3-6在一个for语句中迭代多个可迭代对象

# -*- coding:utf-8 -*-

"""
实际案例:
1.某班学生期末成绩,语文,数学,英语分别存储在3个列表中,
同时迭代三个列表,计算每个学生的总分.(并行)

2.某年及有4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,统计全学年成绩高于90分人数.(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,它能将多哥可迭代对象连接.
"""

from random import randint
from itertools import chain

chinese = [randint(60, 100) for _ in range(40)]
math = [randint(60, 100) for _ in range(40)]
english = [randint(60, 100) for _ in range(40)]

total = []
for i in range(len(math)):
total.append(chinese[i] + math[i] + english[i])
# 局限:如果chinese, math, english不是列表,而是生成器的话,将不能用索引访问

total1 = []
# zip([1,2,3,4], ('a','b','c'))
# 在前后两个对象的size不同时,生成的是[(1,'a'),(2,'b'),(3,'c')]
for c, m, e in zip(chinese, math, english):     # 对元组进行'拆包'迭代
total1.append(c + m + e)
print(total1)

e1 = [randint(60, 100) for _ in range(40)]
e2 = [randint(60, 100) for _ in range(40)]
e3 = [randint(60, 100) for _ in range(40)]
e4 = [randint(60, 100) for _ in range(40)]
count = 0
for x in chain(e1, e2, e3, e4):
if x > 90:
count += 1

print(count)