使用 Python 正则表达式按换行符或句点划分字符串
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Divide string by line break or period with Python regular expressions
提问by David Y. Stephenson
I have a string:
我有一个字符串：
"""Hello. It's good to meet you.
My name is Bob."""
I'm trying to find the best way to split this into a list divided by periods and linebreaks:
我正在尝试找到将其拆分为按句点和换行符划分的列表的最佳方法：
["Hello", "It's good to meet you", "My name is Bob"]
I'm pretty sure I should use regular expressions, but, having no experience with them, I'm struggling to figure out how to do this.
我很确定我应该使用正则表达式，但是，由于没有使用它们的经验，我正在努力弄清楚如何做到这一点。
采纳答案by falsetru
You don't need regex.
你不需要正则表达式。
>>> txt = """Hello. It's good to meet you.
... My name is Bob."""
>>> txt.split('.')
['Hello', " It's good to meet you", '\nMy name is Bob', '']
>>> [x for x in map(str.strip, txt.split('.')) if x]
['Hello', "It's good to meet you", 'My name is Bob']
回答by zhangyangyu
>>> s = """Hello. It's good to meet you.
... My name is Bob."""
>>> import re
>>> p = re.compile(r'[^\s\.][^\.\n]+')
>>> p.findall(s)
['Hello', "It's good to meet you", 'My name is Bob']
>>> s = "Hello. #It's good to meet you # .'"
>>> p.findall(s)
['Hello', "#It's good to meet you # "]
回答by Tim Pietzcker
For your example, it would suffice to split on dots, optionally followed by whitespace (and to ignore empty results):
对于您的示例，在点上拆分就足够了，可以选择后跟空格（并忽略空结果）：
>>> s = """Hello. It's good to meet you.
... My name is Bob."""
>>> import re
>>> re.split(r"\.\s*", s)
['Hello', "It's good to meet you", 'My name is Bob', '']
In real life, you'd have to handle Mr. Orange, Dr. Greeneand George W. Bush, though...
在现实生活中，您必须处理Mr. Orange,Dr. Greene和George W. Bush，但是...
回答by Casimir et Hippolyte
You can use this split
您可以使用此拆分
re.split(r"(?<!^)\s*[.\n]+\s*(?!$)", s)
回答by eyquem
Mine:
矿：
re.findall('(?=\S)[^.\n]+(?<=\S)',su)
