用其他内容替换字符串中第 N 次出现的字符

Question

Consider a = paste(1:10,collapse=", ") which results in

a = "1, 2, 3, 4, 5, 6, 7, 8, 9, 10"

I would like to replace every n-th (say 4-th) occurrences of "," and replace it with something else (say "\n"). The desired output would be:

"1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"

I am looking for a code that uses gsub (or something equivalent) and some form of regular expression to achieve this goal.

Answer 1

You can replace ((?:\\d+, ){3}\\d), with \\1\\n

You basically capture everything till fourth comma in group1 and comma separately and replace it with \\1\\n which replaces matched text with group1 text and newline, giving you the intended results.

Regex Demo

R Code demo

gsub("((?:\\d+, ){3}\\d),", "\\1\n", "1, 2, 3, 4, 5, 6, 7, 8, 9, 10")

Prints,

[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"

Edit:

To generalize above solution to any text, we can change \\d to [^,]

New R code demo

gsub("((?:[^,]+, ){3}[^,]+),", "\\1\n", "1, 2, 3, 4, 5, 6, 7, 8, 9, 10")
gsub("((?:[^,]+, ){3}[^,]+),", "\\1\n", "a, bb, ccc, dddd, 500, 600, 700, 800, 900, 1000")

Output,

[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"
[1] "a, bb, ccc, dddd\n 500, 600, 700, 800\n 900, 1000"

Answer 2

Using both regex and gsub .

a = paste(1:10,collapse=", ")
x <- gsub("([^,]*,[^,]*,[^,]*,[^,]*),", '\\1\n', a)
x
#> [1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"

Answer 3

regex is the best alternative, nontheless here's another approach without regex

> str_vec <- strsplit(a, " ")[[1]] 
> where <- seq_along(str_vec) %% 4 == 0
> str_vec[where] <- sub(",", "\n", str_vec[where])
> paste(str_vec, collapse=" ")
[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"

Answer 4

regmatches as yet another alternative:

a <- "1, 2, 3, 4, 5, 6, 7, 8, 9, 10"

fn <- ","
rp <- "\n"
n <- 4

regmatches(a, gregexpr(fn, a)) <- list(c(rep(fn,n-1),rp))
a
#[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"

As a function:

a <- "1, 2, 3, 4, 5, 6, 7, 8, 9, 10"

replN <- function(x, fn, rp, n) {
    regmatches(x, gregexpr(fn, x)) <- list(c(rep(fn,n-1),rp))
    x
}
replN(a, ",", "\n", 4)
#[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10

You could even extend this to be vectorised over the replacement argument:

a = "1, 2, 3, 4, 5, 6, 7, 8, 9, 10"

replN <- function(x,fn,rp,n) {
    sel <- rep(fn, n*length(rp))
    sel[seq_along(rp)*n] <- rp
    regmatches(x, gregexpr(fn, x)) <- list(sel)
    x
}
replN(a, fn=",", rp=c("1st","2nd"), n=4)
#[1] "1, 2, 3, 41st 5, 6, 7, 82nd 9, 10"

Answer 5

This one can be replace with a string instead of a character. I did a function that you can use easily:)

A demo here to understand the regex

> a = paste(1:10,collapse=", ")
> a
[1] "1, 2, 3, 4, 5, 6, 7, 8, 9, 10"
> # if you want the 2nd occurence
> gsub("(.*?,.*?),(.*)", "\\1\n\\2", a)
[1] "1, 2\n 3, 4, 5, 6, 7, 8, 9, 10"
> # if you want the 3rd occurence
> gsub("(.*?,.*?,.*?),(.*)", "\\1\n\\2", a)
[1] "1, 2, 3\n 4, 5, 6, 7, 8, 9, 10"
> # if you want the 4rd occurence
> gsub("(.*?,.*?,.*?,.*?),(.*)", "\\1\n\\2", a)
[1] "1, 2, 3, 4\n 5, 6, 7, 8, 9, 10"
> # if you want the last occurence
> gsub("(.*,.*),(.*)", "\\1\n\\2", a)
[1] "1, 2, 3, 4, 5, 6, 7, 8, 9\n 10"
> 
> 
> replace.occurence <- function(x, pattern, replacement, which.occu) {
+   if( which.occu == "last" ) {
+     gsub(paste0("(.*", pattern, ".*)", pattern, "(.*)"), paste0("\\1", replacement, "\\2"), x)
+   } else {
+     gsub(paste0("(.*?", paste0(rep(paste0(pattern, ".*?"), which.occu - 1), collapse = ""), ")", pattern, "(.*)"), paste0("\\1", replacement, "\\2"), x)
+   }
+ }
> 
> replace.occurence(a, pattern = ",", replacement = "\n", which.occu = 2)
[1] "1, 2\n 3, 4, 5, 6, 7, 8, 9, 10"
> replace.occurence(a, pattern = ",", replacement = "\n", which.occu = 3)
[1] "1, 2, 3\n 4, 5, 6, 7, 8, 9, 10"
> replace.occurence(a, pattern = ",", replacement = "\n", which.occu = 4)
[1] "1, 2, 3, 4\n 5, 6, 7, 8, 9, 10"
> replace.occurence(a, pattern = ",", replacement = "\n", which.occu = "last")
[1] "1, 2, 3, 4, 5, 6, 7, 8, 9\n 10"
> 
> replace.occurence(a, pattern = ", 3, 4,", replacement = ", 4, 3,", which.occu = 1)
[1] "1, 2, 4, 3, 5, 6, 7, 8, 9, 10"