具有特定規則的拆分列表

Question

我想基於循環創建子列表，但仍然找不到如何做的邏輯？

''' source list'''
list = [1,2,3,4,5,6,7,8,9]
''' sublist goals'''
list_1 = [1,4,7]
list_2 = [2,5,8]
list_3 = [3,6,9]

Answer 1

list = [1,2,3,4,5,6,7,8,9]

list_1 = []
list_2 = []
list_3 = []

for j in range(1,4):
    for i in range(j,len(list)+1,3):
        if j == 1:
            list_1.append(i)
        if j == 2:
            list_2.append(i)
        if j == 3:
            list_3.append(i)

print (list_1)
print (list_2)
print (list_3)

輸出：

[1, 4, 7]
[2, 5, 8]
[3, 6, 9]

Answer 2

只需創建一個 3x3 列表，然后根據條件將項目附加到列表中

li = [1,2,3,4,5,6,7,8,9]

#Create a 3x3 list
res = [ [] for _ in range(3)]

for idx in range(len(li)):
    #Append elements accordingly
    index = int(idx%3)
    res[index].append(li[idx])

print(res)

輸出看起來像

[[1, 4, 7], 
[2, 5, 8], 
[3, 6, 9]]

Answer 3

除了其他人發布的內容之外，您還可以創建一個以列表名稱為鍵、以列表為值的字典，如下所示：

>>> for i in range(3):
...   d["list_{}".format(i)] = [list[i], list[i+3], list[i+6]]
... 
>>> d
{'list_2': [3, 6, 9], 'list_1': [2, 5, 8], 'list_0': [1, 4, 7]}```

Answer 4

有沒有人考慮過這一點？

>>> list=[1,2,3,4,5,6,7,8,9]
>>> list_1 = list[0::3]
>>> list_2 = list[1::3]
>>> list_3 = list[2::3]
>>> list_1
[1, 4, 7]
>>> list_2
[2, 5, 8]
>>> list_3
[3, 6, 9]

一個循環看起來像這樣

for i in range(0,3):
    list_i = list[i::3]
    print(list_i)