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 text.split() if word not in (stopwords.words('english'))]) 

I am processing 6 mil of such strings so speed is important. Profiling my code, the slowest part is the lines above, is there a better way to do this? I'm thinking of using something like regex's re.sub but I don't know how to write the pattern for a set of words. Can someone give me a hand and I'm also happy to hear other possibly faster methods.

Note: I tried someone's suggest of wrapping stopwords.words('english') with set() but that made no difference.

Thank you.

Answer 1

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 2

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.