Accept and return @something but reject first@last.
r'@([A-Z][A-Z0-9_]*[A-Z0-9])
The above regexp will accept @something (starts with letter, ends with letter or number, may have underscore in middle, atleast 2 characters long) and returns the part after the @
symbol.
I do not want to return strings which contain some letters or number A-Z0-9
before the @
symbol.
Spaces, new lines, special characters, etc before @
is allowed.
CODE:
re.findall(r'@([A-Z][A-Z0-9_]*[A-Z0-9])', text, re.I)
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Answer
Use
re.findall(r'(?<![A-Z0-9])@([A-Z][A-Z0-9_]*[A-Z0-9])', text, re.I)
See regex proof.
EXPLANATION
-------------------------------------------------------------------------------- (?<! 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