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

注意,difflib。SequenceMatcher只找到最长的连续匹配子序列,这通常不是我们想要的,例如:

>>> a1 = "Apple"
>>> a2 = "Appel"
>>> a1 *= 50
>>> a2 *= 50
>>> SequenceMatcher(None, a1, a2).ratio()
0.012  # very low
>>> SequenceMatcher(None, a1, a2).get_matching_blocks()
[Match(a=0, b=0, size=3), Match(a=250, b=250, size=0)]  # only the first block is recorded

寻找两个字符串之间的相似性与生物信息学中成对序列比对的概念密切相关。有许多专门的库,包括生物马拉松。这个例子实现了Needleman Wunsch算法:

>>> from Bio.Align import PairwiseAligner
>>> aligner = PairwiseAligner()
>>> aligner.score(a1, a2)
200.0
>>> aligner.algorithm
'Needleman-Wunsch'

使用biopython或其他生物信息学包比python标准库的任何部分都更灵活,因为有许多不同的评分方案和算法可用。此外,你可以得到匹配的序列来可视化正在发生的事情:

>>> alignment = next(aligner.align(a1, a2))
>>> alignment.score
200.0
>>> print(alignment)
Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-Apple-
|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-|||-|-
App-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-elApp-el

其他回答

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

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

BLEUscore

BLEU,即双语评估替补,是一个用于比较的分数 文本到一个或多个参考译文的候选翻译。 完全匹配的结果是1.0,而完全不匹配的结果是1.0 结果得分为0.0。 虽然它是为翻译而开发的,但也可以用来评估文本 为一套自然语言处理任务生成。

代码:

import nltk
from nltk.translate import bleu
from nltk.translate.bleu_score import SmoothingFunction
smoothie = SmoothingFunction().method4

C1='Text'
C2='Best'

print('BLEUscore:',bleu([C1], C2, smoothing_function=smoothie))

示例:通过更新C1和C2。

C1='Test' C2='Test'

BLEUscore: 1.0

C1='Test' C2='Best'

BLEUscore: 0.2326589746035907

C1='Test' C2='Text'

BLEUscore: 0.2866227639866161

你也可以比较句子的相似度:

C1='It is tough.' C2='It is rough.'

BLEUscore: 0.7348889200874658

C1='It is tough.' C2='It is tough.'

BLEUscore: 1.0

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])