Reputation: 5281
I am trying to split strings by using the first white space coming after 3 characters. Here is my code:
string <- c("Le jour la nuit", "Les jours les nuits")
part1 <- sub("(\\S{3,})\\s?(.*)", "\\1", string)
part2 <- sub("(\\S{3,})\\s?(.*)", "\\2", string)
# output
> part1
[1] "Le jour" "Les"
> part2
[1] "Le la nuit" "jours les nuits"
For the first part, it works exactly as desired. However, it is not the case for the second part: part2[1]
should be la nuit
instead of Le la nuit
.
I am not sure how achieve this and would be thankful for some guidance.
Upvotes: 3
Views: 97
Reputation: 43169
Not sure what you really want but per your requirements, you could use
^(.{3,}?)(?:(?<!,)\\s)+(.*)
This says:
^ # start of the string
(.{3,}?) # capture 3+ characters lazily, up to...
(?:(?<!,)\\s)+ # 1+ whitespaces that must not be preceeded by a comma
(.*) # capture the rest of the string
In R
:
string <- c("Le jour la nuit", "Les jours les nuits", "les, jours les nuits")
(part1 <- sub("^(.{3,}?)(?:(?<!,)\\s)+(.*)", "\\1", string, perl = T))
(part2 <- sub("^(.{3,}?)(?:(?<!,)\\s)+(.*)", "\\2", string, perl = T))
Yielding
[1] "Le jour" "Les" "les, jours"
and
[1] "la nuit" "jours les nuits" "les nuits"
dataframe
as a result, if so, you could define yourself a little function (using sapply
and some logic):
make_df <- function(text) {
parts <- sapply(text, function(x) {
m <- regexec("^(.{3,}?)(?:(?<!,)\\s)+(.*)", x, perl = T)
groups <- regmatches(x, m)
c(groups[[1]][2], groups[[1]][3])
}, USE.NAMES = F)
(setNames(as.data.frame(t(parts), stringsAsFactors = F), c("part1", "part2")))
}
(df <- make_df(string))
This would yield for string <- c("Le jour la nuit", "Les jours les nuits", "les, jours les nuits", "somejunk")
:
part1 part2
1 Le jour la nuit
2 Les jours les nuits
3 les, jours les nuits
4 <NA> <NA>
Upvotes: 3