Match the last number in a string

I was trying to match the last number in a string, such as "34" in

3.1 General definitions 34

I am using Python-style regex, and I have tried:

(.*?)(\d)

so that later I can use \1 to refer "3.1 General definitions " and \2 to refer "34".

But \2 matches "4" instead of "34". 开发者_JAVA百科So how shall I do?

Thanks and regards!

You're currently only matching a single digit. Try

(.*?)(\d+)

to match at least one digit. This should be all you need, as you've already made the first part of the match reluctant (non-greedy).

Depending on how you're performing the match, you may need an "end of string" anchor ($) at the end, to make sure it doesn't match the digits at the start of the string.

Your original approach:

>>> re.match(r'(.*?)(\d)', '3.1 General definitions 34').groups()
('', '3')

Jon Skeet's answer (edit: I forgot to inculde Jon's suggestion to anchor to end of string, example is now corrected):

>>> re.match(r'(.*?)(\d+$)', '3.1 General definitions 34').groups()
('3.1 General definitions ', '34')

What I think is correct: new: If the string doesn't end with a number, I would do it this way:

>>> re.match(r'(.*)(?<=\D)(\d+)', '3.1 General definitions 34 more text').groups()
('3.1 General definitions ', '34')

.* needs to be greedy to get the beginning of the string that you wanted, but we can't have it slurp up the first digit of 34 so we provide (?<=\D), a lookbehind assertion, to stop it from happening

You could try:

(.+?)(\d+$)

as it will giive you:

>>> r.groups()
(u'3.1 General definitions ', u'34')

You need to make it greedy at the begining.