我认为这可以从前视或后视引用中受益。python引用是 https://docs.python.org/3/library/re.html ，我经常引用的一个通用正则表达式站点是 https://www.regular-expressions.info/lookaround.html 。

您的数据：

words = ["don't",
         "'George",
         "ma'am",
         "end.'",
         "didn't.'",
         "'Won't",]

现在，我将使用正则表达式和它们的替代品来定义一个元组。

In [230]: apo = (
    (re.compile("(?<=[A-Za-z])'(?=[A-Za-z])"), "<apostrophe>",),
    (re.compile("(?<![A-Za-z])'(?=[A-Za-z])"), "<opensingle>",),
    (re.compile("(?<=[.A-Za-z])'(?![A-Za-z])"), "<closesingle>", ),
    (re.compile("(?<=[A-Za-z])\\.(?![A-Za-z])"), "<period>",),
     ...:      ...:      ...:      ...:      ...:      ...: 
In [231]: words = ["don't",
         "'George",
         "ma'am",
         "end.'",
         "didn't.'",
         "'Won't",]
     ...:      ...:      ...:      ...:      ...:      ...: 
In [232]: reduce(lambda w2,x: [ x[0].sub(x[1], w) for w in w2], apo, words)
Out[232]: 
['don<apostrophe>t',
 '<opensingle>George',
 'ma<apostrophe>am',
 'end<period><closesingle>',
 'didn<apostrophe>t<period><closesingle>',
 '<opensingle>Won<apostrophe>t']

下面是正则表达式的情况：

1. (?<=[A-Za-z]) 表示仅匹配(但不消耗)前面的字符是letter.
2. (?=[A-Za-z]) 是前视(仍然不消耗)如果后面的字符是letter.
3. (?<![A-Za-z]) 是负向后视，这意味着如果前面有一个字母，则它不会match.
4. (?![A-Za-z]) 是负向前视。

请注意，我在 <closesingle> 中添加了一个 . 检查，并且 apo 中的顺序很重要，因为您可能会用 <period> 替换 . ...

这是对单个单词的操作，但也应该适用于句子。

In [233]: onelong = """
don't
'George
ma'am
end.'
didn't.'
'Won't
     ...:      ...:      ...:      ...:      ...:      ...:      ...: 
In [235]: print(
    reduce(lambda sentence,x: x[0].sub(x[1], sentence), apo, onelong)
     ...:      ...: