如何獲得 python 列表的唯一嵌套列表？

Question

我的問題與嵌套在主列表中的列表的唯一性有關。 具體來說，我有以下結構，它是一個包含列表的lists ：

lst = [  [
            [14], ['walk'], [69], ['walk'], [7]
         ],
         [
           [14], ['walk'], [69], ['walk'], [7]
          ],
        [
          [30], [948], [75], [947], [10]
        ],
        [
          [14]
        ]
   ]

為了獲得列表的獨特元素，我嘗試了 το 實現以下內容：

unique_data = [list(x) for x in set(tuple(x) for x in lst)]但我收到以下錯誤：

TypeError: unhashable type: 'list' 。

預期的 output 如下：

[[[14], ['walk'], [69], ['walk'], [7]],

[[30], [948], [75], [947], [10]],

[[14]]]

有什么建議么？

Answer 1

嘗試：

lst = [
    [[14], ["walk"], [69], ["walk"], [7]],
    [[14], ["walk"], [69], ["walk"], [7]],
    [[30], [948], [75], [947], [10]],
    [[14]],
]

out, seen = [], set()
for subl in lst:
    t = tuple(tuple(i) for i in subl)
    if t not in seen:
        seen.add(t)
        out.append(subl)

print(out)

印刷：

[
    [[14], ["walk"], [69], ["walk"], [7]],
    [[30], [948], [75], [947], [10]],
    [[14]],
]

Answer 2

如果你創建一個字典，並將值保存為鍵，那么如果順序很重要，你可以解決這個問題

lst = [[[14], ['walk'], [69], ['walk'], [7]], [[14], ['walk'], [69], ['walk'], [7]], [[30], [948], [75], [947], [10]], [[14]]]
 
unique = {}
for sublist in lst:
    key = ''
    for sub_sublist in sublist:
            for value in sub_sublist:
                    key += '_' + str(value)
    unique[key] = sublist
 
print(unique)
# output -> {'_14_walk_69_walk_7': [[14], ['walk'], [69], ['walk'], [7]], '_30_948_75_947_10': [[30], [948], [75], [947], [10]], '_14': [[14]]}

result = list(unique.values())
print(result)
# output. -> [[[14], ['walk'], [69], ['walk'], [7]], [[30], [948], [75], [947], [10]], [[14]]]

Answer 3

為了使用set() ，您需要將所有列表（和子列表）轉換為元組。

試試這個代碼：

unique = tuple([tuple([tuple(x) for x in y]) for y in lst])
unique = set(unique)