Make a subgroup reference (\g<1>) optional in re.sub

Question

How can I make a subgroup reference ( \\g<1> ) optional in re.sub() ? For example with:

import re

regexp = re.compile(r'^http://(lists\.|www\.)?example\.com/')
regexp.sub(
    r'https://\g<1>example.com/',
    r'http://example.com/helllo-there'
)

I would like \\g<1> to be replaced with nothing, the optional subgroup isn't matched (and not raise an exception).

I know I can use regexp.match(..).groups() to check which groups are present, but this seems like a lot of work to me (we would need a bunch of replacement patterns, since some examples go up to \\g<6> ). It's also not very fast since we need to do a match and a replace .

For example in JavaScript, I can use $1 , if it's not matched it's just ignored:

'http://example.com/helllo-there'.replace(
    RegExp('^http://(lists\.|www\.)?example\.com/'),
    'https://$1example.com/')
// Outputs: "https://example.com/helllo-there"

Answer 1

Another option is to provide an explicit empty alternative:

 regexp = re.compile(r'^http://(lists\.|www\.|)example\.com/')

Also, you can use just \\1 instead of \\g<1> .

Answer 2

如果我理解正确，只需做x(y?)z而不是x(y)?z

Answer 3

I would do like this. Just put the pattern inside a non-capturing group and make it as optional. Now include that optional non-capturing group inside a capturing group.

>>> re.sub(r'^http://((?:lists\.|www\.)?)example\.com/',r'https://\g<1>example.com/', 'http://example.com/helllo-there')
'https://example.com/helllo-there'