Python 正则表达式：将大文本文件拆分为较小的部分

Question

考虑以下文本文件。

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| First Block of text   |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

----------------------- Monday 8 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| Second Block of text  |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

----------------------- Friday 12 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| 3rd Block of text     |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~
 
----------------------- Friday 19 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| 4th Block of text     |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

如何提取第二个、第三个和第四个块并根据上面给出的日期保存它们？例如，我需要提取

中的所有行

     ~~~~~~~~~~~~~~~~~~~~~~~
    |                       |
    | Second Block of text  |
    |                       |
     ~~~~~~~~~~~~~~~~~~~~~~~

然后将其保存到名称为 Monday 8 August 2021 的文件或变量中。

使用以下正则表达式，我可以找到包含日期的行：https://regex101.com/r/nKW1W4/1

-(?P<date>.*?)-

Answer 1

您可以使用以下表达式匹配您的块并使用第一组作为文件名：

^
-+([^-]+)-+$
(.+?(?=^--|\Z))

见a demo on regex101.com（注意修饰符）。

Answer 2

你可以使用：

input_text = """
 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| First Block of text   |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

----------------------- Monday 8 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| Second Block of text  |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

----------------------- Friday 12 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| 3rd Block of text     |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~
 
----------------------- Friday 19 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| 4th Block of text     |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~
"""

a = re.split(r'-+(.*?)-+', a)

for k, v in enumerate(a):
    a[k] = a[k].strip()

print(a)

列出理解哪个更简洁suggested by @fsimonjetz

input_text = """
 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| First Block of text   |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

----------------------- Monday 8 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| Second Block of text  |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~

----------------------- Friday 12 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| 3rd Block of text     |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~
 
----------------------- Friday 19 August 2021 -----------------------

 ~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| 4th Block of text     |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~
"""

result = [x.strip() for x in re.split(r'-+(.*?)-+', input_text)]

Answer 3

在您的模式中，您只在左侧和右侧匹配一个 -，而 .*? 匹配 0+ 个字符而不是换行符非贪婪。

这会给你很多部分匹配而不是匹配整行。

---

您也可以使用匹配，并将捕获组 1 用作文件名，将捕获组 2 用作数据。

^-+([^-]+)-+((?:\n(?!--).*)*)

说明

• ^ 字符串开始
• -+匹配1+次-
• ([^-]+) 捕获 group 1 作为日期部分，匹配除- 之外的所有字符
• -+匹配1+次-
• ( 为数据部分捕获 group 2
  • (?:\n(?!--).*)* 例如匹配所有不以-- 开头的行
• )关闭第二组

Regex demo

例如

import re

pattern = r"^-+([^-]+)-+((?:\n(?!--).*)*)"

s = (" ~~~~~~~~~~~~~~~~~~~~~~~\n"
    "|                      |\n"
    "| First Block of text   |\n"
    "|                      |\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n\n"
    "----------------------- Monday 8 August 2021 -----------------------\n\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n"
    "|                      |\n"
    "| Second Block of text  |\n"
    "|                      |\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n\n"
    "----------------------- Friday 12 August 2021 -----------------------\n\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n"
    "|                      |\n"
    "| 3rd Block of text     |\n"
    "|                      |\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n"
    " \n"
    "----------------------- Friday 19 August 2021 -----------------------\n\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n"
    "|                      |\n"
    "| 4th Block of text     |\n"
    "|                      |\n"
    " ~~~~~~~~~~~~~~~~~~~~~~~\n")

matches = re.findall(pattern, s, re.M)
if matches:
    filename = matches[0][0].strip();
    data = matches[0][1].strip();
    
    print(filename)
    print(data)

输出

Monday 8 August 2021
~~~~~~~~~~~~~~~~~~~~~~~
|                       |
| Second Block of text  |
|                       |
 ~~~~~~~~~~~~~~~~~~~~~~~