描述
zip()函数用于将可迭代的对象作为参数，将对象中对应的元素打包成一个个元组，然后返回由这些元组组成的对象，这样做的好处是节约了不少的内存。
可以使用list()转换来输出列表。
若各个迭代器的元素个数不一致，则返回列表长度与最短的对象相同，利用*号操作符，可以将元祖解压为列表。
例1
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| a = [1,2,3]
c = [1,2,3,4]
zipped = zip(a,c)
l_zipped = list(zipped)
b = 3
print(type(zipped),type(l_zipped))
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输出结果为：
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| <class 'zip'> <class 'list'>
[(1, 1), (2, 2), (3, 3)]
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例2
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| a = [
[1,2],
[3,4],
[5,6]
]
c = [
[5,6],
[7,8],
[9,10]
]
zipped = zip(a,c)
l_zipped = list(zipped)
b = 3
print(type(zipped),type(l_zipped))
print(l_zipped)
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输出结果：
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| <class 'zip'> <class 'list'>
[([1, 2], [5, 6]), ([3, 4], [7, 8]), ([5, 6], [9, 10])]
|
例3
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| a = [1,2,3]
c = [1,2,3,4]
zipped = zip(a,c)
print(zipped,type(zipped))
accmulated = 1*[0]
accmulated = [x_c[0] + x_c[1] for x_c in zip(a,c)]
for i in zipped:
print(i,type(i))
l_zipped = list(zipped) # 转换为列表
#print(l_zipped)
print("accmulated:",accmulated)
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输出结果：
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| <zip object at 0x7f3128d8d5c8> <class 'zip'>
(1, 1) <class 'tuple'>
(2, 2) <class 'tuple'>
(3, 3) <class 'tuple'>
accmulated: [2, 4, 6]
|
例4
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| a = [
[1,2],
[3,4],
[5,6]
]
c = [
[5,6],
[7,8],
[9,10]
]
zipped = zip(a,c)
l_zipped = list(zipped)
accmulated = 1 *[0]
# 注：此处将zip(a,c)换为zipped则显示为空
for i in zip(a,c):
print(i,type(i))
accmulated = [x_c[0] + x_c[1] for x_c in zip(a,c)]
print("accmulated:",accmulated)
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输出结果：
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| ([1, 2], [5, 6]) <class 'tuple'>
([3, 4], [7, 8]) <class 'tuple'>
([5, 6], [9, 10]) <class 'tuple'>
accmulated: [[1, 2, 5, 6], [3, 4, 7, 8], [5, 6, 9, 10]]
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例5：矩阵的行列互换
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| a = [[1,2,3],[3,4,5]]
a1 = zip(*a)
print(list(a1))
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输出结果：
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| [(1, 3), (2, 4), (3, 5)]
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