>>> r=[1,2,3,4,5]
>>> m=[2,4]
>>> list(set(r).intersection(set(m)))
[2, 4]
>>> a=[1,2,3,4,5]
>>> b=[2,4,6,8]
>>> list(set(r).intersection(set(m)))
[2, 4]
>>> list(set(a).union(set(b)))
[1, 2, 3, 4, 5, 6, 8]
>>> list(set(b).difference(set(a)))  # b中有而a中没有的
[8, 6]
>>> list(set(a).difference(set(b)))  # a中有而b中没有的
[1, 3, 5]
>>> a=[3,4,2,5,1]
>>> list(set(r).intersection(set(m)))
[2, 4]
>>> list(set(a).union(set(b)))
[1, 2, 3, 4, 5, 6, 8]
>>> list(set(b).difference(set(a)))  # b中有而a中没有的
[8, 6]
>>> list(set(a).difference(set(b)))  # a中有而b中没有的
[1, 3, 5]
>>> #两个list比较的话，利用了set的属性，对各自元素的顺序无关，但是会过滤掉相同的元素

>>> a=set(a)
>>> a
set([1, 2, 3, 4, 5])
>>> b=set(b)
>>> b
set([8, 2, 4, 6])
>>> a | b  #求并集
set([1, 2, 3, 4, 5, 6, 8])
>>> a - b #求差集，a中有而b中无
set([1, 3, 5])
>>> b - a  #求差集，b中有而a中没有的
set([8, 6])
>>> a & b #求交集
set([2, 4])
>>> a in b
False
>>> a not in b
True
>>> a==b
False
>>> a!=b
True
>>> len(a)
5
>>> a=set([1, 4, 3, 2, 5])
>>> a
set([1, 2, 3, 4, 5])
>>> a - b #求差集，a中有而b中无
set([1, 3, 5])
>>> #两个set比较，对各自元素的顺序无关，但是会过滤掉重复的元素
>>> 

>>> atuple
(1, 2, 3, \'d\', 2, 3, \'n\', \'d\')
>>> btuple
(2, 3, \'n\', \'d\')
>>> cmp(btuple,btuple)
0
>>> tupleb=(2, 3, \'d\', \'n\')
>>> cmp(btuple,tupleb)
1
>>> cmp(tupleb,btuple)
-1
>>> #tuple比较与元素的顺序有关
>>> abtuple=atuple+btuple  #组合
>>> abtuple
(1, 2, 3, \'d\', 2, 3, \'n\', \'d\', 2, 3, \'n\', \'d\')
>>> aatuple=atuple*2  #复制
>>> aatuple
(1, 2, 3, \'d\', 2, 3, \'n\', \'d\', 1, 2, 3, \'d\', 2, 3, \'n\', \'d\')
>>> aatuple
(1, 2, 3, \'d\', 2, 3, \'n\', \'d\', 1, 2, 3, \'d\', 2, 3, \'n\', \'d\')
>>> 

>>> adict={\'id\':1,\'name\':\'lucy\',\'age\':23,\'birth\':\'20160909\'}
>>> bdict={\'name\':\'lucy\',\'id\':1,\'age\':23,\'birth\':\'20160909\'}
>>> adict
{\'age\': 23, \'id\': 1, \'birth\': \'20160909\', \'name\': \'lucy\'}
>>> bdict
{\'age\': 23, \'name\': \'lucy\', \'birth\': \'20160909\', \'id\': 1}
>>> adict.keys()
[\'age\', \'id\', \'birth\', \'name\']
>>> bdict.keys()
[\'age\', \'name\', \'birth\', \'id\']
TypeError: unsupported operand type(s) for &: \'list\' and \'list\'
>>> adict.items()
[(\'age\', 23), (\'id\', 1), (\'birth\', \'20160909\'), (\'name\', \'lucy\')]
>>> bdict.items()
[(\'age\', 23), (\'name\', \'lucy\'), (\'birth\', \'20160909\'), (\'id\', 1)]
>>> adict.update(bdict)
>>> adict
{\'name\': \'lucy\', \'age\': 23, \'birth\': \'20160909\', \'id\': 1}
>>> abdict=adict.items()+bdict.items()#组合
>>> abdict
[(\'name\', \'lucy\'), (\'age\', 23), (\'birth\', \'20160909\'), (\'id\', 1), (\'age\', 23), (\'name\', \'lucy\'), (\'birth\', \'20160909\'), (\'id\', 1)]
>>> abdict=dict(abdict) #转换为字典后去除了重复的键
>>> abdict
{\'age\': 23, \'name\': \'lucy\', \'birth\': \'20160909\', \'id\': 1}
>>> cdict={\'name\': \'kate\', \'age\': 23, \'birth\': \'20160909\', \'id\': 2}
>>> acdict=adict.items()+cdict.items()
>>> acdict
[(\'name\', \'lucy\'), (\'age\', 23), (\'birth\', \'20160909\'), (\'id\', 1), (\'age\', 23), (\'name\', \'kate\'), (\'birth\', \'20160909\'), (\'id\', 2)]
>>> acdict=dict(acdict)  #转换为字典后去除了重复的键，对于相同的键，选择最后合并进来的元素
>>> acdict
{\'age\': 23, \'name\': \'kate\', \'birth\': \'20160909\', \'id\': 2}
>>>