具有自定义匹配功能的Python序列匹配器

您自己的实现

在情况1)中,您可以查看this question
并阅读了一些书籍,例如CLRS或Robert Sedgewick的书籍.

破解difflib模块

在情况2)中,请查看source code：get_matching_blocks在line 479处调用find_longest_match.在find_longest_match的核心中,您具有b2j字典,该字典将列表a的元素映射到列表b中的索引.如果覆盖此词典,则可以实现所需的功能.这是标准版本：

>>> import difflib
>>> from difflib import SequenceMatcher
>>> list3 = ["orange","grape"]
>>> list4 = ["pears","grapes"]
>>> s = SequenceMatcher(None,list3,list4)
>>> s.get_matching_blocks()
[Match(a=1,b=2,size=1),Match(a=4,b=6,size=0)]
>>> [(b.a+i,b.b+i,list3[b.a+i],list4[b.b+i]) for b in s.get_matching_blocks() for i in range(b.size)]
[(1,'apple')]

这是被黑的版本：

>>> s = SequenceMatcher(None,list4)
>>> s.b2j
{'pears': [0],'oranges': [1],'apple': [2],'lemon': [3],'cherry': [4],'grapes': [5]}
>>> s.b2j = {**s.b2j,'orange':s.b2j['oranges'],'lemons':s.b2j['lemon'],'grape':s.b2j['grapes']}
>>> s.b2j
{'pears': [0],'grapes': [5],'orange': [1],'lemons': [3],'grape': [5]}
>>> s.get_matching_blocks()
[Match(a=0,b=1,size=3),Match(a=3,b=5,list4[b.b+i]) for b in s.get_matching_blocks() for i in range(b.size)]
[(0,'oranges'),(1,'apple'),(2,'lemon'),(3,'grape','grapes')]

这并不难实现自动化,但是我不建议您使用该解决方案,因为有一个非常简单的解决方法.

解决方法

想法是按家庭对单词进行分组：

families = [{"pears","peras"},{"orange","naranjas"},{"apple","manzana"},{"lemons","limón"},{"cherry","cereza"},{"grape","grapes"}]

现在,很容易创建一个字典,将家庭中的每个单词映射到其中一个单词(我们称其为主要单词)：

>>> d = {w:main for main,*alternatives in map(list,families) for w in alternatives}
>>> d
{'pears': 'peras','orange': 'naranjas','oranges': 'naranjas','manzana': 'apple','lemon': 'lemons','limón': 'lemons','cherry': 'cereza','grape': 'grapes'}

请注意,map(list,family)中的main *替代词使用star运算符将家庭分解成一个主词(列表的第一个)和备选列表：

>>> head,*tail = [1,4,5]
>>> head
1
>>> tail
[2,5]

然后,您可以将列表转换为仅使用主要词：

>>> list3=["orange","grape"]
>>> list4=["pears","grapes"]
>>> list5=["peras","uvas"]
>>> [d.get(w,w) for w in list3]
['naranjas','limón','grapes']
>>> [d.get(w,w) for w in list4]
['peras','naranjas','cereza',w) for w in list5]
['peras','uvas']

如果w是键,则表达式d.get(w,w)将返回d [w],否则返回w本身.因此,属于一个家庭的单词将转换为该家庭的主要单词,而其他单词则保持不变.

这些列表很容易与difflib进行比较.

重要提示：与diff算法相比,列表转换的时间复杂度可忽略不计,因此您不会看到任何区别.

完整代码

另外,完整代码如下：

def match_seq(list1,list2):
    """A generator that yields matches of list1 vs list2"""
    s = SequenceMatcher(None,list2)
    for block in s.get_matching_blocks():
        for i in range(block.size):
            yield block.a + i,block.b + i # you don't need to store the matches,just yields them

def create_convert(*families):
    """Return a converter function that converts a list
    to the same list with only main words"""
    d = {w:main for main,families) for w in alternatives}
    return lambda L: [d.get(w,w) for w in L]

families = [{"pears","grapes","uvas"}]
convert = create_convert(*families)

list3=["orange","uvas"]

print ("list3 vs list4")
for a,b in match_seq(convert(list3),convert(list4)):
    print(a,list3[a],list4[b])

#  list3 vs list4
# 0 1 orange oranges
# 1 2 apple apple
# 2 3 lemons lemon
# 3 5 grape grapes

print ("list3 vs list5")
for a,convert(list5)):
    print(a,list5[b])

# list3 vs list5
# 0 1 orange naranjas
# 1 2 apple manzana
# 2 3 lemons limón
# 3 5 grape uvas