一些regular的tips：
1 非贪婪flag
>>> re.findall(r"a(\d+?)", "a23b")
[\'2\']
>>> re.findall(r"a(\d+)", "a23b")
[\'23\']
[\'2\']
>>> re.findall(r"a(\d+)", "a23b")
[\'23\']
注意比较这种情况：
>>> re.findall(r"a(\d+)b", "a23b")
[\'23\']
>>> re.findall(r"a(\d+?)b", "a23b")
[\'23\']
[\'23\']
>>> re.findall(r"a(\d+?)b", "a23b")
[\'23\']
2 如果你要多行匹配，那么加上re.S和re.M标志
re.S：.将会匹配换行符，默认.不会匹配换行符
>>> re.findall(r"a(\d+)b.+a(\d+)b", "a23b\na34b")
[]
>>> re.findall(r"a(\d+)b.+a(\d+)b", "a23b\na34b", re.S)
[(\'23\', \'34\')]
>>>
[]
>>> re.findall(r"a(\d+)b.+a(\d+)b", "a23b\na34b", re.S)
[(\'23\', \'34\')]
>>>
re.M：^$标志将会匹配每一行，默认^和$只会匹配第一行
>>> re.findall(r"^a(\d+)b", "a23b\na34b")
[\'23\']
>>> re.findall(r"^a(\d+)b", "a23b\na34b", re.M)
[\'23\', \'34\']
[\'23\']
>>> re.findall(r"^a(\d+)b", "a23b\na34b", re.M)
[\'23\', \'34\']
但是，如果没有^标志，
>>> re.findall(r"a(\d+)b", "a23b\na23b")
[\'23\', \'23\']
[\'23\', \'23\']
可见，是无需re.M
原文地址:http://www.juyimeng.com/python-multi-line-non-greedy-regular-expression-sample.html