I know that for parsing I should ideally remove all spaces and linebreaks but I was just doing this as a quick fix for something I was trying and I can't figure out why its not working.. I have wrapped different areas of text in my document with the wrappers like "####1" and am trying to parse based on this but its just not working no matter what I try, I think I am using multiline correctly.. any advice is appreciated
This returns no results at all:
string='
####1
ttteest
####1
ttttteeeestt
####2
ttest
####2'
import re
pattern = '.*?####(.*?)####'
returnmatch = re.compile(pattern, re.MULTILINE).findall(string)
return returnmatch
Pattern. MULTILINE or (? m) tells Java to accept the anchors ^ and $ to match at the start and end of each line (otherwise they only match at the start/end of the entire string). Pattern.
To match a character having special meaning in regex, you need to use a escape sequence prefix with a backslash ( \ ). E.g., \. matches "." ; regex \+ matches "+" ; and regex \( matches "(" . You also need to use regex \\ to match "\" (back-slash).
A Regular Expression (RegEx) is a sequence of characters that defines a search pattern. For example, ^a...s$ The above code defines a RegEx pattern. The pattern is: any five letter string starting with a and ending with s . A pattern defined using RegEx can be used to match against a string.
Multiline doesn't mean .
will match line return, it means that ^
and $
are limited to lines only
re.M re.MULTILINE
When specified, the pattern character '^' matches at the beginning of the string and at the >beginning of each line (immediately following each newline); and the pattern character '$' >matches at the end of the string and at the end of each line (immediately preceding each >newline). By default, '^' matches only at the beginning of the string, and '$' only at the >end of the string and immediately before the newline (if any) at the end of the string.
re.S
or re.DOTALL
makes .
match even new lines.
Source
http://docs.python.org/
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