Question

r'(^|^A)(\S+)(B$|$)'

results to matches everything, which actually equals to ^\S$.

How to write one matches “begins with A or ends with B, may both but not neither?”

PS: I also need refer to group (\S+) in the substring module.

Example:

Match Aanything, anythingB, and refer anything group in the replace.

Answer 1

Problem solved.

I use this regex in python, I found this in the Python manual:

(?(id/name)yes-pattern|no-pattern)
Will try to match with yes-pattern if
the group with given id or name
exists, and with no-pattern if it
doesn’t. no-pattern is optional and
can be omitted. For example,
(<)?(\w+@\w+(?:.\w+)+)(?(1)>) is a
poor email matching pattern, which
will match with ” as
well as ‘[e-mail]’, but not with
‘

New in version 2.4.

So my final answer is:

r'(?P<prefix>A)?(?P<key>\S+)(?(prefix)|B)'

Commands:

>>>re.sub(r'(?P<prefix>A)?(?P<key>\S+)(?(prefix)|B)','\g<key>',"Aanything")

‘anything’

>>>re.sub(r'(?P<prefix>A)?(?P<key>\S+)(?(prefix)|B)','\g<key>',"anythingB")

‘anything’

While AanythingB give me anythingB back, but I don’t care anyway.

>>>re.sub(r'(?P<prefix>A)?(?P<key>\S+)(?(prefix)|B)','\g<key>',"AanythingB")

‘anythingB’