I have been trying to teach myself Regexes in python and I decided to print out all the sentences of a text. I have been tinkering with the regular expressions for the past 3 hours to no avail.

I just tried the following but couldn't do anything.

p = open('anan.txt')
process = p.read()
regexMatch = re.findall('^[A-Z].+\s+[.!?]$',process,re.I)
print regexMatch
p.close()

My input file is like this:

OMG is this a question ! Is this a sentence ? My.
name is.

This prints no outputs. But when I remove "My. name is.", it prints OMG is this a question and Is this a sentence together as if it only reads the first line.

What is the best solution of regex that can find all sentences in a text file - regardless if the sentence carries to new line or so - and also 开发者_StackOverflow中文版reads the entire text? Thanks.

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Something like this works:

## pattern: Upercase, then anything that is not in (.!?), then one of them
>>> pat = re.compile(r'([A-Z][^\.!?]*[\.!?])', re.M)
>>> pat.findall('OMG is this a question ! Is this a sentence ? My. name is.')
['OMG is this a question !', 'Is this a sentence ?', 'My.']

Notice how name is. is not in the result because it does not start with a uppercase letter.

Your problem comes from the use of the ^$ anchors, they work on the whole text.

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There are two issues in your regex:

1. Your expression is anchored by ^ and $, which are the "start of line" and "end of line" anchors, respectively. That means that your pattern is looking to match an entire line of your text.
2. You are searching for \s+ before your punctuation character, which specifies one or more whitespace character. If you don't have whitespace before your punctuation, the expression will not match.

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Edited: now it will work with multiline sentences too.

>>> t = "OMG is this a question ! Is this a sentence ? My\n name is."
>>> re.findall("[A-Z].*?[\.!?]", t, re.MULTILINE | re.DOTALL )
['OMG is this a question !', 'Is this a sentence ?', 'My\n name is.']

Only one thing left to explain - re.DOTALL makes . match newline as described here

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Thank you cji and Jochen Ritzel.

sentence=re.compile("[A-Z].*?[\.!?] ", re.MULTILINE | re.DOTALL )

I think this is the best, just add a space at the end.

 SampleReport='I image from 08/25 through 12. The patient image 1.2, 23, 34, 45 and 64 from serise 34. image look good to have a tumor in this area.  It has been resected during the interval between scans.  The'

if use

pat = re.compile(r'([A-Z][^\.!?]*[\.!?])', re.M)
pat.findall(SampleReport)

The result will be:

['I image from 08/25 through 12.',
'The patient image 1.',
 'It has been resected during the interval between scans.']

The bug is it can't handle digit like 1.2. But this one works perfectly.

sentence.findall(SampleReport)

Result

['I image from 08/25 through 12. ',
'The patient image 1.2, 23, 34, 45 and 64 from serise 34. ',
 'It has been resected during the interval between scans. ']

---

Try the other way around: Split the text at sentence boundaries.

lines = re.split(r'\s*[!?.]\s*', text)

If that doesn't work, add a \ before the ..

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You can try:

p = open('a')
process = p.read()
print process
regexMatch = re.findall('[^.!?]+[.!?]',process)
print regexMatch
p.close()

The regex used here is [^.!?]+[.!?] which tries to match one or more non-sentence delimiter followed by a sentence delimiter.

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I tried on Notepad++, and I got this :

.*$

And activate the multiline option :

re.MULTILINE

Cheers