I have the following text (just an example): </i>5 <i></i><span class
I'd like to remove this space, so I tried:
re.sub(r'</i>.* <i></i><span class', '</i>%02d<i></i><span class' %, text)
But this did not work. How can I catch the "thing" which is found in ".*"? %02d is obviously incorrect...
Thanks for the help :)
You can use a capturing group:
re.sub(r'</i>(.*) <i></i><span class', r'</i>\1<i></i><span class', text)
This capturing group, (.*), captures the "5", and it is placed in the \1 in the replacement text. Note the presence of r before the second string: that tells Python it's a raw string (see here for more details)
As David mentioned, a capturing group is what you need. To elaborate further:
Round brackets capture whatever it is that they match to. This is called a 'capturing group', and a 'backreference' is created to whatever is caught. Each subsequent backreference can be referred to by \1. So:
(.)b\1
matches 'aba' and 'mnm', but not 'abc'.
Similarly,
(.)(.)b\1\2
matches 'abbab', 'xybxy'
and
(.)(.)b\2\1
matches 'abbba', 'xybyx'
This can be then used to check a palindrome (Not that it's advised, regex's cannot match palindromes of unspecified length limits):
(.?)(.)(.)\3?\2\1
is a regex which will match a palindrome of length 3 or less.
