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)
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
You can use
\B@([A-Z][A-Z0-9_]*[A-Z0-9])
The pattern matches:
\B
Assert a position where a word boundary does not match @
Match literally (
Capture group 1
[AZ][A-Z0-9_]*[A-Z0-9]
)
Close group 1 import re
text = "Accept and return @something but reject first@last."
print(re.findall(r'\B@([A-Z][A-Z0-9_]*[A-Z0-9])', text, re.I))
Output
['something']
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