Extract values between '\r'

One way would be to use strsplit:

x <- "\r Joel Corry feat. MNEK\r Head & Heart\r"    

df$interpret <- trimws(strsplit(x, split = "\\r")[[1]])[2]
df$song<- trimws(strsplit(x, split = "\\r")[[1]])[3]

library(data.table)
dat = data.table(raw_string =  c(
  "\r Joel Corry feat. MNEK\r Head & Heart\r ",
  "\r Intepret 2\r Song 2\r ",
  "\r Intepret 3\r Song 3\r ")
)
dat[,c("interpret","song"):= tstrsplit(str_remove_all(raw_string,"(^\r\\s?)|(\r\\s?$)"),"\r")][]

Output:

                                   raw_string             interpret          song
                                       <char>                <char>        <char>
1: \r Joel Corry feat. MNEK\r Head & Heart\r  Joel Corry feat. MNEK  Head & Heart
2:                  \r Intepret 2\r Song 2\r             Intepret 2        Song 2
3:                  \r Intepret 3\r Song 3\r             Intepret 3        Song 3

Use extract to define a regex that (i) exhaustively describes the string but (ii) wraps into capture groups (...) only what you want to extract into the two new columns (and thereby effectively removes anything not inside a capture group). Note that \ needs to escaped by double backslash (in R):

library(tidyr)
df %>%
  extract(str,
          into = c("interpret", "song"),
          regex = "\\\r (.*)\\\r (.*)\\\r")
              interpret         song
1 Joel Corry feat. MNEK Head & Heart

Data:

df <- data.frame(
  str = "\r Joel Corry feat. MNEK\r Head & Heart\r "
)

Please find below another possibility using the stringr library

Reprex

• Code

library(stringr)

setNames(as.data.frame(str_extract_all(my_text, "(?<=\\r ).+", simplify = TRUE)), c("interpret", "song"))

• Output

#>               interpret         song
#> 1 Joel Corry feat. MNEK Head & Heart

• Data

my_text <- "\r Joel Corry feat. MNEK\r Head & Heart\r "

^{Created on 2022-02-21 by the reprex package (v2.0.1)}