那么，蛮力的答案是：

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]]

如果你想要一个填充值，你可以在列表解析之前完成：

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]]

您可以使用itertools文档中的方法中的grouper函数：

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    """Collect data into fixed-length chunks or blocks.

    >>> grouper('ABCDEFG', 3, 'x')
    ['ABC', 'DEF', 'Gxx']
    """
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

不如

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

参见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，或者类似的方法。

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