How to match a paragraph using regex

Question

I have been struggling with python regex for a while trying to match paragraphs within a text, but I haven\'t been successful. I need to obtain the start and end positions o

Answer 1

What is the newline symbol? Let us suppose the newline symbol is '\r\n', if you want to match the paragraphs starting with Lorem, you can do like this:

pattern = re.compile('\r\nLorem.*\r\n')
str = '...'    # your source text
matchlist = re.findall(pattern, str)

The matchlist will contain all the paragragh start with Lorem. And the other two words are the same.

Answer 2

i tried to use the recommended RegEx with the default Java RegEx engine. That gave me several times a StackOverflowException, so in the end i rewrote the RegEx and optimized it a little more.

So this is working fine for me in Java:

(?s)(.*?[^\:\-\,])(?:$|\n{2,})

This also handles the end of document without new lines and tries to concat lines which ends with ':', '-' or ',' to the next paragraph.

And to avoid that trailing blanks (whitespace or tabs) breaks the above described feature i am stripping them before with following regex:

(?m)[[:blank:]]+$

Answer 3

Using split is one way, you can do so with regular expression also like this:

paragraphs = re.search('(.+?\n\n|.+?$)',TEXT,re.DOTALL)

The .+? is a lazy match, it will match the shortest substring that makes the whole regex matched. Otherwise, it will just match the whole string.

So basically here we want to find a sequence of characters (.+?) which ends by a blank line (\n\n) or the end of string ($). The re.DOTALL flag makes the dot to match newline also (we also want to match a paragraph consisting of three lines without blank lines within)

Answer 4

Try

^(.+?)\n\s*\n

or

^(.+?)\r\n\s*\r\n

just do not forget append extra new line at the end of text

Answer 5

You can split on double-newline like this:

paragraphs = re.split(r"\n\n", DATA)

Edit: To capture the paragraphs as matches, so you can get their start and end points, do this:

for match in re.finditer(r'(?s)((?:[^\n][\n]?)+)', DATA):
   print match.start(), match.end()

# Prints:
# 0 214
# 215 298
# 299 589