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)

#输出：<zip object at 0x7f5a22651c08>

for i in zip1:

...     print(i)

#输出：

(1, 4, 7)

(2, 5, 8)

(3, 6, 9)

例2

 zip2=zip(a1,a2,a4)

 print(zip2)

#输出：<zip object at 0x7f5a22651d48>

 for j in zip2:

...     print(j)

#输出：

(1, 4, 'a')

(2, 5, 'b')

(3, 6, 'c')

例3

 zip3=zip(a4)

 print(zip3)

#输出：<zip object at 0x7f5a22651d08>

 for i in zip3:

...     print(i)

#输出

('a',)

('b',)

('c',)

('d',)

例4

 zip4=zip(*a4 *3)

 print(zip4)

#输出：<zip object at 0x7f5a22651f08>

 for j in zip4:

...     print(j)
#输出：('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)