How can I remove text within parentheses with a regex?

s/\([^)]*\)//

So in Python, you'd do:

re.sub(r'\([^)]*\)', '', filename)

The pattern that matches substrings in parentheses having no other ( and ) characters in between (like (xyz 123) in Text (abc(xyz 123)) is

\([^()]*\)

Details:

\( - an opening round bracket (note that in POSIX BRE, ( should be used, see sed example below)
[^()]* - zero or more (due to the * Kleene star quantifier) characters other than those defined in the negated character class/POSIX bracket expression, that is, any chars other than ( and )
\) - a closing round bracket (no escaping in POSIX BRE allowed)

Removing code snippets:

JavaScript: string.replace(/$[^()]*$/g, '')
PHP: preg_replace('~$[^()]*$~', '', $string)
Perl: $s =~ s/$[^()]*$//g
Python: re.sub(r'$[^()]*$', '', s)
C#: Regex.Replace(str, @"$[^()]*$", string.Empty)
VB.NET: Regex.Replace(str, "$[^()]*$", "")
Java: s.replaceAll("\$[^()]*\$", "")
Ruby: s.gsub(/$[^()]*$/, '')
R: gsub("\$[^()]*\$", "", x)
Lua: string.gsub(s, "%([^()]*%)", "")
Bash/sed: sed 's/([^()]*)//g'
Tcl: regsub -all {$[^()]*$} $s "" result
C++ std::regex: std::regex_replace(s, std::regex(R"($[^()]*$)"), "")
Objective-C:
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"\$[^()]*\$" options:NSRegularExpressionCaseInsensitive error:&error]; NSString *modifiedString = [regex stringByReplacingMatchesInString:string options:0 range:NSMakeRange(0, [string length]) withTemplate:@""];
Swift: s.replacingOccurrences(of: "\$[^()]*\$", with: "", options: [.regularExpression])
Google BigQuery: REGEXP_REPLACE(col, "\$[^()]*\$" , "")

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