python for遍历元素时，取n个元素

def chunks(s, step):
    return [s[i:i+step] for i in range(0, len(s), step)]

chunks(range(50), 10)

返回值
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [20, 21, 22, 23, 24, 25, 26, 27, 28, 29], [30, 31, 32, 33, 34, 35, 36, 37, 38, 39], [40, 41, 42, 43, 44, 45, 46, 47, 48, 49]]

f=lambda s, step:[s[i:i+step] for i in range(0, len(s), step)]

f([1,2,3,4,5,6],2)

返回值
[[1, 2], [3, 4], [5, 6]]

from itertools import groupby, count

def chunks(it, step):
    return (list(g) for k, g in groupby(it, key=lambda x, c=count(): next(c) // step))

list(chunks(range(50), 10))

返回值

[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [20, 21, 22, 23, 24, 25, 26, 27, 28, 29], [30, 31, 32, 33, 34, 35, 36, 37, 38, 39], [40, 41, 42, 43, 44, 45, 46, 47, 48, 49]]