Python 如何按索引对列表进行分组？

我有一个列表，需要使用用户输入对元素进行分组（请参见代码中的

split

variable）并创建新列表。 我已经尝试过了，但不是分组，而是单独连接元素

split = 3 # user input 
data = [[1,2], [3,4], [5,6], [7,8], [9,10], [11,12], [13,14], [15,16], [17,18]]
z = [] ; y = []
for i,d in enumerate(data):
    z.append(d)
    if (i+1)%split==0:
        y.append(z)
        z = []
xx = (y+[z])
print(xx)
[[[1, 2], [3, 4], [5, 6]], [[7, 8], [9, 10], [11, 12]], [[13, 14], [15, 16], [17, 18]], []]
# ^____________________^    ^_______________________^  this needs to be merged

输入：

data = [[1,2], [3,4], [5,6], [7,8], [9,10], [11,12], [13,14], [15,16], [17,18]]

预期产出：

[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18]]

[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18]]

使用numpy：

data = [[1,2], [3,4], [5,6], [7,8], [9,10], [11,12], [13,14], [15,16], [17,18]]
import numpy as np
print np.array(data).reshape((split,-1)).tolist()

输出：

[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18]]

你可以试试这个->

split = 3
data = [[1,2], [3,4], [5,6], [7,8], [9,10], [11,12], [13,14], [15,16], [17,18]]
z=[]
x=[]
for i,j in enumerate(data):
    if i!=0 and i%split==0: 
        z.append(x)
        x=[]
    for k in j:
        x.append(k)
z.append(x)
print z

---

下面是使用列表理解的一种方法：

>>> sp = 3
>>> fragment = len(data)//sp
>>> [[t for item in data[i:i+fragment] for t in item] for i in range(0, len(data), fragment)]
[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18]]

下面是一个基于intertools的配方：

>>> from itertools import islice, chain
>>> 
>>> [list(chain.from_iterable(islice(data, i, i+fragment))) for i in range(0, len(data), fragment)]
[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18]]

---

您可以只使用

z.extend（d）

而不是

append

您可以在基本Python中这样做：

import itertools
flatlist = [*itertools.chain(*data)]
groupsize = int(len(flatlist) / split)

data2 = [flatlist[i:i+groupsize] for i in range(0, len(flatlist), groupsize)]
print(data2)

输出是