from random import randint a1 = [randint(10, 50) for _ in range(5)] a2 =
[randint(10, 50) for _ in range(5)] a3 = [randint(10, 50) for _ in range(5)] a4
= []

<>方法一：直接用for循环引用
#弊端：只能支持引索操作：a1[]，若操作对象是生成器，则不能实现； for i in range(5): t = a1[i] + a2[i] +
a3[i] a4.append(t) print(a4) #输出：[84, 67, 85, 88, 82]
<>方法二：用内置函数zip()
for x, y, z in zip(a1, a2, a3): a4.append(x + y + z) print(a4) #输出：[44, 72,
73, 94, 130]
**例二：**川行操作：在一个for循环中实现多个列表的川行迭代；

**方案：**使用标准库itertools.chain，它能使多个迭代对象连接

from itertools import chain b1 = [1, 2, 3, 4] b2 = ['a', 'b', 'c'] b3 =
list(chain(b1, b2)) print(b3) #输出：[1, 2, 3, 4, 'a', 'b', 'c']

for x in chain(b1, b2): print(x) #输出：1 2 3 4 a b c
**案例：**对4个列表进行迭代操作，筛选出目标数据（大于40的个数）：
from itertools import chain from random import randint a1 = [randint(10, 50)
for _ in range(40)] a2 = [randint(10, 50) for _ in range(41)] a3 = [randint(10,
50) for _ in range(42)] a4 = [randint(10, 50) for _ in range(43)] count = 0 for
x in chain(a1, a2, a3, a4): if x >=40: count += 1 print(count)