从大字符串中提取子字符串

Question

我有一个字符串：

string="(2021-07-02 01:00:00 AM BST)  
---  
syl.hs has joined the conversation  
  
  

(2021-07-02 01:00:23 AM BST)  
---  
e.wang  
Good Morning
How're you?
  
  
  

(2021-07-02 01:05:11 AM BST)  
---  
wk.wang  
Hi, I'm Good.  
  
  

(2021-07-02 01:08:01 AM BST)  
---  
perter.derrek   
we got the update on work. 
It will get complete by next week.

(2021-07-15 08:59:41 PM BST)  
---  
ad.ft has left the conversation  
  
  
  
  
---  
  
* * *"

我只想提取对话文本（名称和时间戳之间的文本）预期输出为：

cmets=['早上好，你好吗？'，'嗨，我很好。'，'我们得到了 工作更新。它将在下周完成。']

我试过的是：

cmets=re.findall(r'---\s*\n(.(?:\n(?!(?:(\s\d{4}-\d {2}-\d{2}\s\d{2}:\d{2}:\d{2}\s*[AP]M\s+GMT\s*)\w+\s*\n )?---).))',string)

Answer 1

您可以使用单个捕获组：

^---\s*\n(?!.* has (?:joined|left) the conversation|\* \* \*)\S.*((?:\n(?!\(\d|---).*)*)

模式匹配：

• ^ 字符串开始
• ---\s*\n 匹配 --- 可选的空白字符和换行符
• (?!.* has (?:joined|left) the conversation|\* \* \*) 断言该行不包含 has joined 或 has left 对话部分，或包含 * * *
• \S.* 在行首和行的其余部分至少匹配一个非空白字符
• ( Capture group 1（这将由 re.findall 返回）
  • (?:\n(?!\(\d|---).*)* 匹配所有不以 ( 开头的行和一个数字或 --
• )关闭第一组

查看regex demo 和Python demo。

例子

pattern = r"^---\s*\n(?!.* has (?:joined|left) the conversation|\* \* \*)\S.*((?:\n(?!\(\d|---).*)*)"
result = [m.strip() for m in re.findall(pattern, s, re.M) if m]
print(result)

输出

["Good Morning\nHow're you?", "Hi, I'm Good.", 'we got the update on work. \nIt will get complete by next week.']

Answer 2

我假设：

• 感兴趣的文本在三行块之后开始：一行包含时间戳，然后是 "---" 行，它可以用空格填充到右侧，然后是由一串字母组成的行，其中包含一个既不在该字符串的开头也不在结尾的句点，并且该字符串可以在右侧用空格填充。
• 感兴趣的文本块可能包含空行，空行是只包含空格和行终止符的字符串。
• 感兴趣的文本块的最后一行不能为空行。

我相信以下正则表达式（设置了多行 (m) 和不区分大小写 (i) 标志）满足这些要求。

^\(\d{4}\-\d{2}\-\d{2} .*\) *\r?\n-{3} *\r?\n[a-z]+\.[a-z]+ *\r?\n((?:.*[^ (\n].*\r?\n| *\r?\n(?=(?: *\r?\n)*(?!\(\d{4}\-\d{2}\-\d{2} .*\)).*[^ (\n]))*)

感兴趣的线块包含在捕获组 1 中。

Start your engine!

表达式的元素如下。

^\(\d{4}\-\d{2}\-\d{2} .*\) *\r?\n  # match timestamp line
-{3} *\r?\n                         # match 3-hyphen line
[a-z]+\.[a-z]+ *\r?\n               # match name
(                                   # begin capture group 1
  (?:                               # begin non-capture group (a)
    .*[^ (\n].*\r?\n                # match a non-blank line
    |                               # or
    \ *\r?\n                        # match a blank line
    (?=                             # begin a positive lookahead
      (?:                           # begin non-capture group (b)
        \ *\r?\n                    # match a blank line
      )*                            # end non-capture group b and execute 0+ times
      (?!                           # begin a negative lookahead
        \(\d{4}\-\d{2}\-\d{2} .*\)  # match timestamp line
      )                             # end negative lookahead
      .*[^ (\n]                     # march a non-blank line
    )                               # end positive lookahead
  )*                                # end non-capture group a and execute 0+ times
)                                   # end capture group 1

Answer 3

这是一个自我记录的正则表达式，它将去除前导和尾随空格：

(?x)(?m)(?s)                                                    # re.X, re.M, re.S (DOTALL)
(?:                                                             # start of non capturing group
 ^\(\d{4}-\d{2}-\d{2}\ \d{2}:\d{2}:\d{2}\ [AP]M\ BST\)\s*\r?\n  # date and time
 (?!---\s*\r?\nad\.ft has)                                      # next lines are not the ---\n\ad.ft etc.
 ---\s*\r?\n                                                    # --- line
 [\w.]+\s*\r?\n                                                 # name line
 \s*                                                            # skip leading whitespace
)                                                               # end of non-capture group
# The folowing is capture group 1. Match characters until you get to the next date-time:
((?:(?!\s*\r?\n\(\d{4}-\d{2}-\d{2}\ \d{2}:\d{2}:\d{2}\ [AP]M\ BST\)).)*)# skip trailing whitespace

See Regex Demo

See Python Demo

import re

string = """(2021-07-02 01:00:00 AM BST)
---
syl.hs has joined the conversation



(2021-07-02 01:00:23 AM BST)
---
e.wang
Good Morning
How're you?




(2021-07-02 01:05:11 AM BST)
---
wk.wang
Hi, I'm Good.



(2021-07-02 01:08:01 AM BST)
---
perter.derrek
we got the update on work.
It will get complete by next week.

(2021-07-15 08:59:41 PM BST)
---
ad.ft has left the conversation




---

* * *"""

regex = r'''(?x)(?m)(?s)                                        # re.X, re.M, re.S (DOTALL)
(?:                                                             # start of non capturing group
 ^\(\d{4}-\d{2}-\d{2}\ \d{2}:\d{2}:\d{2}\ [AP]M\ BST\)\s*\r?\n  # date and time
 (?!---\s*\r?\nad\.ft has)                                      # next lines are not the ---\n\ad.ft etc.
 ---\s*\r?\n                                                    # --- line
 [\w.]+\s*\r?\n                                                 # name line
 \s*                                                            # skip leading whitespace
)                                                               # end of non-capture group
# The folowing is capture group 1. Match characters until you get to the next date-time:
((?:(?!\s*\r?\n\(\d{4}-\d{2}-\d{2}\ \d{2}:\d{2}:\d{2}\ [AP]M\ BST\)).)*)# skip trailing whitespace
'''

matches = re.findall(regex, string)
print(matches)

打印：

["Good Morning\nHow're you?", "Hi, I'm Good.", 'we got the update on work.\nIt will get complete by next week.']