使用 re.sub 后从 backrefrence 中删除字符

Question

I have a string with a list of usernames mentioned for instance:

s = '@romeo went to @juliet and said hi, I'm @romeo'

I want to replace that username patter with links to user profile which should become <a href="/u/username">@username</a> I am now able to replace the patterns, however, I cannot seem to get rid of the @ in the href with using backreferences.

print(re.sub(r"(^|[^@\\w])@(\\w{1,31})", r'<a href="/u/\\g<0>">\\g<0></a>', s))

That right now prints:

<a href="/u/@romeo">@romeo</a> went to<a href="/u/ @juliet"> @juliet</a> and said hi, Im<a href="/u/ @romeo"> @romeo</a>

Which now you can see the extra space and @ I cannot seem to get rid of that after using the regex

Answer 1

You need to use

print(re.sub(r"\B(?<!@)@(\w{1,31})", r'<a href="/u/\1">\g<0></a>', s))

See the Python demo and the regex demo .

Regex

• \\B@ - @ char either at the start of string, or when immediately preceded with a non-word char
• (?<!@) - the preceding @ should not be immediately preceded with @
• (\\w{1,31}) - Capturing group 1 ( \\1 ): one to thirty-one word chars.

The \\1 in r'<a href="/u/\\1">\\g<0></a>' stands for the Group 1 value. \\g<0> stands for the whole match.