Removing/Replacing Brackets from R String Using Gsub

Removing/replacing brackets from R string using gsub

Using the correct regex works:

gsub("[()]", "", "(abc)")

The additional square brackets mean "match any of the characters inside".

Remove square brackets from a string vector

You could gsub out the brackets like so:

n = c("[Dave]", "[Tony]", "[Sara]")

gsub("\\[|\\]", "", n)
[1] "Dave" "Tony" "Sara"

Using Gsub in R to remove a string containing brackets

We can by placing the brackets inside the square brackets as () is a metacharacter

gsub('[(]4:4aCO[)]','', '(5:3)(4:4)(5:3)(4:4)(4:4aCO)(6:2)_(4:4a)')

Or with fixed = TRUE to evaluate the literal meaning of that character

gsub('(4:4aCO)','', '(5:3)(4:4)(5:3)(4:4)(4:4aCO)(6:2)_(4:4a)', fixed = TRUE)

Remove trailing brackets in a string

Another, relatively easy, regex solution is this:

data.frame(df) %>%
  mutate(df = gsub("\\[\\W+|\\W+\\]", "", df))
                     df
1           Mamie Smith
2 Screamin' Jay Hawkins

Here we remove any non-alphanumeric character (\\W+) occurring one or more times on the condition that it be preceded OR (|) followed by a square bracket.

Alternatively, to borrow from @TaerJae but greatly simplified:

library(stringr)
data.frame(df) %>%
  mutate(df = str_extract(df, '\\w.*\\w'))

Here we simply focus on the alphanumeric characters (\\w) on either side of the string, while allowing for any characters (.*) to occur in-between them thus capturing, for example, the apostrophe in Screamin'and the whitespaces.

r - How can I remove a single pair of parentheses with text in a character?

Use sub :

sub('\\(.*?\\)\\s', '', value)
#[1] "This is a (keep) test sentence."

• () are metacharacters and need to be escaped with \\.

• .*? is to match as few characters possible till a closing bracket ()) is encountered.

Removing parenthesis & dollar sign using gsub() in R?

I think the easiest way is to step through what is replaced. I'm inferring that you do not want to lose the negative-ness suggested by the parens, so we'll do two steps:

s <- c("($2,345)", "$3,500", "$5,600", "($3,234)")
gsub("[$),]", "", s)
# [1] "(2345" "3500"  "5600"  "(3234"

This removes most of the junk (that we did not want/need to keep), now let's deal with the leading left-paren:

gsub("^\\s*[(]", "-", gsub("[$),]", "", s))
# [1] "-2345" "3500"  "5600"  "-3234"

From here, if needed, you can convert to numeric:

as.numeric(gsub("^\\s*[(]", "-", gsub("[$),]", "", s)))
# [1] -2345  3500  5600 -3234

This is not very robust to mal-formed strings. For instance, though I look for (and remove) leading space for the left-paren, anything else there will be a problem.

How to remove square brackets [] from R data frame without regex?

Somewhat mysterious, but

gsub("[][]","", x)

will do it (i.e., remove [ and ]). The outer square brackets are interpreted as defining a set of characters; the inner square brackets are the set to remove. Putting the inner square brackets in the 'correct' order ([[]]) doesn't work presumably because the inner [] is interpreted as part of a special like [:alpha:]?

This regex does not work in stringr::str_remove() ("Missing closing bracket on a bracket expression"), but

stringr::str_remove_all(x,"[\\[\\]]")

does. (The double-backslashes are required because R uses single-backslash to identify special characters like newline (\n) etc..)

In principle it should be possible to use the relatively new "raw strings" feature in R to simplify the expression (i.e. use regex pattern r"[\[\]]" instead of "[\\[\\]]", but I seem to be discovering lots of weird interactions between that and the regex machinery (unless I'm doing something wrong/being sloppy, which is always a possibility ...)

If you really don't want to use regular expressions then

gsub(fixed = TRUE, "[", "",
   gsub(fixed = TRUE, "]", "", x)
)

or

x |> str_remove_all(fixed("[")) |> str_remove_all(fixed("]"))

should work.

R - Clear brackets and anything inside it with gsub

s1<-as.character("Norway(19)")
s1<-gsub("\\(.*\\)", "", s1)

should get you there, . is the wildcard in regex and * repeats something 0 or more times.

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