如何在python中按n个元素对元素进行分组

Question

给定一个列表 [1,2,3,4,5,6,7,8,9,10,11,12] 和指定的块大小（比如 3），我怎样才能得到一个块列表[[1,2,3],[4,5,6],[7,8,9],[10,11,12]]？

Answer 1

好吧，强力答案是：

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

其中 N 是组大小（在您的例子中为 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]]

如果您想要一个填充值，您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 2

您可以使用 itertools 文档中 recipes 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)

Answer 3

怎么样

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

Answer 4

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

您需要“grouper”方法或类似的方法。

Answer 5

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