如何按n个元素对python中的元素进行分组？

How to group elements in python by n elements?

提问by TheOne

Possible Duplicate:
How do you split a list into evenly sized chunks in Python?

可能的重复：
如何在 Python 中将列表拆分为大小均匀的块？

I'd like to get groups of size n elements from a list l:

我想从列表 l 中获取大小为 n 个元素的组：

ie:

IE：

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

回答by Adam Vandenberg

See examples at the bottom of the itertools docs: http://docs.python.org/library/itertools.html?highlight=itertools#module-itertools

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

You want the "grouper" method, or something like it.

您需要“石斑鱼”方法或类似方法。

回答by Mike DeSimone

Well, the brute force answer is:

好吧，蛮力答案是：

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

where Nis the group size (3 in your case):

N组大小在哪里（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]]

If you want a fill value, you can do this right before the list comprehension:

如果你想要一个填充值，你可以在列表理解之前这样做：

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

Example:

例子：

>>> 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]]

回答by Mark Byers

You can use grouper from the recipeson the itertools documentation page:

您可以使用itertools 文档页面上的配方中的 grouper：

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

回答by sizzzzlerz

How about

怎么样

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

回答by inspectorG4dget

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