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Refactor word_occurrence to use dict#15186

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Refactor word_occurrence to use dict #15186

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refactor: simplify word_occurrence using dict
Miladkhoshdel Sep 4, 2026
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[pre-commit.ci] auto fixes from pre-commit.com hooks
pre-commit-ci[bot] Sep 4, 2026
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13 changes: 6 additions & 7 deletions strings/word_occurrence.py
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Original file line numberDiff line numberDiff line change
@@ -1,23 +1,22 @@
# Created by sarathkaul on 17/11/19
# Modified by Arkadip Bhattacharya(@darkmatter18) on 20/04/2020
from collections import defaultdict


def word_occurrence(sentence: str) -> dict:
def word_occurrence(sentence: str) -> dict[str, int]:
"""
>>> from collections import Counter
>>> SENTENCE = "a b A b c b d b d e f e g e h e i e j e 0"
>>> occurence_dict = word_occurrence(SENTENCE)
>>> all(occurence_dict[word] == count for word, count
>>> occurrence_dict = word_occurrence(SENTENCE)
>>> all(occurrence_dict[word] == count for word, count
... in Counter(SENTENCE.split()).items())
True
>>> dict(word_occurrence("Two spaces"))
{'Two': 1, 'spaces': 1}
"""
occurrence: defaultdict[str, int] = defaultdict(int)
# Creating a dictionary containing count of each word
occurrence: dict[str, int] = {}

for word in sentence.split():
occurrence[word] += 1
occurrence[word] = occurrence.get(word, 0) + 1

@cclausscclaussSep 4, 2026

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Python's default_dict is in the standard library to accelerate use cases like this one. There is a function call overhead in the proposed solution that does not exist in the current solution.

Create a https://docs.python.org/3/library/timeit.html#timeit.timeit benchmark and compare the two on a big dataset.

@MiladkhoshdelMiladkhoshdelSep 5, 2026

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Thanks for the feedback. I’ve kept the existing defaultdict implementation unchanged and moved the type-hint and typo fixes into a separate, focused PR: #15194.

return occurrence


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