从字符串创建两个列表，排除和包括括号之间的字符串

Question

假设我们有这样一个字符串：

s = u'apple banana lemmon (hahaha) dog cat whale (hehehe) red blue black'

我想创建以下列表：

including = ['hahaha', 'hehehe']
excluding = ['apple banana lemmon (', ') dog cat whale (', ') red blue black']

第一个列表直接使用正则表达式：

including = re.findall('\((.*?)\)',s)

但我无法为其他列表获得类似的东西。你可以帮帮我吗？提前谢谢你！

Answer 1

使用正则表达式

a = re.findall('\)?[^()]*\(?', s)
excluded = a[::2]
included = a[1::2]
print(included, excluded, sep='\n')

['hahaha', 'hehehe', '']
['apple banana lemmon (', ') dog cat whale (', ') red blue black']

注意空字符串

a = re.findall('\)?[^()]*\(?', s)
excluded = [*filter(bool, a[::2])]
included = [*filter(bool, a[1::2])]
print(included, excluded, sep='\n')

['hahaha', 'hehehe']
['apple banana lemmon (', ') dog cat whale (', ') red blue black']

---

没有正则表达式

from itertools import cycle

def f(s):
  c = cycle('()')
  a = {'(': 1, ')': 0}
  while s:
    d = next(c)
    i = s.find(d)
    if i > -1:
      j = a[d]
      yield d, s[:i + j]
      s = s[i + j:]
    else:
      yield d, s
      break

included = []
excluded = []

for k, v in f(s):
  if k == '(':
    excluded.append(v)
  else:
    included.append(v)

print(included, excluded, sep='\n')

['hahaha', 'hehehe']
['apple banana lemmon (', ') dog cat whale (', ') red blue black']

---

同样的想法没有覆盖s

from itertools import cycle

def f(s):
  c = cycle('()')
  a = {'(': 1, ')': 0}
  j = 0
  while True:
    d = next(c)
    i = s.find(d, j)
    if i > -1:
      k = a[d]
      yield d, s[j:i + k]
      j = i + k
    else:
      yield d, s[j:]
      break

included = []
excluded = []

for k, v in f(s):
  if k == '(':
    excluded.append(v)
  else:
    included.append(v)

print(included, excluded, sep='\n')

['hahaha', 'hehehe']
['apple banana lemmon (', ') dog cat whale (', ') red blue black']

Answer 2

您可以使用正向后视和正向前瞻来拆分括号之间的单词：

>>> re.split(r'(?<=\().*?(?=\))', s)
['apple banana lemmon (', ') dog cat whale (', ') red blue black']

Answer 3

excluding = re.split('|'.join(including), s)

对于您知道包含信息不包含特殊字符或正则表达式定义的简单情况。

如果您不确定是否会出现这种情况：

re.split('|'.join(map(re.escape, including)), s)

这将转义特殊的正则表达式字符，否则会导致 re.split 函数功能障碍