Python Regular Expression Matching: ## ##

Question

I\'m searching a file line by line for the occurrence of ##random_string##. It works except for the case of multiple #...

pattern=\'##(.*?)##\'
prog=re.compile(p         


        

Answer 1

Try the "block comment trick": /##((?:[^#]|#[^#])+?)##/

Answer 2

Your problem is with your inner match. You use ., which matches any character that isn't a line end, and that means it matches # as well. So when it gets ###hey##, it matches (.*?) to #hey.

The easy solution is to exclude the # character from the matchable set:

prog = re.compile(r'##([^#]*)##')

Protip: Use raw strings (e.g. r'') for regular expressions so you don't have to go crazy with backslash escapes.

Trying to allow # inside the hashes will make things much more complicated.

EDIT: If you do not want to allow blank inner text (i.e. "####" shouldn't match with an inner text of ""), then change it to:

prog = re.compile(r'##([^#]+)##')

+ means "one or more."

Answer 3

>>> import re
>>> text= 'lala ###hey## there'
>>> matcher= re.compile(r"##[^#]+##")
>>> print matcher.sub("FOUND", text)
lala #FOUND there
>>>

Answer 4

To match at least two hashes at either end:

pattern='##+(.*?)##+'

Answer 5

have you considered doing it non-regex way?

>>> string='lala ####hey## there'
>>> string.split("####")[1].split("#")[0]
'hey'

Answer 6

'^#{2,}([^#]*)#{2,}' -- any number of # >= 2 on either end

be careful with using lazy quantifiers like (.*?) because it'd match '##abc#####' and capture 'abc###'. also lazy quantifiers are very slow