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
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))
其他回答
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
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))
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])
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."))
