gsub unable to remove empty brackets in R

Question

I have the following string in R

A<-"A (23) 56 hh()"

I want to get the following output

"A (23) 56 hh"

I tried the following code

B<-gsub(pattern = "()", replacement = "", x = A)

That didnt yield the desired result. How can I accomplish the same

See - [How do I deal with special characters like \^$.?*|+()[{ in my regex?](https://stackoverflow.com/questions/27721008/how-do-i-deal-with-special-characters-like-in-my-regex) — Ronak Shah, Jul 30 '21 at 12:38

score 5 · Answer 1 · answered Jul 30 '21 at 12:20

5

Try fixed = TRUE in gsub

> gsub("()", "", A, fixed = TRUE)
[1] "A (23) 56 hh"

answered Jul 30 '21 at 12:20

ThomasIsCoding

score 5 · Answer 2 · answered Jul 30 '21 at 17:02

5

Using str_remove

library(stringr)
str_remove_all(A, fixed("()"))

-ouptut

[1] "A (23) 56 hh"

answered Jul 30 '21 at 17:02

akrun

1

Hi master akrun. Need your help. How could I regex `(, 1, 0, )` to `10` – TarJae Jul 30 '21 at 17:24
2

Try with `gsub("\\D+", "", str1)# [1] "10"` – akrun Jul 30 '21 at 17:25

score 4 · Answer 3 · answered Jul 30 '21 at 12:20

4

try B<-gsub(pattern = "\$\$", replacement = "", x = A)

\\ indicates that it is a specific character - not the regex expression in brackets

answered Jul 30 '21 at 12:20

dy_by

score 4 · Answer 4 · answered Jul 30 '21 at 12:49

4

dy_by and ThomasIsCoding have good answers. Here is a modification of dy_by's answer

gsub(pattern = "\\()", replacement = "", x = A)

[1] "A (23) 56 hh"

answered Jul 30 '21 at 12:49

TarJae

score 3 · Answer 5 · answered Jul 30 '21 at 19:39

3

Another option defining removal of two consecutive parenthesis chars, which obviates the need for fixed=TRUE:

library(stringr)

A %>% str_remove("[()]{2}")

[1] "A (23) 56 hh"

answered Jul 30 '21 at 19:39

GuedesBF

5 Answers5