y.1" "z"   "z.1"
Improve this answer

Comments

7

One option is to rbind the two vectors and collapse them into a single vector to get them in the required order.

c(rbind(vec, paste0(vec, ".1")))
#[1] "x"   "x.1" "y"   "y.1" "z"   "z.1"
Improve this answer

2 Comments

With the nice alternative c(cbind(vec, paste0(vec, ".1"))) giving "x" "y" "z" "x.1" "y.1" "z.1"
Small efficiency gain possible with dim(...) <- NULL instead of c() to flatten.
6

For the sake of completeness.

If your real data is well represented by your toy data, i.e. you want to repeat an element of vec only once and have already executed

vec = c("x", "y", "z")
vec = c(vec, paste0(vec,".1"))

Then you can achieve what you are after by sorting.

> sort(vec)
[1] "x"   "x.1" "y"   "y.1" "z"   "z.1"

However, I strongly recommend ThomasIsCoding's answer.

Improve this answer

2 Comments

If data is enormous sorting is expensive.
@s_baldur Obviously. I do not think performance matters here, especially I do not think OP deals with 35^4 unique charactors.
4 
vec <- c("x", "y", "z")
suf <- c(".1", "")

data.table::CJ(vec, suf)[, paste0(vec, suf)]
#> [1] "x"   "x.1" "y"   "y.1" "z"   "z.1"

Created on 2025-04-05 with reprex v2.1.1

Improve this answer

Comments

4

If you wanted:

"x.1" "x.2" "y.1" "y.2" "z.1" "z.2"

(which has a more balanced look, and for names of data frames, can facilitate reshaping),

then the makeunique package could be used. It has just one function, make_unique, that asserts to "Make all elements of a character vector unique."

It differs from base R's make.unique function in that the non-duplicated characters are also given a suffix, starting from 1.

vec <- c("x", "y", "z")

library(makeunique)

make_unique(rep(vec, each=2), sep = ".", wrap = FALSE)
[1] "x.1" "x.2" "y.1" "y.2" "z.1" "z.2"
Improve this answer

Comments

3

Here’s another perspective on this problem—just offering it as an extra option alongside the more streamlined answers already here:

vec <- c("x", "y", "z")

mapply(c,vec,paste0(vec, ".1"), SIMPLIFY = F, USE.NAMES = F) |> unlist()

"x"   "x.1" "y"   "y.1" "z"   "z.1"
Improve this answer

3 Comments

Although unelegantly It's almost done with c(vec, paste0(vec,".1")), just sorting is needed. Why would one wrap mapply() (multivariate looping) and unlist() around it? At best, this is misleading. Please note, you are doing ~ unlist(Map(c, vec, paste0(vec, ".1"))).
c(vec, paste0(vec, '.1')) is definitely a simpler and more concise way to achieve a similar result. My goal with the mapply() approach was to offer an alternative that explicitly pairs up the elements step-by-step before flattening them, which might be useful in cases where the pairing logic needs to be more controlled or extended (e.g., if someone wanted to apply a custom function to each pair). I agree it’s less elegant for this specific case, and I appreciate you pointing out the Map() equivalence—mapply() with SIMPLIFY = FALSE is indeed just a more verbose cousin here.
There is simply no need for multivariate looping; neither for univariate looping since full vectorised solutions are out there. For the future, you are aware of mapply and Map side by side which is good to hear!
1

A bit verbose but aiming to do the least work using base R.

foo <- function(v) {
  # Updated version, uses seq.int() instead of seq()
  n <- length(v)
  out <- character(n * 2L)
  out[seq.int(1L, n * 2L, by = 2L)] <- v
  out[seq.int(2L, n * 2L, by = 2L)] <- paste0(v, ".1")
  out
}
foo(vec)
# [1] "x"   "x.1" "y"   "y.1" "z"   "z.1"

Benchmark:

# Generate 35^4 ~ 1.5 million unique characters
vec <- do.call(data.table::CJ, rep(list(c(letters, 1:9)), 4))[, do.call(paste0, .SD)] 
bench::mark(
  Tho = make.unique(rep(vec, each = 2)),
  Ron = c(rbind(vec, paste0(vec, ".1"))),
  Tim = {paste0(rep(vec, each = 2), c("", ".1"))},
  Tar = as.vector(vapply(vec, \(x) c(x, paste0(x, ".1")), character(2))),
  Fri = {x = c(vec, paste0(vec,".1"));sort(x)},
  `M--` = {suf <- c(".1", "");data.table::CJ(vec, suf)[, paste0(vec, suf)]},
  SAL = {mapply(c,vec,paste0(vec, ".1"), SIMPLIFY = F, USE.NAMES = F) |> unlist()},
  s_b = foo(vec)
  # Edwards not included as solution doesn't match exactly expected output
)
#   expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc
#   <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>
# 1 Tho        984.11ms 984.11ms    1.02     100.7MB     2.03     1     2
# 2 Ron         239.3ms 264.44ms    3.78      57.2MB     1.89     2     1
# 3 Tim        477.86ms 482.61ms    2.07      45.8MB     0        2     0
# 4 Tar           4.83s    4.83s    0.207     57.3MB     2.49     1    12
# 5 Fri          13.46s   13.46s    0.0743    68.7MB     0        1     0
# 6 M--        822.16ms 822.16ms    1.22      80.2MB     1.22     1     1
# 7 SAL           1.86s    1.86s    0.538     45.8MB     3.23     1     6
# 8 s_b        208.96ms 254.67ms    4.00      45.8MB     1.33     3     1
Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.