Python match at least 3 words in a set

Question

I have str with a phrase, for example:

phrase = "My cat has two eyes and like to catch rats"

And I have a set of words that I would like to matche at least 3 of theses words in the phrase.

words = set(["eyes", "like", "cat"])

Currently I have the following code

found = bool(set(phrase.lower().split()) & words)

But it matches if any of the words are in the phrase, and I want at least 3 words matching.

What I can do to achieve this? I don't want to use regex .

Answer 1

You can check if the length of the intersection is at least 3.

found = len(set(phrase.lower().split()).intersection(words)) >= 3

Answer 2

You can do something like the following:

from typing import Set


def words_matcher(phrase: str, words: Set[str], threshold: int = 3) -> bool:
    phrase_as_set = set(phrase.lower().split())
    common_words = phrase_as_set.intersection(words)
    return len(common_words) >= threshold

Answer 3

You are almost there. & performs intersection on the set objects. But instead of doing bool , you need to get the length and check whether it is >=3. Hence use this:

>>> phrase = "My cat has two eyes and like to catch rats"
>>> words = set(["eyes", "like", "cat"])

>>> len(set(phrase.lower().split()) & words) >= 3
True

Answer 4

If you want to check if all words in your set appear in phrase you might check if it is subset ie

phrase = "My cat has two eyes and like to catch rats"
words = set(["eyes", "like", "cat"])
print(words.issubset(phrase.lower().split()))  # True