zip函数接受任意多个可迭代对象作为参数,将对象中对应的元素打包成一个tuple,然后返回一个可迭代的zip对象.
这个可迭代对象可以使用循环的方式列出其元素
若多个可迭代对象的长度不一致,则所返回的列表与长度最短的可迭代对象相同.
用法1:用两个列表生成一个zip对象
例1：
>>> a1=[1,2,3] >>> a2=[4,5,6] >>> a3=[7,8,9] >>> a4=["a","b","c","d"] >>> zip1=zip(a1,a2,a3) >>> print(zip1) <zip object at 0x7f5a22651c08> >>> for i in zip1: ... print(i) ... (1, 4, 7) (2, 5, 8) (3, 6, 9)
例2：
>>> zip2=zip(a1,a2,a4) >>> print(zip2) <zip object at 0x7f5a22651d48> >>> for j in zip2: ... print(j) ... (1, 4, 'a') (2, 5, 'b') (3, 6, 'c')
例3：
>>> zip3=zip(a4)
>>> print(zip3)
<zip object at 0x7f5a22651d08>
>>> for i in zip3:
... print(i)
...
('a',)
('b',)
('c',)
('d',)
例4：
>>> zip4=zip(*a4 *3)
>>>
>>> print(zip4)
<zip object at 0x7f5a22651f08>
>>> for j in zip4:
... print(j)
...
('a', 'b', 'c', 'd', 'a', 'b', 'c', 'd', 'a', 'b', 'c', 'd')
用法2:二维矩阵变换(矩阵的行列互换)
>>> l1=[[1,2,3],[4,5,6],[7,8,9]] >>> print([[j[i] for j in l1] for i in range(len(l1[0])) ]) [[1, 4, 7], [2, 5, 8], [3, 6, 9]] >>> zip(*l1) <zip object at 0x7f5a22651f88> >>> for i in zip(*l1): ... print(i) ... (1, 4, 7) (2, 5, 8) (3, 6, 9)
原文地址：http://blog.51cto.com/12332406/2105267
时间： 2024-10-29 21:28:48