Python:如何使用regex将句子拆分为新行，然后使用空格将标点符号和单词分开？

Question

我有以下意见：

input = "I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"

首先，每句话都应该移到一个新的行。然后，除"/“、”‘“、"-”、"+“和"$”外，所有标点符号都应与单词分开。

因此，产出应该是：

"I love programming with Python-3 . 3 ! 
Do you ?  
It's great . . . 
I give it a 10/10 . 
It's free-to-use , no $$$ involved !"

我使用了以下代码：

>>> import re
>>> re.sub(r"([\w/'+$\s-]+|[^\w/'+$\s-]+)\s*", r"\1 ", input)
"I love programming with Python-3 . 3 ! Do you ? It's great ... I give it a 10/10 . It's free-    to-use , no $$$ involved ! "

但问题是它并没有把句子分成新的行。在创建标点符号和字符之间的空白之前，我如何使用正则表达式来实现这一点？

Answer 1

([!?.])(?=\s*[A-Z])\s*

您可以使用此正则表达式在regex.See demo.Replace by \1\n之前创建句子。

https://regex101.com/r/sH8aR8/5

x="I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"
print re.sub(r"([!?.])(?=\s*[A-Z])",r"\1\n",x)

编辑：

(?<![A-Z][a-z])([!?.])(?=\s*[A-Z])\s*

尝试不同数据集的this.See演示。

https://regex101.com/r/sH8aR8/9

Answer 2

有点像

>>> import re
>>> from string import punctuation
>>> print re.sub(r'(?<=['+punctuation+'])\s+(?=[A-Z])', '\n', input)
I love programming with Python-3.3!
Do you?
It's great...
I give it a 10/10.
It's free-to-use, no $$$ involved!