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4c1f876567
* Solving the `Top k most frequent words` problem using a max-heap
* Mentioning Python standard library solution in `Top k most frequent words` docstring
* ruff --fix .
* updating DIRECTORY.md
---------
Co-authored-by: Amos Paribocci <aparibocci@gmail.com>
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
102 lines
3.1 KiB
Python
102 lines
3.1 KiB
Python
"""
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Finds the top K most frequent words from the provided word list.
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This implementation aims to show how to solve the problem using the Heap class
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already present in this repository.
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Computing order statistics is, in fact, a typical usage of heaps.
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This is mostly shown for educational purposes, since the problem can be solved
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in a few lines using collections.Counter from the Python standard library:
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from collections import Counter
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def top_k_frequent_words(words, k_value):
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return [x[0] for x in Counter(words).most_common(k_value)]
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"""
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from collections import Counter
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from functools import total_ordering
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from data_structures.heap.heap import Heap
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@total_ordering
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class WordCount:
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def __init__(self, word: str, count: int) -> None:
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self.word = word
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self.count = count
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def __eq__(self, other: object) -> bool:
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"""
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>>> WordCount('a', 1).__eq__(WordCount('b', 1))
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True
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>>> WordCount('a', 1).__eq__(WordCount('a', 1))
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True
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>>> WordCount('a', 1).__eq__(WordCount('a', 2))
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False
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>>> WordCount('a', 1).__eq__(WordCount('b', 2))
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False
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>>> WordCount('a', 1).__eq__(1)
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NotImplemented
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"""
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if not isinstance(other, WordCount):
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return NotImplemented
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return self.count == other.count
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def __lt__(self, other: object) -> bool:
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"""
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>>> WordCount('a', 1).__lt__(WordCount('b', 1))
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False
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>>> WordCount('a', 1).__lt__(WordCount('a', 1))
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False
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>>> WordCount('a', 1).__lt__(WordCount('a', 2))
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True
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>>> WordCount('a', 1).__lt__(WordCount('b', 2))
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True
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>>> WordCount('a', 2).__lt__(WordCount('a', 1))
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False
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>>> WordCount('a', 2).__lt__(WordCount('b', 1))
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False
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>>> WordCount('a', 1).__lt__(1)
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NotImplemented
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"""
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if not isinstance(other, WordCount):
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return NotImplemented
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return self.count < other.count
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def top_k_frequent_words(words: list[str], k_value: int) -> list[str]:
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"""
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Returns the `k_value` most frequently occurring words,
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in non-increasing order of occurrence.
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In this context, a word is defined as an element in the provided list.
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In case `k_value` is greater than the number of distinct words, a value of k equal
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to the number of distinct words will be considered, instead.
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>>> top_k_frequent_words(['a', 'b', 'c', 'a', 'c', 'c'], 3)
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['c', 'a', 'b']
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>>> top_k_frequent_words(['a', 'b', 'c', 'a', 'c', 'c'], 2)
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['c', 'a']
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>>> top_k_frequent_words(['a', 'b', 'c', 'a', 'c', 'c'], 1)
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['c']
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>>> top_k_frequent_words(['a', 'b', 'c', 'a', 'c', 'c'], 0)
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[]
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>>> top_k_frequent_words([], 1)
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[]
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>>> top_k_frequent_words(['a', 'a'], 2)
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['a']
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"""
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heap: Heap[WordCount] = Heap()
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count_by_word = Counter(words)
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heap.build_max_heap(
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[WordCount(word, count) for word, count in count_by_word.items()]
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)
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return [heap.extract_max().word for _ in range(min(k_value, len(count_by_word)))]
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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