还添加了Spacy NLP库;

@profile
def main():
    str1= "Mar 31 09:08:41  The world is beautiful"
    str2= "Mar 31 19:08:42  Beautiful is the world"
    print("NLP Similarity=",nlp(str1).similarity(nlp(str2)))
    print("Diff lib similarity",SequenceMatcher(None, str1, str2).ratio()) 
    print("Jellyfish lib similarity",jellyfish.jaro_distance(str1, str2))

if __name__ == '__main__':

    #python3 -m spacy download en_core_web_sm
    #nlp = spacy.load("en_core_web_sm")
    nlp = spacy.load("en_core_web_md")
    main()

使用Robert Kern的line_profiler运行

kernprof -l -v ./python/loganalysis/testspacy.py

NLP Similarity= 0.9999999821467294
Diff lib similarity 0.5897435897435898
Jellyfish lib similarity 0.8561253561253562

然而，时间的启示

Function: main at line 32

Line #      Hits         Time  Per Hit   % Time  Line Contents
==============================================================
    32                                           @profile
    33                                           def main():
    34         1          1.0      1.0      0.0      str1= "Mar 31 09:08:41  The world is beautiful"
    35         1          0.0      0.0      0.0      str2= "Mar 31 19:08:42  Beautiful is the world"
    36         1      43248.0  43248.0     99.1      print("NLP Similarity=",nlp(str1).similarity(nlp(str2)))
    37         1        375.0    375.0      0.9      print("Diff lib similarity",SequenceMatcher(None, str1, str2).ratio()) 
    38         1         30.0     30.0      0.1      print("Jellyfish lib similarity",jellyfish.jaro_distance(str1, str2))

出于我的目的，我有自己的quick_ratio()，它比difflib SequenceMatcher的quick_ratio()快2倍，同时提供类似的结果。A和b是字符串:

    score = 0
    for letters in enumerate(a):
        score = score + b.count(letters[1])

我想你们可能在寻找一种描述字符串之间距离的算法。这里有一些你可以参考的:

汉明距离 Levenshtein距离 Damerau-Levenshtein距离 Jaro-Winkler距离

这是内置的。

from difflib import SequenceMatcher

def similar(a, b):
    return SequenceMatcher(None, a, b).ratio()

使用它:

>>> similar("Apple","Appel")
0.8
>>> similar("Apple","Mango")
0.0

包装距离包括Levenshtein距离:

import distance
distance.levenshtein("lenvestein", "levenshtein")
# 3

这是我想到的:

import string

def match(a,b):
    a,b = a.lower(), b.lower()
    error = 0
    for i in string.ascii_lowercase:
            error += abs(a.count(i) - b.count(i))
    total = len(a) + len(b)
    return (total-error)/total

if __name__ == "__main__":
    print(match("pple inc", "Apple Inc."))