Get rid of stopwords and punctuation

Cappai picture Cappai · Apr 4, 2011 · Viewed 33.4k times · Source

I'm struggling with NLTK stopword.

Here's my bit of code.. Could someone tell me what's wrong?

from nltk.corpus import stopwords

def removeStopwords( palabras ):
     return [ word for word in palabras if word not in stopwords.words('spanish') ]

palabras = ''' my text is here '''

Answer

JHSaunders picture JHSaunders · Apr 4, 2011

Your problem is that the iterator for a string returns each character not each word.

For example:

>>> palabras = "Buenos dias"
>>> [c for c in palabras]
['B', 'u', 'e', 'n', 'a', 's', ' ', 'd', 'i', 'a', 's']

You need to iterate and check each word, fortunately the split function already exists in the python standard library under the string module. However you are dealing with natural language including punctuation you should look here for a more robust answer that uses the re module.

Once you have a list of words you should lowercase them all before comparison and then compare them in the manner that you have shown already.

Buena suerte.

EDIT 1

Okay try this code, it should work for you. It shows two ways to do it, they are essentially identical but the first is a bit clearer while the second is more pythonic.

import re
from nltk.corpus import stopwords

scentence = 'El problema del matrimonio es que se acaba todas las noches despues de hacer el amor, y hay que volver a reconstruirlo todas las mananas antes del desayuno.'

#We only want to work with lowercase for the comparisons
scentence = scentence.lower() 

#remove punctuation and split into seperate words
words = re.findall(r'\w+', scentence,flags = re.UNICODE | re.LOCALE) 

#This is the simple way to remove stop words
important_words=[]
for word in words:
    if word not in stopwords.words('spanish'):
        important_words.append(word)

print important_words

#This is the more pythonic way
important_words = filter(lambda x: x not in stopwords.words('spanish'), words)

print important_words 

I hope this helps you.