Answer 1

Answer 2

Answer 3

Answer 4

正如我所评论的那样，我不想对a对象和匹配的文本重用b和Match。我会选择一个函数来取出比赛，如下所示：

>>> def text_or_none(v): return v.group() if v is not None else None
>>> a = text_or_none(re.search("\d", "foo"))
None
>>> b = text_or_none(re.search("\w+", "foo"))
foo

Answer 5