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nvogorov
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Nov 22, 2023
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| for word in string.lower().split(): | ||
| if word not in self.uniquewords: | ||
| self.uniquewords.append(word) |
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Здесь получается довольно высокая сложность О(N^2)
Можно сделать оптимальнее, если хранить уникальные слова не в массиве, а в какой-то другой структуре
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| for word in self.uniquewords: | ||
| result.append(string.lower().count(word)) |
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Второй раз приводим в нижний регистр те же данные
К тому же, тут может быть баг, если мы передадим к примеру строку "bla blablabla", то в результате получим [4, 1] так как bla найдется 4 раза
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Реализация класса CountVectorizer, имеющего