需要 Python Regex 多模式匹配和反模式匹配

尽管您的要求有点模糊，但考虑到您的特定输入字符串，合理的尝试似乎是在任何以文字.开头且后跟文字(letter)模式的空格上进行拆分。

import re

s = "(a) be given by XXX for its election per scenario 3.3(a). (b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially"

print(re.split(r"(?<=\.) (?=\([a-z]\))", s))

输出：

['(a) be given by XXX for its election per scenario 3.3(a).', 
 '(b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially']

我会谨慎地在大型或复杂的输入上使用它，因为误报的可能性很高。

另一个想法：如果您保证提取字母表中的每个字母，确保每个字母最终都会按顺序显示，并且想要将任何乱序视为正常内容，您可以尝试构建一个庞大的正则表达式：

import re
from string import ascii_lowercase

s = "(a) be given by XXX for its election per scenario 3.3(a). (b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially. (c) blah blah  (c) blah blah (d) asd ad(a) (b) (e) ee (b) (a) (d) (f) (f) fff f ff (g) (a) gggg (h) hhhh (b) (i) iii i i (i) i (j) jjj (k) k (l) ll (a) (b) (x) (m) mm (n) nn (o) oo) () () (p) ppp (A) (B) (Q) (q) qq (r) rr (s) ss (t) tt( u ) (u) uu (v) vvv (ww) (w) ww (x) xx (y) yy (z) zzz"

pattern = "".join([f"((?: |^)\({l}\) .+)" for l in ascii_lowercase])

for result in re.findall(pattern, s)[0]:
    print(result.strip())

输出：

(a) be given by XXX for its election per scenario 3.3(a).
(b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially. (c) blah blah
(c) blah blah
(d) asd ad(a) (b)
(e) ee (b) (a) (d) (f)
(f) fff f ff
(g) (a) gggg
(h) hhhh (b) (i) iii i i
(i) i
(j) jjj
(k) k
(l) ll (a) (b) (x)
(m) mm
(n) nn
(o) oo) () ()
(p) ppp (A) (B) (Q)
(q) qq
(r) rr
(s) ss
(t) tt( u )
(u) uu
(v) vvv (ww)
(w) ww
(x) xx
(y) yy
(z) zzz

这仍然对输入做出了一些全面的假设，但可能值得一试；将其视为概念证明。

换行符是另一个需要考虑的问题，如果存在的话（以及许多其他事情）。长话短说，手工编写解析器可能比正则表达式更好。