regex - Find last match with python regular expression

Question

I want to match the last occurrence of a simple pattern in a string, e.g.

list = re.findall(r"w+ AAAA w+", "foo bar AAAA foo2 AAAA bar2")
print "last match: ", list[len(list)-1]

However, if the string is very long, a huge list of matches is generated. Is there a more direct way to match the second occurrence of " AAAA ", or should I use this workaround?

Answer 1

you could use $ that denotes end of the line character:

>>> s = """foo bar AAAA
foo2 AAAA bar2"""
>>> re.findall(r"w+ AAAA w+$", s)
['foo2 AAAA bar2']

Also, note that list is a bad name for your variable, as it shadows built-in type. To access the last element of a list you could just use [-1] index:

>>> lst = [2, 3, 4]
>>> lst[-1]
4