Convert a list of lists to a character vector

Question

I have a list of lists of characters. For example:

l <- list(list("A"),list("B"),list("C","D"))

So as you can see some elements are lists of length > 1.

I want to convert this list of lists to a character vector, but I'd like the lists with length > 1 to appear as a single element in the character vector.

the unlist function does not achieve that but rather:

> unlist(l)
[1] "A" "B" "C" "D"

Is there anything faster than:

sapply(l,function(x) paste(unlist(x),collapse=""))

To get my desired result:

"A"  "B"  "CD"

Answer 1

You can skip the unlist step. You already figured out that paste0 needs collapse = TRUE to "bind" sequential elements of a vector together:

> sapply( l, paste0, collapse="")
[1] "A"  "B"  "CD"

Answer 2

Here's a variation of @thela's suggestion, if you don't mind a multi-line approach:

x <- lengths(l)                                     ## Get the lengths of each list
l[x > 1] <- lapply(l[x > 1], paste0, collapse = "") ## Paste only those together
unlist(l, use.names = FALSE)                        ## Unlist the result
# [1] "A"  "B"  "CD"

Alternatively, if you don't mind using a package, look at the "stringi" package, specifically stri_flatten, as suggested by @Jota.

---

Here's a performance comparison:

l <- list(list("A"), list("B"), list("B"), list("B"), list("B"),
          list("C","D"), list("E","F", "G", "H"), 
          as.list(rep(letters,10)), as.list(rep(letters,2)))
l <- unlist(replicate(1000, l, FALSE), recursive = FALSE)

funop <- function() sapply(l,function(x) paste(unlist(x),collapse=""))
fun42 <- function() sapply(l, paste0, collapse="")
funv  <- function() vapply(l, paste0, character(1L), collapse = "")
funam <- function() {
  x <- lengths(l)
  l[x > 1] <- lapply(l[x > 1], paste0, collapse = "")
  unlist(l, use.names = FALSE)
}
funj <- function() sapply(l, stri_flatten)
funamj <- function() {
  x <- lengths(l)
  l[x > 1] <- lapply(l[x > 1], stri_flatten)
  unlist(l, use.names = FALSE)
}

library(microbenchmark)
microbenchmark(funop(), fun42(), funv(), funam(), funj(), times = 20)
# Unit: milliseconds
#      expr      min       lq     mean   median       uq      max neval   cld
#   funop() 78.21822 84.79588 85.30055 85.36399 86.90540 90.48321    20     e
#   fun42() 56.16938 57.35735 61.60008 58.04969 65.82836 81.46482    20    d 
#    funv() 54.64101 56.23245 60.07896 57.26049 63.96815 78.58043    20    d 
#   funam() 45.89760 46.89890 48.99810 47.29617 48.28764 56.92544    20   c  
#    funj() 28.73405 29.94041 32.00676 30.56711 31.11448 39.93765    20  b   
#  funamj() 18.64829 19.01328 21.05989 19.12468 19.52516 32.87569    20 a 

---

Note: The relative efficiency of this approach would depend on how many list items are going to have length(x) > 1. If most of them are going to be > 1 anyway, then just go with @42-'s approach. stri_flatten only improves performance if you have long character vectors to paste together as in sample list used for the above benchmark, otherwise, it doesn't help.