>>> name = ("lilei","lihua","zhang")

>>> age = ("","","")

>>> for a,n in zip(name,age):

...     print a,n

...

lilei 21

lihua 14

zhang 34

>>>

>>> all={"hell":20,"helo":23,"sony":15,"lei":19}

>>> for i in all.keys():

...     print i,all[i]

...

hell 20

helo 23

lei 19

sony 15

>>>

>>> aa = ("","","")

>>> bb = ("ee","tt")

>>> for a,n in zip(aa,bb):

...     print a,n

...

1 ee

3 tt

>>> zip(aa,bb)

[('', 'ee'), ('', 'tt')]

>>>

>>> aa

('', '', '')

>>> bb

('ee', 'tt')

>>> ad = zip(aa,bb)

>>> zip(*ad)

[('', ''), ('ee', 'tt')]

>>> zip(ad)

[(('', 'ee'),), (('', 'tt'),)]

>>>

* 二维矩阵变换（矩阵的行列互换）

比如我们有一个由列表描述的二维矩阵

a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

通过python列表推导的方法，我们也能轻易完成这个任务

print [ [row[col] for row in a] for col in range(len(a[0]))]

[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

另外一种让人困惑的方法就是利用zip函数：

>>> a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

>>> zip(*a)

[(1, 4, 7), (2, 5, 8), (3, 6, 9)]

>>> map(list,zip(*a))

[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

zip函数接受任意多个序列作为参数，将所有序列按相同的索引组合成一个元素是各个序列合并成的tuple的新序列，新的序列的长度以参数中最短的序列为准。另外(*)操作符与zip函数配合可以实现与zip相反的功能，即将合并的序列拆成多个tuple。

①tuple的新序列

>>>>x=[1,2,3],y=['a','b','c']

>>>zip(x,y)

[(1,'a'),(2,'b'),(3,'c')]

②新的序列的长度以参数中最短的序列为准.

>>>>x=[1,2],y=['a','b','c']

>>>zip(x,y)

[(1,'a'),(2,'b')]

③(*)操作符与zip函数配合可以实现与zip相反的功能,即将合并的序列拆成多个tuple。

>>>>x=[1,2,3],y=['a','b','c']

>>>>zip(*zip(x,y))

[(1,2,3),('a','b','c')]

1.zip打包解包列表和倍数

>>> a = [1, 2, 3]

>>> b = ['a', 'b', 'c']

>>> z = zip(a, b)

>>> z

[(1, 'a'), (2, 'b'), (3, 'c')]

>>> zip(*z)

[(1, 2, 3), ('a', 'b', 'c')]

2. 使用zip合并相邻的列表项

>>> a = [1, 2, 3, 4, 5, 6]

>>> zip(*([iter(a)] * 2))

[(1, 2), (3, 4), (5, 6)]

>>> group_adjacent = lambda a, k: zip(*([iter(a)] * k))

>>> group_adjacent(a, 3)

[(1, 2, 3), (4, 5, 6)]

>>> group_adjacent(a, 2)

[(1, 2), (3, 4), (5, 6)]

>>> group_adjacent(a, 1)

[(1,), (2,), (3,), (4,), (5,), (6,)]

>>> zip(a[::2], a[1::2])

[(1, 2), (3, 4), (5, 6)]

>>> zip(a[::3], a[1::3], a[2::3])

[(1, 2, 3), (4, 5, 6)]

>>> group_adjacent = lambda a, k: zip(*(a[i::k] for i in range(k)))

>>> group_adjacent(a, 3)

[(1, 2, 3), (4, 5, 6)]

>>> group_adjacent(a, 2)

[(1, 2), (3, 4), (5, 6)]

>>> group_adjacent(a, 1)

[(1,), (2,), (3,), (4,), (5,), (6,)]

3.使用zip和iterators生成滑动窗口 (n -grams)

>>> from itertools import islice

>>> def n_grams(a, n):

...     z = (islice(a, i, None) for i in range(n))

...     return zip(*z)

...

>>> a = [1, 2, 3, 4, 5, 6]

>>> n_grams(a, 3)

[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]

>>> n_grams(a, 2)

[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]

>>> n_grams(a, 4)

[(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6)]

4.使用zip反转字典

>>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}

>>> m.items()

[('a', 1), ('c', 3), ('b', 2), ('d', 4)]

>>> zip(m.values(), m.keys())

[(1, 'a'), (3, 'c'), (2, 'b'), (4, 'd')]

>>> mi = dict(zip(m.values(), m.keys()))

>>> mi

{1: 'a', 2: 'b', 3: 'c', 4: 'd'}