python - 如何在 python 中使用正则表达式忽略以特定模式开头的字符串？

Question

接受并返回@something 但拒绝 first@last。

r'@([A-Z][A-Z0-9_]*[A-Z0-9])

上面的正则表达式将接受@something(以字母开头，以字母或数字结尾，中间可以有下划线，至少2个字符长)并返回@之后的部分。符号。

我不想返回包含一些字母或数字的字符串 A-Z0-9在 @ 之前符号。

@ 之前的空格、换行符、特殊字符等是允许的。

代码:

re.findall(r'@([A-Z][A-Z0-9_]*[A-Z0-9])', text, re.I)

Answer 1

使用

re.findall(r'(?<![A-Z0-9])@([A-Z][A-Z0-9_]*[A-Z0-9])', text, re.I)

参见 regex proof .

解释

--------------------------------------------------------------------------------
  (?<!                     look behind to see if there is not:
--------------------------------------------------------------------------------
    [A-Z0-9]                 any character of: 'A' to 'Z', '0' to '9'
--------------------------------------------------------------------------------
  )                        end of look-behind
--------------------------------------------------------------------------------
  @                        '@'
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    [A-Z]                    any character of: 'A' to 'Z'
--------------------------------------------------------------------------------
    [A-Z0-9_]*               any character of: 'A' to 'Z', '0' to
                             '9', '_' (0 or more times (matching the
                             most amount possible))
--------------------------------------------------------------------------------
    [A-Z0-9]                 any character of: 'A' to 'Z', '0' to '9'
--------------------------------------------------------------------------------
  )                        end of \1