Question

可能重复：
How do you split a list into evenly sized chunks in Python?

我想从列表l：

中获取n个元素的组

即：

[1,2,3,4,5,6,7,8,9] -> [[1,2,3], [4,5,6],[7,8,9]] where n is 3

Answer 1

您可以在itertools文档页面上使用recipes中的grouper：

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Answer 2

蛮力的回答是：

subList = [theList[n:n+N] for n in range(0, len(theList), N)]

其中N是组大小（在您的情况下为3）：

>>> theList = list(range(10))
>>> N = 3
>>> subList = [theList[n:n+N] for n in range(0, len(theList), N)]
>>> subList
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

如果您想要填充值，可以在列表理解之前执行此操作：

tempList = theList + [fill] * N
subList = [tempList[n:n+N] for n in range(0, len(theList), N)]

示例：

>>> fill = 99
>>> tempList = theList + [fill] * N
>>> subList = [tempList[n:n+N] for n in range(0, len(theList), N)]
>>> subList
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 99, 99]]

Answer 3

请参阅itertools文档底部的示例：http://docs.python.org/library/itertools.html?highlight=itertools#module-itertools

你想要“石斑鱼”方法，或类似的东西。

Answer 4

怎么样

a = range(1,10)
n = 3
out = [a[k:k+n] for k in range(0, len(a), n)]

Answer 5

answer = [L[3*i:(3*i)+3] for i in range((len(L)/3) +1)]
if not answer[-1]:
    answer = answer[:-1]

如何用n个元素对python中的元素进行分组？

5 个答案: