不均匀子集genetarion的Python代码优化

Question

我想在优化代码方面寻求帮助。我列出了26个元素的列表：

indata = [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42, 30, 16, 14, 85, 44, 89, 26, 0, 67, 67, 23, 0, 0]

仅供进一步阅读:当我提到“子集”=>是数据的子‘集合’，而不是数据类型。我在找“分名单”。

我正在准备一个函数，对这个列表的子集进行进一步的计算。问题是，如果子集是在不均匀的数字上生成的，有时相同的元素会两次或更多地进入不同的子集。我要找的子集是：

• 子集1 =>原始数据子集2&3
• 子集2&3 =>
• 子集4-7 =>的第一、第二、第三和第四组数据
• 子集8- 15 =>下1/8。

f 210

我在函数体中想出了一个相当草率而又冗长的解决方案，它是这样的：

for i in iterate:
    if i == 0:
        subset = indata
    elif i == 1:
        subset = indata[0:int(len(indata)/2)]
    elif i == 2:
        subset = indata[int(len(indata)/2):]
    elif i == 3:
        subset = indata[0:int(len(indata)/4)]
    elif i == 4:
        subset = indata[int(len(indata)/4):int(round((len(indata)/4)*2,0))]
    elif i == 5:
        subset = indata[int(round((len(indata)/4)*2,0)):int(round((len(indata)/4)*3,0))]
    elif i == 6:
        subset = indata[int(round((len(indata)/4)*3,0)):]
    elif i == 7:
        subset = indata[0:int(len(indata)/8)]        
    elif i == 8:
        subset = indata[int(len(indata)/8):int(round((len(indata)/8)*2,0))]        
    elif i == 9:
        subset = indata[int(len(indata)/8)*2:int(round((len(indata)/8)*3,0))]        
    elif i == 10:
        subset = indata[int((len(indata)/8)*3+0.25):int(round((len(indata)/8)*4,0))]        
    elif i == 11:
        subset = indata[int((len(indata)/8)*4+0.25):int(round((len(indata)/8)*5,0))]        
    elif i == 12:
        subset = indata[int((len(indata)/8)*5+0.25):int(round((len(indata)/8)*6,0))]
    elif i == 13:
        subset = indata[int((len(indata)/8)*6+0.5):int(round((len(indata)/8)*7,0))]
    elif i == 14:
        subset = indata[int((len(indata)/8)*7+0.5):]
    else:
        subset = indata[int((len(indata)/8)*7+0.5):] 

-here go further instruction on the subset, then loop go back and repeat.

它做它应该做的事情(添加的0.25和0.5部分是为了避免将相同的元素包含到两个或更多个子集中，假设子集的长度为3.25)。然而，肯定有更好的方法来做到这一点。我不介意有不平衡的集合，让我们说，当除以4有2 7元素列表和2 6元素列表。只要元素是不同的。

谢谢你的帮助。

Answer 1

您可以使用列表理解来获取以下子集：

indata = [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42, 30, 16, 14, 85, 
         44, 89, 26, 0, 67, 67, 23, 0, 0]

subsets = [indata[p*size:(p+1)*size] 
           for parts in (1,2,4,8) 
           for size in [len(indata)//parts] 
           for p in range(parts)]

输出：

for i,subset in enumerate(subsets,1): print(i,subset)

1 [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42, 30, 16, 14, 85, 44, 
   89, 26, 0, 67, 67, 23, 0, 0]

2 [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42]
3 [30, 16, 14, 85, 44, 89, 26, 0, 67, 67, 23, 0, 0]

4 [0, 0, 50, 0, 32, 35]
5 [151, 163, 9, 1, 3, 3]
6 [42, 30, 16, 14, 85, 44]
7 [89, 26, 0, 67, 67, 23]

8 [0, 0, 50]
9 [0, 32, 35]
10 [151, 163, 9]
11 [1, 3, 3]
12 [42, 30, 16]
13 [14, 85, 44]
14 [89, 26, 0]
15 [67, 67, 23]

请注意，当列表的大小不是分区数目的倍数时(例如，26/4和26/8)，这将删除项。有几种方法可以处理这个问题(更多的子集、较大的块、不同的子集大小来均匀或随机地分布项、添加到第一个子集、添加到最后一个子集、.)但你必须指明你想要哪一个。

例如，该变体将额外的项分散到前几组(每组不超过1项)：

subsets = [indata[p*size+min(p,spread):(p+1)*size+min(p+1,spread)]
           for parts in (1,2,4,8)
           for size,spread in [divmod(len(indata),parts)]
           for p in range(parts)]

for i,subset in enumerate(subsets,1): print(i,subset,len(subset))

1 [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42, 30, 16, 14, 
   85, 44, 89, 26, 0, 67, 67, 23, 0, 0] 26

2 [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42] 13
3 [30, 16, 14, 85, 44, 89, 26, 0, 67, 67, 23, 0, 0] 13

4 [0, 0, 50, 0, 32, 35, 151] 7
5 [163, 9, 1, 3, 3, 42, 30] 7
6 [16, 14, 85, 44, 89, 26] 6
7 [0, 67, 67, 23, 0, 0] 6

8 [0, 0, 50, 0] 4
9 [32, 35, 151, 163] 4
10 [9, 1, 3] 3
11 [3, 42, 30] 3
12 [16, 14, 85] 3
13 [44, 89, 26] 3
14 [0, 67, 67] 3
15 [23, 0, 0] 3

Answer 2

def divide_data(data, chunks):
    idx = 0
    sizes = [len(data) // chunks + int(x < len(data)%chunks) for x in range(chunks)]
    for size in sizes:
        yield data[idx:idx+size]
        idx += size

data = list(range(26))  # or whatever, e.g. [0, 0, 50, ...]
for num_subsets in (1, 2, 4, 8):
    print(f'num subsets: {num_subsets}')
    for subset in divide_data(data, num_subsets):
        print(subset)

num subsets: 1
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
num subsets: 2
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
[13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
num subsets: 4
[0, 1, 2, 3, 4, 5, 6]
[7, 8, 9, 10, 11, 12, 13]
[14, 15, 16, 17, 18, 19]
[20, 21, 22, 23, 24, 25]
num subsets: 8
[0, 1, 2, 3]
[4, 5, 6, 7]
[8, 9, 10]
[11, 12, 13]
[14, 15, 16]
[17, 18, 19]
[20, 21, 22]
[23, 24, 25]

归功于this answer的灵感

Answer 3

您可以使用np.array_split +列表理解：

sublists = [arr.tolist() for num in [1,2,4,8] for arr in np.array_split(np.array(indata), num)]

输出：

[[0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42, 30, 16, 14, 85, 44, 89, 26, 0, 67, 67, 23, 0, 0],
 [0, 0, 50, 0, 32, 35, 151, 163, 9, 1, 3, 3, 42],
 [30, 16, 14, 85, 44, 89, 26, 0, 67, 67, 23, 0, 0],
 [0, 0, 50, 0, 32, 35, 151],
 [163, 9, 1, 3, 3, 42, 30],
 [16, 14, 85, 44, 89, 26],
 [0, 67, 67, 23, 0, 0],
 [0, 0, 50, 0],
 [32, 35, 151, 163],
 [9, 1, 3],
 [3, 42, 30],
 [16, 14, 85],
 [44, 89, 26],
 [0, 67, 67],
 [23, 0, 0]]