fastest way to split strings into fixed-length elements in R - r

Question

How to split a string into elements of fixed length in R is a commonly asked question to which typical answers either rely on substring(x) or strsplit(x, sep="") followed by paste(y, collapse = "").
For instance, one would slit the string "azertyuiop" into "aze", "rty","uio", "p" by specifying a fixed length of 3 characters.
I'm looking for the fastest way possible.
After some testing with long strings (> 1000 chars), I have found that substring() is way too slow. The strategy is hence to split the string into individual characters, and them paste them back into groups of the desired length, by applying some cleverness.
Here is the fastest function I could come up with. The idea is to split the string into individual chars, then have a separator interspersed in the character vector at the right positions, collapse the characters (and separators) back into a string, then split the string again, but this time specifying the separator.
splitInParts <- function(string, size) { #can process a vector of strings. "size" is the length of desired substrings
chars <- strsplit(string,"",T)
lengths <- nchar(string)
nFullGroups <- floor(lengths/size) #the number of complete substrings of the desired size
#here we prepare a vector of separators (comas), which we will replace by the characters, except at the positions that will have to separate substring groups of length "size". Assumes that the string doesn't have any comas.
seps <- Map(rep, ",", lengths + nFullGroups) #so the seps vector is longer than the chars vector, because there are separators (as may as they are groups)
indices <- Map(seq, 1, lengths + nFullGroups) #the positions at which separators will be replaced by the characters
indices <- lapply(indices, function(x) which(x %% (size+1) != 0)) #those exclude the positions at which we want to retain the separators (I haven't found a better way to generate such vector of indices)
temp <- function(x,y,z) { #a fonction describing the replacement, because we call it in the Map() call below
x[y] <- z
x
}
res <- Map(temp, seps, indices, chars) #so now we have a vector of chars with separators interspersed
res <- sapply(res, paste, collapse="", USE.NAMES=F) #collapses the characters and separators
res <- strsplit(res, ",", T) #and at last, we can split the strings into elements of the desired length
}
This looks quite tedious, but I have tried to simply put the chars vector into a matrix with the adequate number of rows, then collapse the matrix columns with apply(mat, 2, paste, collapse=""). This is MUCH slower. And splitting the character vector with split() into a list of vectors of the right length, so as to collapse elements, is even slower.
So if you can find something faster, let me know. If not, well my function may be of some use. :)

Answer 1

Was fun reading the updates, so I benchmarked:
> nchar(mystring)
[1] 260000
My idea was near the same as #akrun's one as str_extract_all use the same function under the hood IIRC)
library(stringr)
tensiSplit <- function(string,size) {
str_extract_all(string, paste0('.{1,',size,'}'))
}
And the results on my machine:
> microbenchmark(splitInParts(mystring,3),akrunSplit(mystring,3),splitInParts2(mystring,3),tensiSplit(mystring,3),gsubSplit(mystring,3),times=3)
Unit: milliseconds
expr min lq mean median uq max neval
splitInParts(mystring, 3) 64.80683 64.83033 64.92800 64.85384 64.98858 65.12332 3
akrunSplit(mystring, 3) 4309.19807 4315.29134 4330.40417 4321.38461 4341.00722 4360.62983 3
splitInParts2(mystring, 3) 21.73150 21.73829 21.90200 21.74507 21.98725 22.22942 3
tensiSplit(mystring, 3) 21.80367 21.85201 21.93754 21.90035 22.00447 22.10859 3
gsubSplit(mystring, 3) 53.90416 54.28191 54.55416 54.65966 54.87915 55.09865 3

Answer 2

We can split by specifying a regex lookbehind to match the position preceded by 'n' characters, For example, if we are splitting by 3 characters, we match the position/boundary preceded by 3 characters ((?<=.{3})).
splitInParts <- function(string, size){
pat <- paste0('(?<=.{',size,'})')
strsplit(string, pat, perl=TRUE)
}
splitInParts(str1, 3)
#[[1]]
#[1] "aze" "rty" "uio" "p"
splitInParts(str1, 4)
#[[1]]
#[1] "azer" "tyui" "op"
splitInParts(str1, 5)
#[[1]]
#[1] "azert" "yuiop"
Or another approach is using stri_extract_all from library(stringi).
library(stringi)
splitInParts2 <- function(string, size){
pat <- paste0('.{1,', size, '}')
stri_extract_all_regex(string, pat)
}
splitInParts2(str1, 3)
#[[1]]
#[1] "aze" "rty" "uio" "p"
stri_extract_all_regex(str1, '.{1,3}')
data
str1 <- "azertyuiop"

Answer 3

Alright, there was a faster solution published here (d'oh!)
Simply
strsplit(gsub("([[:alnum:]]{size})", "\\1 ", string)," ",T)
Here using a space as separator.
(didn't think about [[:allnum::]]{}).
How can I mark my own question as a duplicate? :(