实际案例：
1.某班学生期末考试成绩，语文，数学，英语分别存储在三个列表中，
同时迭代三个列表，计算每个学生的总分（并行）
2.某年级有四个班，某次考试每班英语成绩分别存储在4个列表中，
依次迭代每个列表，统计全年级成绩高于90分的人数（串行）
解决方案：
并行：使用内置函数zip，它能将多个可迭代对象合并，每次迭代返回一个元组
串行：使用标准库中的itertools.chain，它能将多个可迭代对象连接
方案一：
zip(iter1 [,iter2 [...]]) --> zip object
|
| Return a zip object whose .__next__() method returns a tuple where
| the i-th element comes from the i-th iterable argument. The .__next__()
| method continues until the shortest iterable in the argument sequence
| is exhausted and then it raises StopIteration.
from random import randint
count = 30 # 班级人数
chinese = [randint(50,100) for _ in range(count)]
math = [randint(50,100) for _ in range(count)]
english = [randint(50,100) for _ in range(count)]
# 方式一：索引 局限性：不适用与迭代器与生成器
for i in range(count):
print(i,chinese[i],math[i],english[i],)
# 方式二：zip和拆包
for c,m,e in zip(chinese,math,english):
print(c,m,e)
方案二：
chain(*iterables) --> chain object
Return a chain object whose .__next__() method returns elements from the
first iterable until it is exhausted, then elements from the next
iterable, until all of the iterables are exhausted.
from itertools import chain
# 随机生成4个班的英语成绩
cls1 = [randint(60,100) for _ in range(30)]
cls2 = [randint(60,100) for _ in range(30)]
cls3 = [randint(60,100) for _ in range(30)]
count = 0
for x in chain(cls1,cls2,cls3):
if x >= 90:
count += 1
print(count)
