正则表达式在python中管理“-”

Question

我正在寻找适合ready-to-work，ready-to-work-today，ready-to的正则表达式。我一直以这种方式放置多个选项：

text=''' ready-to-work ready-to ready-to-work-today'''

re.findall(r'\b[a-zA-Z0-9]+?-[a-zA-Z0-9]+?-[a-zA-Z0-9]+?\b', text)

我设法找到了最多三个术语的表达式，例如['ready-to-work']，但是我无法捕获ready-to和ready-to-work-today。我希望['ready-to-work','ready-to','ready-to-work-today']。

感谢您的帮助。

Answer 1

尝试使用此正则表达式\b\w+(-\w+)+\b。

Answer 2

如果要将连字符和字符类的部分匹配1-3次，则可以将非捕获组与量词{1,3}配合使用。要匹配1次以上，请使用+。

\b[a-zA-Z0-9]+(?:-[a-zA-Z0-9]+){1,3}\b

• \b字边界
• [a-zA-Z0-9]+匹配列出的1次以上任何时间
• (?:非捕获组
  • -[a-zA-Z0-9]+匹配-，并且是列出的任意项的1倍以上
• ){1,3}关闭非捕获组并重复1-3次
• \b字边界

Regex demo

Answer 3

我猜测此表达式可能返回所需的输出：

\b(?:\w+-\w+(?:[\w-])*)\b

使用re.findall

进行测试

import re

regex = r"\b(?:\w+-\w+(?:[\w-])*)\b"

test_str = "ready-to-work ready-to-work-today ready-to ready-to-work ready-to ready-to-work-today ready-to ready-to-work-todayready-to ready-to-work-today"

print(re.findall(regex, test_str))

输出

['ready-to-work', 'ready-to-work-today', 'ready-to', 'ready-to-work', 'ready-to', 'ready-to-work-today', 'ready-to', 'ready-to-work-todayready-to', 'ready-to-work-today']

演示

如果要浏览/简化/修改该表达式，请在this demo的右上角进行解释。