I have a file with such data:
Sentence[0].Sentence[1].Sentence[2].'/n'
Sentence[0].Sentence[1].Sentence[2].'/n'
Sentence[0].Sentence[1].Sentence[2].'/n'
What I want to print out are all Sentences0. This is what I have done, but it prints out a blank list.
from nltk import *
import codecs
f=codecs.open('topon.txt','r+','cp1251')
text = f.readlines()
first=[sentence for sentence in text if re.findall('\.\n^Abc',sentence)]
print first
You don't need NLTK for this (nor are you using it). Unless I misunderstand the question, this should do the trick:
with open('topon.txt') as infile:
for line in infile:
print line.split('.', 1)[0]
In addition to @inspectorG4dget 's answer, you can do it by regexes:
from nltk import *
import codecs
f = codecs.open('a.txt', 'r+', 'cp1251')
text = f.readlines()
print [re.findall('^[^.]+', sentence) for sentence in text]
Splitting a paragraph at periods works only if every sentence ends with a period, and periods are used for nothing else. If you have a lot of real text, neither of these is even close to true. Abbreviations, questions? exclamations! etc. will trip you up a lot. So, use the tool that the nltk provides for this purpose: the function sent_tokenize(). It's not perfect, but it's a whole lot better than looking for periods. If text is your list of paragraphs, you use it like this:
first = [ ]
for par in text:
sentences = nltk.sent_tokenize(par)
first.append(sentences[0])
You could fold the above into a list comprehension, but it's not going to be very readable...
Abcwhen you say you want the first of the three sentences? Please clarify your task.