Python re.sub use non-greedy mode (.*?) with end of string ($) it comes greedy!

Code:

str = '<br><br />A<br />B'
print(re.sub(r开发者_如何转开发'<br.*?>\w$', '', str))

It is expected to return <br><br />A, but it returns an empty string ''!

Any suggestion?

Greediness works from left to right, but not otherwise. It basically means "don't match unless you failed to match". Here's what's going on:

1. The regex engine matches <br at the start of the string.
2. .*? is ignored for now, it is lazy.
3. Try to match >, and succeeds.
4. Try to match \w and fails. Now it's interesting - the engine starts backtracking, and sees the .*? rule. In this case, . can match the first >, so there's still hope for that match.
5. This keep happening until the regex reaches the slash. Then >\w can match, but $ fails. Again, the engine comes back to the lazy .* rule, and keeps matching, until it matches<br><br />A<br />B

Luckily, there's an easy solution: By replacing <br[^>]*>\w$ you don't allow matching outside of your tags, so it should replace the last occurrence.
Strictly speaking, this doesn't work well for HTML, because tag attributes can contain > characters, but I assume it's just an example.

The non-greediness won't start later on like that. It matches the first <br and will non-greedily match the rest, which actually need to go to the end of the string because you specify the $.

To make it work the way you wanted, use

/<br[^<]*?>\w$/

but usually, it is not recommended to use regex to parse HTML, as some attribute's value can have < or > in it.