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4kyu_Most_frequently_used_words_in_a_text.py
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4kyu_Most_frequently_used_words_in_a_text.py
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"""
Write a function that, given a string of text (possibly with punctuation and line-breaks), returns an array of the top-3 most occurring words, in descending order of the number of occurrences.
Assumptions:
A word is a string of letters (A to Z) optionally containing one or more apostrophes (') in ASCII. (No need to handle fancy punctuation.)
Matches should be case-insensitive, and the words in the result should be lowercased.
Ties may be broken arbitrarily.
If a text contains fewer than three unique words, then either the top-2 or top-1 words should be returned, or an empty array if a text contains no words.
Examples:
top_3_words("In a village of La Mancha, the name of which I have no desire to call to
mind, there lived not long since one of those gentlemen that keep a lance
in the lance-rack, an old buckler, a lean hack, and a greyhound for
coursing. An olla of rather more beef than mutton, a salad on most
nights, scraps on Saturdays, lentils on Fridays, and a pigeon or so extra
on Sundays, made away with three-quarters of his income.")
# => ["a", "of", "on"]
top_3_words("e e e e DDD ddd DdD: ddd ddd aa aA Aa, bb cc cC e e e")
# => ["e", "ddd", "aa"]
top_3_words(" //wont won't won't")
# => ["won't", "wont"]
Bonus points (not really, but just for fun):
Avoid creating an array whose memory footprint is roughly as big as the input text.
Avoid sorting the entire array of unique words.
"""
def top_3_words(text):
for i in text:
if (not i.isalpha()) and (i != "'"):
text = text.replace(i, " ")
while " ''" in text: text = text.replace(" ''", " ")
if " ' " in text: text = text.replace(" ' ", " ")
lst = text.lower().split()
if not lst:
return []
most = max(set(lst), key=lst.count)
if len(set(lst)) < 2:
return [most]
second = max(set(lst)-set([most]), key=lst.count)
if len(set(lst)) < 3:
return [most, second]
third = max(set(lst)-set([most, second]), key=lst.count)
return [most, second, third]