字符串的Sklearn余弦相似度，Python

Question

I am writing an algorithm that checks how much a string is equal to another string. I am using Sklearn cosine similarity.

My code is:

from sklearn.feature_extraction.text import TfidfVectorizer
from sklearn.metrics.pairwise import cosine_similarity

example_1 = ("I am okey", "I am okeu")
example_2 = ("I am okey", "I am crazy")

tfidf_vectorizer = TfidfVectorizer()
tfidf_matrix = tfidf_vectorizer.fit_transform(example_1)
result_cos = cosine_similarity(tfidf_matrix[0:1], tfidf_matrix)
print(result_cos[0][1])

Running this code for example_1, prints 0.336096927276. Running it for example_2, it prints the same score. The result is the same in both cases because there is only one different word.

What I want is to get a higher score for example_1 because the different words "okey vs okeu" have only one different letter. In contrast in example_2 there are two completely different words "okey vs crazy".

How can my code take in consideration that in some cases the different words are not completely different?

Answer 1

For short strings, Levenshtein distance will probably yield better results than cosine similarity based on words. The algorithm below is adapted from Wikibooks . Since this is a distance metric, smaller score is better.

def levenshtein(s1, s2):
    if len(s1) < len(s2):
        s1, s2 = s2, s1

    if len(s2) == 0:
        return len(s1)

    previous_row = range(len(s2) + 1)
    for i, c1 in enumerate(s1):
        current_row = [i + 1]
        for j, c2 in enumerate(s2):
            insertions = previous_row[j + 1] + 1
            deletions = current_row[j] + 1
            substitutions = previous_row[j] + (c1 != c2)
            current_row.append(min(insertions, deletions, substitutions))
        previous_row = current_row

    return previous_row[-1]/float(len(s1))

example_1 = ("I am okey", "I am okeu")
example_2 = ("I am okey", "I am crazy")

print(levenshtein(*example_1))
print(levenshtein(*example_2))