Word counts in Python using regular expression

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Solution 1

Using \w+ won't correctly count words containing apostrophes or hyphens, eg "can't" will be counted as 2 words. It will also count numbers (strings of digits); "12,345" and "6.7" will each count as 2 words ("12" and "345", "6" and "7").

Solution 2

This seems to work as expected.

>>> import re
>>> words=re.findall('\w+', open('/usr/share/dict/words').read().lower())
>>> len(words)
234936
>>> 
bash-3.2$ wc /usr/share/dict/words
  234936  234936 2486813 /usr/share/dict/words

Why are you lowercasing your words? What does that have to do with the count?

I'd submit that the following would be more efficient:

words=re.findall(r'\w+', open('/usr/share/dict/words').read())
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Zhe Li
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Zhe Li

Playing with Scala, Python, Kubernetes and stuff.

Updated on June 29, 2022

Comments

  • Zhe Li
    Zhe Li almost 2 years

    What is the correct way to count English words in a document using regular expression?

    I tried with:

    words=re.findall('\w+', open('text.txt').read().lower())
    len(words)
    

    but it seems I am missing few words (compares to the word count in gedit). Am I doing it right?

    Thanks a lot!