我想尋求幫助優化代碼。我有一個讓我們說 26 個元素的串列：
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]
僅供進一步閱讀：當我提到“子集”時 => 是資料的子“集”，而不是資料型別。我正在尋找“子串列”。
我正在準備一個函式，該函式將對該串列的子集執行進一步的計算。問題是，如果子集是在奇數上生成的，有時相同的元素會進入不同的子集兩次或更多次。我正在尋找的子集是：
- 子集 1 => 原始資料
- 子集 2 & 3 => 資料的前半部分和后半部分
- 子集 4 - 7 => 1/4 資料的第一個、第二個、第三個和第四個
- 子集 8 - 15 => 下一個 1/8 的集合。
我在函式體內想出了一個相當草率且冗長的解決方案，如下所示：
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.
它做了它應該做的事情（添加的 0.25 和 0.5 部分是為了避免將相同的元素包含到兩個或多個子集，假設子集的長度為 3.25）。但是，肯定有更好的方法來做到這一點。我不介意有不均勻的集合，比如說，當除以 4 有 2 個 7 元素串列和 2 個 6 元素串列時。只要元素是不同的。
謝謝你的幫助。
uj5u.com熱心網友回復：
您可以使用串列理解來獲取這些子集：
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)]
輸出：
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]
請注意，當串列的大小不是磁區數的倍數（例如 26/4 和 26/8）時，這將洗掉專案。有幾種方法可以處理這個問題（更多的子集、更大的塊、不同的子集大小以均勻或隨機地分布專案、添加到第一個子集、添加到最后一個，...）但您必須指定您想要的.
例如，此變體將額外的專案分散到前幾組（每組不超過 1 個額外專案）：
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
uj5u.com熱心網友回復：
您可以使用np.array_split 串列理解：
sublists = [arr.tolist() for num in [1,2,4,8] for arr in np.array_split(np.array(indata), num)]
輸出：
[[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]]
uj5u.com熱心網友回復：
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]
歸功于此答案的靈感
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