[Python] zip()函数

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

提示Tips:本文讲解的是python3的zip()函数，该函数返回的结果是zip对象，如需展示列表，需手动 list()转换；展示字典，需手动dict()转换

语法

zip([iterable, ...])

参数 

一个或多个迭代器(其中iterable,... 表示多个列表、元组、字典、集合、字符串，甚至还可以为 range()区间)

返回值

返回一个对象 

实例

nums1 = [1,2,3]
nums2 = [4,5,6]
zipped = zip(nums1,nums2)
# 返回一个对象
type(zipped) # zip object
print(zipped) # 
# list()转换为列表
list(zipped) # [(1, 4), (2, 5), (3, 6)]

names = ['Andy','Rita','Lee','Odin']
ages = [18,19,20,21]
'''
并行迭代
Andy is 18 years old
Rita is 19 years old
Lee is 20 years old
Odin is 21 years old
'''
for name, age in zip(names,ages):
    print(name,'is',age,'years old')

如果可迭代对象的长度不相同将按短的序列为准 

nums1 = [1,2,3]
nums3 = [4,5,6,7,8]
# 元素个数与最短的列表一致
list(zip(nums1,nums3)) # [(1, 4), (2, 5), (3, 6)]

# [(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)]
list(zip(range(5),range(100000000)))

只要是可迭代的对象皆可相互使用，例如zip(dict,list)

注意:zip(dict)得到的只有key没有value 

tuple1 = (1,2,3,4)
dict1 = {"a": 1, "b": 2, "c": 3}
dict2 = {4: "d", 5: "e", 6: "f"}
# [('a',), ('b',), ('c',)]
print([i for i in zip(dict1)])
# [(1, 'a'), (2, 'b'), (3, 'c')]
print([i for i in zip(tuple1, dict1)])
# [(1, 4), (2, 5), (3, 6)]
print([i for i in zip(tuple1, dict2)])

扩展补充:两个列表转换成字典 

keys = ['a','b','c']
values = [1,2,3]
# {'a': 1, 'b': 2, 'c': 3}
dict_nums = dict(zip(keys,values))