Given a string:
#symbol 1# / 7 - #symbol 2#
I want to return:
func('#symbol 1#') / 7 - func('#symbol 2#')
I've tried:
re.sub('[#*#]', 'func(\'', f2)
which gives:
func(symbol 1func( / 7 - func(symbol 2func(
where I don't want the func(' for the second hash.
Is there a way to get the end of the hash replaced with ') using the re.sub?
You may use a regex like
#[^\s#](?:[^#]*[^\s#])?#
Replace with func('\g<0>') where \g<0> refers to the whole match. See the regex demo.
Details
# - a # char[^\s#] - a char other than whitespace and #(?:[^#]*[^\s#])? - an optional non-capturing group matching 1 or 0 occurrences of
[^#]* - 0 or more chars other than #[^\s#] - a char other than whitespace and ## - a # char.See the Python demo:
import re
text = '#symbol 1# / 7 - #symbol 2#'
print( re.sub(r'#[^\s#](?:[^#]*[^\s#])?#', r"func('\g<0>')", text) )
# => func('#symbol 1#') / 7 - func('#symbol 2#')
#[\w\s]+# or #[^#]+# or #[^\s#](?:[^#]*[^\s#])?#. It depends on what the rule for what can be in between the delimiting #s. Please add the requirement to the question, I will fix the regex accordingly.