正则表达式在 Python 中拆分单词

Question

我正在设计一个正则表达式来拆分给定文本中的所有实际单词：

输入示例：

"John's mom went there, but he wasn't there. So she said: 'Where are you'"

预期输出：

["John's", "mom", "went", "there", "but", "he", "wasn't", "there", "So", "she", "said", "Where", "are", "you"]

我想到了这样的正则表达式：

"(([^a-zA-Z]+')|('[^a-zA-Z]+))|([^a-zA-Z']+)"

在 Python 中拆分后，结果包含None 项和空格。

如何去掉 None 项？为什么空格不匹配？

编辑：
在空格上拆分，将给出如下项目：["there."]
并且拆分非字母，将给出如下项目：["John","s"]
除了' 之外的非字母拆分，将给出如下项目：["'Where","you'"]

Answer 1

您可以使用字符串函数来代替正则表达式：

to_be_removed = ".,:!" # all characters to be removed
s = "John's mom went there, but he wasn't there. So she said: 'Where are you!!'"

for c in to_be_removed:
    s = s.replace(c, '')
s.split()

但是，在您的示例中，您不想删除 John's 中的撇号，但您希望删除 you!!' 中的撇号。所以字符串操作在那个时候失败了，你需要一个微调的正则表达式。

编辑：可能一个简单的正则表达式可以解决您的问题：

(\w[\w']*)

它将捕获所有以字母开头的字符并继续捕获，而下一个字符是撇号或字母。

(\w[\w']*\w)

第二个正则表达式是针对一个非常特殊的情况......第一个正则表达式可以捕获像you' 这样的词。这将避免这种情况，并且仅在 is 在单词内（不在开头或结尾）时才捕获撇号。但是在那一点上，出现了一种情况，您无法使用第二个正则表达式捕获撇号Moss' mom。您必须决定是否在以 s 结尾的名称中捕获尾随撇号并定义所有权。

例子：

rgx = re.compile("([\w][\w']*\w)")
s = "John's mom went there, but he wasn't there. So she said: 'Where are you!!'"
rgx.findall(s)

["John's", 'mom', 'went', 'there', 'but', 'he', "wasn't", 'there', 'So', 'she', 'said', 'Where', 'are', 'you']

更新 2：我在我的正则表达式中发现了一个错误！它不能捕获单个字母后跟像A' 这样的撇号。修复了全新的正则表达式：

(\w[\w']*\w|\w)

rgx = re.compile("(\w[\w']*\w|\w)")
s = "John's mom went there, but he wasn't there. So she said: 'Where are you!!' 'A a'"
rgx.findall(s)

["John's", 'mom', 'went', 'there', 'but', 'he', "wasn't", 'there', 'So', 'she', 'said', 'Where', 'are', 'you', 'A', 'a']

Answer 2

你的正则表达式中有太多的捕获组；使它们不被捕获：

(?:(?:[^a-zA-Z]+')|(?:'[^a-zA-Z]+))|(?:[^a-zA-Z']+)

演示：

>>> import re
>>> s = "John's mom went there, but he wasn't there. So she said: 'Where are you!!'"
>>> re.split("(?:(?:[^a-zA-Z]+')|(?:'[^a-zA-Z]+))|(?:[^a-zA-Z']+)", s)
["John's", 'mom', 'went', 'there', 'but', 'he', "wasn't", 'there', 'So', 'she', 'said', 'Where', 'are', 'you', '']

只返回 一个 为空的元素。

Answer 3

这个正则表达式只允许一个结束撇号，后面可以再跟一个字符：

([\w][\w]*'?\w?)

演示：

>>> import re
>>> s = "John's mom went there, but he wasn't there. So she said: 'Where are you!!' 'A a'"
>>> re.compile("([\w][\w]*'?\w?)").findall(s)
["John's", 'mom', 'went', 'there', 'but', 'he', "wasn't", 'there', 'So', 'she', 'said', 'Where', 'are', 'you', 'A', "a'"]

Answer 4

我是 python 新手，但我想我已经弄明白了

import re
s = "John's mom went there, but he wasn't there. So she said: 'Where are you!!'"
result = re.findall(r"(.+?)[\s'\",!]{1,}", s)
print(result)

结果 ['John', 's', 'mom', 'went', 'there', 'but', 'he', 'wasn', 't', 'there.', 'So', 'she', '说：'，'在哪里'，'在'，'你']