如何在Python中获得一个字符串与另一个字符串相似的概率?

我想要得到一个十进制值,比如0.9(意思是90%)等等。最好是标准的Python和库。

e.g.

similar("Apple","Appel") #would have a high prob.

similar("Apple","Mango") #would have a lower prob.

当前回答

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."))

其他回答

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])

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

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."))

内置的SequenceMatcher在大输入时非常慢,下面是如何用diff-match-patch完成的:

from diff_match_patch import diff_match_patch

def compute_similarity_and_diff(text1, text2):
    dmp = diff_match_patch()
    dmp.Diff_Timeout = 0.0
    diff = dmp.diff_main(text1, text2, False)

    # similarity
    common_text = sum([len(txt) for op, txt in diff if op == 0])
    text_length = max(len(text1), len(text2))
    sim = common_text / text_length

    return sim, diff

你可以创建这样一个函数:

def similar(w1, w2):
    w1 = w1 + ' ' * (len(w2) - len(w1))
    w2 = w2 + ' ' * (len(w1) - len(w2))
    return sum(1 if i == j else 0 for i, j in zip(w1, w2)) / float(len(w1))