Question

任何人都知道字符串相似性方法会给出下面的正确结果吗？我正在处理字母数字ID，其中：

弦的早期部分的变化比后者更重要。我想我可以做ngrams？虽然在一个字符串有前缀的情况下可能会出现问题？
将“a”改为“b”时，改变“a”到“b”的角色的差异很重要，而不是将其改为“c”。

Levenstein和Jaro-Winkler似乎没有做正确的事。

见下面的例子。

import jellyfish
t1="100"
t21=["100a","a100"] # case 1. expecting: similar, not similar
t22=["101","105","200"] # case 2. expecting: similar, less similar, least similar

fun = jellyfish.levenshtein_distance
print([fun(t1, t) for t in t21]) # all the same
print([fun(t1, t) for t in t22]) # all the same

fun = jellyfish.jaro_winkler
print([fun(t1, t) for t in t21]) # all the same
print([fun(t1, t) for t in t22]) # all the same

为了增加乐趣，第一个字符串的前缀基本上与作为ID的字符串无关，但会混淆字符串相似性。

t1="pre-100"
t21=["100a","a100"] # expecting: similar, not similar
t22=["101","105","200"] # expecting: similar, less similar, least similar

fun = jellyfish.levenshtein_distance
print([fun(t1, t) for t in t21]) # picks the wrong one
print([fun(t1, t) for t in t22]) # all the same

fun = jellyfish.jaro_winkler
print([fun(t1, t) for t in t21]) # picks the wrong one
print([fun(t1, t) for t in t22]) # picks the right one

字符串相似性，其中ascii代码的顺序和差异很重要

0 个答案: