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The simplest way to extract the string between two parentheses is to use slicing and string. find() . First, find the indices of the first occurrences of the opening and closing parentheses. Second, use them as slice indices to get the substring between those indices like so: s[s.
By placing part of a regular expression inside round brackets or parentheses, you can group that part of the regular expression together. This allows you to apply a quantifier to the entire group or to restrict alternation to part of the regex. Only parentheses can be used for grouping.
[] denotes a character class. () denotes a capturing group. (a-z0-9) -- Explicit capture of a-z0-9 . No ranges.
If your problem is really just this simple, you don't need regex:
s[s.find("(")+1:s.find(")")]
Use re.search(r'\((.*?)\)',s).group(1):
>>> import re
>>> s = u'abcde(date=\'2/xc2/xb2\',time=\'/case/test.png\')'
>>> re.search(r'\((.*?)\)',s).group(1)
u"date='2/xc2/xb2',time='/case/test.png'"
If you want to find all occurences:
>>> re.findall('\(.*?\)',s)
[u"(date='2/xc2/xb2',time='/case/test.png')", u'(eee)']
>>> re.findall('\((.*?)\)',s)
[u"date='2/xc2/xb2',time='/case/test.png'", u'eee']
Building on tkerwin's answer, if you happen to have nested parentheses like in
st = "sum((a+b)/(c+d))"
his answer will not work if you need to take everything between the first opening parenthesis and the last closing parenthesis to get (a+b)/(c+d), because find searches from the left of the string, and would stop at the first closing parenthesis.
To fix that, you need to use rfind for the second part of the operation, so it would become
st[st.find("(")+1:st.rfind(")")]
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