python for遍历元素时，取n个元素
浏览数：23 /
时间：2015年06月08日
1.
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]]
2.
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]]
3.
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]]
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