1 >>> name=(‘jack‘,‘beginman‘,‘sony‘,‘pcky‘) 2 >>> age=(2001,2003,2005,2000) 3 >>> for a,n in zip(name,age): 4     print a,n 5  6 输出： 7 jack 2001 8 beginman 2003 9 sony 200510 pcky 2000

1 all={"jack":2001,"beginman":2003,"sony":2005,"pcky":2000}2 for i in all.keys():3     print i,all[i]4 5 输出：6 sony 20057 pcky 20008 jack 20019 beginman 2003

1 >>> type(sorted(s))2 <type ‘list‘>3 >>> type(zip(s))4 <type ‘list‘>5 >>> type(reversed(s))6 <type ‘listreverseiterator‘>7 >>> type(enumerate(s))8 <type ‘enumerate‘>

1 >>> help(zip)2 Help on built-in function zip in module __builtin__:3 4 zip(...)5     zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]6     7     Return a list of tuples, where each tuple contains the i-th element8     from each of the argument sequences.  The returned list is truncated9     in length to the length of the shortest argument sequence.

 1 >>> z1=[1,2,3] 2 >>> z2=[4,5,6] 3 >>> result=zip(z1,z2) 4 >>> result 5 [(1, 4), (2, 5), (3, 6)] 6 >>> z3=[4,5,6,7] 7 >>> result=zip(z1,z3) 8 >>> result 9 [(1, 4), (2, 5), (3, 6)]10 >>> 

1 >>> zip(*result)2 [(1, 2, 3), (4, 5, 6)]

* 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‘}