>>> a_list = ['The', 'sixth', 'sick', "sheik's", 'sixth', "sheep's", 'sick']
>>> set(a_list)
{'sixth', 'The', "sheep's", 'sick', "sheik's"}
# 利用set的唯一属性,的确是个不错的主意,呵呵
>>> a_string = 'EAST IS EAST'
>>> set(a_string)
{'A', ' ', 'E', 'I', 'S', 'T'}
>>> words = ['SEND', 'MORE', 'MONEY']
>>> '.join(words)
'SENDMOREMONEY'
>>> set('.join(words))
{'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}
>>>
>>> re.findall('[A-Z]+', 'SEND + MORE == MONEY')
['SEND', 'MORE', 'MONEY']
#findall()
函数返回所有匹配该模式的子字符串的列表
>>>