按换行符和大写字母的正则表达式拆分

Question

我一直在努力用 Python 中的正则表达式拆分我的字符串。

我有一个要加载的文本文件，格式为：

"Peter went to the gym; \nhe worked out for two hours \nKyle ate lunch 
 at Kate's house. Kyle went home at 9. \nSome other sentence 
 here\n\u2022Here's a bulleted line"

我想得到以下输出：

['Peter went to the gym; he worked out for two hours','Kyle ate lunch 
at Kate's house. He went home at 9.', 'Some other sentence here', 
'\u2022Here's a bulleted line']

我希望用 Python 中的新行和大写字母或项目符号来分割我的字符串。

我已经尝试解决问题的前半部分，只用一个新行和大写字母来分割我的字符串。

这是我目前所拥有的：

print re.findall(r'\n[A-Z][a-z]+',str,re.M)

这只是给了我：

[u'\nKyle', u'\nSome']

这只是第一个词。我已经尝试过该正则表达式的变体，但我不知道如何获得该行的其余部分。

我假设也按项目符号拆分，我将只包含一个 OR 正则表达式，其格式与按大写字母拆分的正则表达式相同。这是最好的方法吗？

我希望这是有道理的，如果我的问题仍然不清楚，我很抱歉。 :)

Answer 1

你可以使用这个split函数：

>>> str = u"Peter went to the gym; \nhe worked out for two hours \nKyle ate lunch at Kate's house. Kyle went home at 9. \nSome other sentence here\n\u2022Here's a bulleted line"
>>> print re.split(u'\n(?=\u2022|[A-Z])', str)

[u'Peter went to the gym; \nhe worked out for two hours ',
 u"Kyle ate lunch at Kate's house. Kyle went home at 9. ",
 u'Some other sentence here',
 u"\u2022Here's a bulleted line"]

Code Demo

Answer 2

您可以在\n 处拆分，以大写字母或项目符号字符开头：

import re
s = """
Peter went to the gym; \nhe worked out for two hours \nKyle ate lunch 
at Kate's house. Kyle went home at 9. \nSome other sentence 
here\n\u2022Here's a bulleted line
"""
new_list = filter(None, re.split('\n(?=•)|\n(?=[A-Z])', s))

输出：

['Peter went to the gym; \nhe worked out for two hours ', "Kyle ate lunch \nat Kate's house. Kyle went home at 9. ", 'Some other sentence \nhere', "•Here's a bulleted line\n"]

或者，不使用项目符号字符的符号：

new_list = filter(None, re.split('\n(?=\u2022)|\n(?=[A-Z])', s))