Answer by root for How to remove all text between the outer parentheses in a...
No Python loops. No recursion. No regexes.Just computing the difference between the cumulative counts of '(' and ')':import numpy as nps = '()a(x(x)x)b(x)c()d()'s_array = np.array(list(s))mask_open =...
View ArticleAnswer by ozdemir for How to remove all text between the outer parentheses in...
I have found a solution here:http://rachbelaid.com/recursive-regular-experession/which says:>>> import regex>>> regex.search(r"^(\((?1)*\))(?1)*$", "()()") is not NoneTrue>>>...
View ArticleAnswer by Wiktor Stribiżew for How to remove all text between the outer...
NOTE: \(.*\) matches the first ( from the left, then matches any 0+ characters (other than a newline if a DOTALL modifier is not enabled) up to the last), and does not account for properly nested...
View ArticleAnswer by bobble bubble for How to remove all text between the outer...
As mentioned before, you'd need a recursive regex for matching arbitrary levels of nesting but if you know there can only be a maximum of one level of nesting have a try with this...
View ArticleAnswer by Bryce Drew for How to remove all text between the outer parentheses...
https://regex101.com/r/kQ2jS3/1'(\(.*\))'This captures the furthest parentheses, and everything in between the parentheses.Your old regex captures the first parentheses, and everything between to the...
View ArticleAnswer by Serge Ballesta for How to remove all text between the outer...
If you are sure that the parentheses are initially balanced, just use the greedy version:re.sub(r'\(.*\)', '', s2)
View ArticleAnswer by alexamici for How to remove all text between the outer parentheses...
re matches are eager so they try to match as much text as possible, for the simple test case you mention just let the regex run:>>> re.sub(r'\(.*\)', '', 'stuff(remove(me))')'stuff'
View ArticleHow to remove all text between the outer parentheses in a string?
When I have a string like this:s1 = 'stuff(remove_me)'I can easily remove the parentheses and the text within using# returns 'stuff'res1 = re.sub(r'\([^)]*\)', '', s1)as explained here.But I sometimes...
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