使用re.sub替换字符-保留一个字符

Question

I'm trying to repair broken email records in a table. There are emails, for example: 'google@google.comyahoo@yahoo.com' but there can be a single email like 'google@google.com'. The best way to make this correct is in my opinion to use re.sub. But there is a little problem. If there is a record:

email = 'google@google.comyahoo@yahoo.com'

I can't simply do replace('.com','.com, ') because it affects both '.com' substrings. So I want to use re.sub('.com\\w', '.com, \\w',email) which replaces only those '.com' substrings, which aren't in the end of the record. The problem is that I want to keep a \\w value there.

print re.sub('.com\\w', '.com, \\w',email)

>>> google@google.com, \wahoo@yahoo.com

instead of

>>> google@google.com, yahoo@yahoo.com

Can anybody give me an advice how to make it work? (I want to separate emails by comma and space)

Answer 1

Use a capturing group and backreference the group inside of the replacement call:

>>> import re
>>> email = 'google@google.comyahoo@yahoo.com'
>>> re.sub(r'\.com(\w)', '.com, \\1', email)
'google@google.com, yahoo@yahoo.com'

Backreferences recall what was matched by a capturing group . A backreference is specified as a backslash ( \\ ); followed by a digit indicating the number of the group to be recalled.

Answer 2

x="google@google.comyahoo@yahoo.com"
print re.sub(r"(?<=\.com)(?=\w)",", ",x)

Output: google@google.com, yahoo@yahoo.com

use lookarounds .See demo.

https://regex101.com/r/sJ9gM7/48

Lookarounds don't consume any of the string. They are just assertions. When you use them, you need not replace the consumed string back like the above answer does.