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)
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