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python - 使用正则表达式匹配成绩单中的名称、对话和动作

问题描述

给定一个如下所示的字符串对话，我需要找到与每个用户对应的句子。

text = 'CHRIS: Hello, how are you...
PETER: Great, you? PAM: He is resting.
[PAM SHOWS THE COUCH]
[PETER IS NODDING HIS HEAD]
CHRIS: Are you ok?'

对于上述对话，我想返回包含三个元素的元组：

  1. 人名

  2. 小写的句子和

  3. 括号内的句子

像这样的东西：

('CHRIS', 'Hello, how are you...', None)

('PETER', 'Great, you?', None)

('PAM', 'He is resting', 'PAM SHOWS THE COUCH. PETER IS NODDING HIS HEAD')

('CHRIS', 'Are you ok?', None)

etc...

我正在尝试使用正则表达式来实现上述目的。到目前为止，我能够使用以下代码获取用户的姓名。我正在努力识别两个用户之间的句子。

actors = re.findall(r'\w+(?=\s*:[^/])',text)

标签： pythonregexstring

解决方案

你可以这样做re.findall：

>>> re.findall(r'\b(\S+):([^:\[\]]+?)\n?(\[[^:]+?\]\n?)?(?=\b\S+:|$)', text)
[('CHRIS', ' Hello, how are you...', ''),
 ('PETER', ' Great, you? ', ''),
 ('PAM',
  ' He is resting.',
  '[PAM SHOWS THE COUCH]\n[PETER IS NODDING HIS HEAD]\n'),
 ('CHRIS', ' Are you ok?', '')]

您将必须弄清楚如何自己删除方括号，而在尝试匹配所有内容的同时，正则表达式无法做到这一点。

正则表达式分解 

\b              # Word boundary
(\S+)           # First capture group, string of characters not having a space
:               # Colon
(               # Second capture group
    [^          # Match anything that is not...
        :       #     a colon
        \[\]    #     or square braces
    ]+?         # Non-greedy match
)
\n?             # Optional newline
(               # Third capture group
    \[          # Literal opening brace
    [^:]+?      # Similar to above - exclude colon from match
    \] 
    \n?         # Optional newlines
)?              # Third capture group is optional
(?=             # Lookahead for... 
    \b          #     a word boundary, followed by  
    \S+         #     one or more non-space chars, and
    :           #     a colon
    |           # Or,
    $           # EOL
)

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