内置函数zip()通过聚合给定的 iterable 返回一个元组或压缩对象，它可能有零个或多个 iterable 作为其参数。

 **zip(*iterables)** #where iterable may be list, string, dict etc 

zip()参数:

它可能使用一个或多个迭代器作为参数。*运算符可用于将列表与zip()一起解压缩。

zip()返回值

zip()没有参数，那么结果将是一个包含空迭代器的 zip 变量。

| 投入 | 返回值 | | 单一可滴定的 | 具有单个元素的元组 | | 多项 | 具有来自所有表的元素的元组 |

Python 中zip()方法的示例

示例zip()在 Python 中是如何工作的

 number_list = [5, 6, 7]
string_list = ['five', 'six', 'seven']

# No iterables are passed
result = zip()

# Converting iterator to list
result_list = list(result)
print(result_list)

# Two iterables are passed
result = zip(number_list, string_list)

# Converting iterator to set
result_set = set(result)
print(result_set) 

输出:

[]
{(7, 'seven'), (6, 'six'), (5, 'five')}

示例 2:不同数量的可迭代元素

 numbers_List = [5, 6, 7]
string_list = ['five', 'six']
numbers_tuple = ('FIVE', 'SIX', 'SEVEN', 'EIGHT')

# Notice, the size of numbersList and numbers_tuple is different
result = zip(numbers_List, numbers_tuple)

# Converting to set
result_set = set(result)
print(result_set)

result = zip(numbers_List, string_list, numbers_tuple)

# Converting to set
result_set = set(result)
print(result_set) 

输出:

{(7, 'SEVEN'), (6, 'SIX'), (5, 'FIVE')}
{(6, 'six', 'SIX'), (5, 'five', 'FIVE')} 

示例 3:如何使用zip()解压缩该值

 coordinates = ['x', 'y', 'z']
values = [1, 2, 3]

result = zip(coordinates, values)
result_list = list(result)
print(result_list)

c, v =  zip(*result_list)
print('c =', c)
print('v =', v) 

输出:

[('x', 1), ('y', 2), ('z', 3)]
c = ('x', 'y', 'z')
v = (1, 2, 3)