Faster way to remove stop words in Python

Question

I am trying to remove stopwords from a string of text:

from nltk.corpus import stopwords
text = \'hello bye the the hi\'
text = \' \'.join([word for word in          


        

Answer 1

Use a regexp to remove all words which do not match:

import re
pattern = re.compile(r'\b(' + r'|'.join(stopwords.words('english')) + r')\b\s*')
text = pattern.sub('', text)

This will probably be way faster than looping yourself, especially for large input strings.

If the last word in the text gets deleted by this, you may have trailing whitespace. I propose to handle this separately.

Answer 2

Try caching the stopwords object, as shown below. Constructing this each time you call the function seems to be the bottleneck.

    from nltk.corpus import stopwords

    cachedStopWords = stopwords.words("english")

    def testFuncOld():
        text = 'hello bye the the hi'
        text = ' '.join([word for word in text.split() if word not in stopwords.words("english")])

    def testFuncNew():
        text = 'hello bye the the hi'
        text = ' '.join([word for word in text.split() if word not in cachedStopWords])

    if __name__ == "__main__":
        for i in xrange(10000):
            testFuncOld()
            testFuncNew()

I ran this through the profiler: python -m cProfile -s cumulative test.py. The relevant lines are posted below.

nCalls Cumulative Time

10000 7.723 words.py:7(testFuncOld)

10000 0.140 words.py:11(testFuncNew)

So, caching the stopwords instance gives a ~70x speedup.

Answer 3

First, you're creating stop words for each string. Create it once. Set would be great here indeed.

forbidden_words = set(stopwords.words('english'))

Later, get rid of [] inside join. Use generator instead.

' '.join([x for x in ['a', 'b', 'c']])

replace to

' '.join(x for x in ['a', 'b', 'c'])

Next thing to deal with would be to make .split() yield values instead of returning an array. I believe regex would be good replacement here. See thist hread for why s.split() is actually fast.

Lastly, do such a job in parallel (removing stop words in 6m strings). That is a whole different topic.

Answer 4

Sorry for late reply. Would prove useful for new users.

• Create a dictionary of stopwords using collections library

• Use that dictionary for very fast search (time = O(1)) rather than doing it on list (time = O(stopwords))

  from collections import Counter
  stop_words = stopwords.words('english')
  stopwords_dict = Counter(stop_words)
  text = ' '.join([word for word in text.split() if word not in stopwords_dict])