用于拆分具有多个捕获组的单词列表的 RegEx

Question

我有以下字符串:

one two three four five six seven eight nine

我正在尝试构建一个正则表达式，将字符串分为三组:

第 1 组:“一二三”

第 2 组:“四五六”

第 3 组:“七八九”

我尝试过 (.*\b(one|two|three)?)(.*\b(four|five|six)?)(.*\b(seven|eight|nine)?) 的变体，但这种模式将完整匹配分为一组包含完整字符串 - the demo can be found here 。

尝试 (.*\b(one|two|three))(.*\b(four|five|six))(.*\b(seven|eight|nine)) 似乎让我更接近我想要的东西，但匹配信息面板显示该模式标识了两个匹配项，每个匹配项包含六个捕获组。

我使用 OR 语句是因为组可以是任意长度，例如 two three four ，将模式应用于此字符串应识别两组 -

第 1 组:“两个”

第 2 组:“三四”。

Answer 1

一个大的正则表达式可能会这样做

(?=.*\b(?:one|two|three|four|five|six|seven|eight|nine)\b)(\b(?:one|two|three)(?:\s+(?:one|two|three))*\b)?.+?(\b(?:four|five|six)(?:\s+(?:four|five|six))*\b)?.+?(\b(?:seven|eight|nine)(?:\s+(?:seven|eight|nine))*\b)?

https://regex101.com/r/rUtkyU/1

可读版本

 (?=
      .* \b 
      (?:
           one
        |  two
        |  three
        |  four
        |  five
        |  six
        |  seven
        |  eight
        |  nine
      )
      \b 
 )
 (                             # (1 start)
      \b   
      (?: one | two | three )

      (?:
           \s+ 
           (?: one | two | three )
      )*
      \b 
 )?                            # (1 end)

 .+? 
 (                             # (2 start)
      \b        
      (?: four | five | six )

      (?:
           \s+ 
           (?: four | five | six )
      )*
      \b     
 )?                            # (2 end)

 .+?   
 (                             # (3 start)
      \b          
      (?: seven | eight | nine )

      (?:
           \s+ 
           (?: seven | eight | nine )
      )*
      \b   
 )?                            # (3 end)