How to keep only specific punctuation mark in a column

Question

In the column text how it is possible to remove all punctuation remarks but keep only the ?

data.frame(id = c(1), text = c("keep<>-??it--!@#"))

expected output

data.frame(id = c(1), text = c("keep??it"))

Answer 1

A more general solution would be to used nested gsub commands that converts ? to a particular unusual string (like "foobar"), gets rid of all punctuation, then writes "foobar" back to ?:

gsub("foobar", "?", gsub("[[:punct:]]", "", gsub("\\?", "foobar", df$text)))
#> [1] "keep??it"

Answer 2

Using gsub you could do:

gsub("(\\?+)|[[:punct:]]","\\1",df$text)

[1] "keep??it"

Answer 3

gsub('[[:punct:] ]+',' ',data) removes all punctuation which is not what you want.

But this is:

library(stringr)
sapply(df, function(x) str_replace_all(x, "<|>|-|!|@|#",""))
     id  text      
[1,] "1" "a"       
[2,] "2" "keep??it"

Better IMO than other answers because no need for nesting, and lets you define whichever characters to sub.

Answer 4

Here's another solution using negative lookahead:

gsub("(?!\\?)[[:punct:]]", "", df$text, perl = T)
[1] "keep??it"

The negative lookahead asserts that the next character is not a ? and then matches any punctuation.

Data:

df <- data.frame(id = c(1), text = c("keep<>-??it--!@#"))