Python 如何有效地发现一个元组是否是另一个元组的子集

我知道

set

实现非常棒。然而，它有一个巨大的局限性：它不能处理一个项目的重复

让我们使用整数：

tuple1 = (100, 100, 375)
layer = [(100, 100, 100, 375), (20, 100, 100, 375), (20, 20, 100, 375), (20, 20, 20, 375), (20, 20, 30, 375), (20, 20, 40, 375), (20, 20, 50, 375), (20, 20, 60, 375), (20, 20, 70, 375), (20, 20, 80, 375), (20, 30, 100, 375), (20, 30, 30, 375), (20, 30, 40, 375), (20, 30, 50, 375), (20, 30, 60, 375), (20, 30, 70, 375), (20, 30, 80, 375), (20, 40, 100, 375), (20, 40, 40, 375), (20, 40, 50, 375), (20, 40, 60, 375), (20, 40, 70, 375), (20, 40, 80, 375), (20, 50, 100, 375), (20, 50, 50, 375), (20, 50, 60, 375), (20, 50, 70, 375), (20, 50, 80, 375), (20, 60, 100, 375), (20, 60, 60, 375), (20, 60, 70, 375), (20, 60, 80, 375), (20, 70, 100, 375), (20, 70, 70, 375), (20, 70, 80, 375), (20, 80, 100, 375), (20, 80, 80, 375), (30, 100, 100, 375), (30, 30, 100, 375), (30, 30, 30, 375), (30, 30, 40, 375), (30, 30, 50, 375), (30, 30, 60, 375), (30, 30, 70, 375), (30, 30, 80, 375), (30, 40, 100, 375), (30, 40, 40, 375), (30, 40, 50, 375), (30, 40, 60, 375), (30, 40, 70, 375), (30, 40, 80, 375), (30, 50, 100, 375), (30, 50, 50, 375), (30, 50, 60, 375), (30, 50, 70, 375), (30, 50, 80, 375), (30, 60, 100, 375), (30, 60, 60, 375), (30, 60, 70, 375), (30, 60, 80, 375), (30, 70, 100, 375), (30, 70, 70, 375), (30, 70, 80, 375), (30, 80, 100, 100), (30, 80, 100, 375), (30, 80, 80, 375), (40, 100, 100, 375), (40, 40, 100, 375), (40, 40, 40, 375), (40, 40, 50, 375), (40, 40, 60, 375), (40, 40, 70, 375), (40, 40, 80, 375), (40, 50, 100, 375), (40, 50, 50, 375), (40, 50, 60, 375), (40, 50, 70, 375), (40, 50, 80, 375), (40, 60, 100, 375), (40, 60, 60, 375), (40, 60, 70, 375), (40, 60, 80, 375), (40, 70, 100, 375), (40, 70, 70, 375), (40, 70, 80, 375), (40, 80, 100, 375), (40, 80, 80, 375), (50, 100, 100, 375), (50, 50, 100, 375), (50, 50, 50, 375), (50, 50, 60, 375), (50, 50, 70, 375), (50, 50, 80, 375), (50, 50, 80, 80), (50, 60, 100, 375), (50, 60, 60, 375), (50, 60, 70, 375), (50, 60, 80, 375), (50, 70, 100, 375), (50, 70, 70, 375), (50, 70, 70, 80), (50, 70, 80, 375), (50, 80, 100, 375), (50, 80, 80, 375), (50, 80, 80, 80), (60, 100, 100, 375), (60, 60, 100, 375), (60, 60, 60, 375), (60, 60, 70, 375), (60, 60, 80, 375), (60, 70, 100, 375), (60, 70, 70, 375), (60, 70, 80, 375), (60, 80, 100, 375), (60, 80, 80, 375), (60, 80, 80, 80), (70, 100, 100, 375), (70, 70, 100, 375), (70, 70, 70, 375), (70, 70, 80, 375), (70, 80, 100, 100), (70, 80, 100, 375), (70, 80, 80, 375), (80, 100, 100, 375), (80, 80, 100, 375), (80, 80, 80, 100), (80, 80, 80, 375)]

我想保留

层

中的项目，这些项目不是由

tuple1

的子集构成的。i、 例如，前两个必须去掉

现在我正在为循环使用

：

new_layer = list()
for elt in layer:
    copy = list(elt)
    for x in tuple1:
        if x in copy:
            copy.remove(x)
    if len(copy) == 1:
        continue
    else:
        new_layer.append(elt)

还有比这更好的解决办法。。。此外，最后一个问题实际上更为复杂

我有5层：


第1层：
len的元素：2

第二层：
len
的元素：3
第3层：
len的元素：4

第4层：
len的元素
：5
第5层：
len的元素
：6

目标是在层
N
中除去从层
N-k
派生的元素（由子集构成）

感谢您的帮助：）
这里有一种方法，即子类化和实现
\uuuuuu包含
（对于
操作符中的
：

输出：

[(100, 100, 100, 375), 
 (20, 100, 100, 375), 
 (30, 100, 100, 375), 
 (40, 100, 100, 375), 
 (50, 100, 100, 375), 
 (60, 100, 100, 375), 
 (70, 100, 100, 375), 
 (80, 100, 100, 375)
]
[(100, 100, 100, 375), 
(20, 100, 100, 375), 
(30, 100, 100, 375), 
(40, 100, 100, 375), 
(50, 100, 100, 375), 
(60, 100, 100, 375), 
(70, 100, 100, 375), 
(80, 100, 100, 375)
]
编辑：注意，这只关心元组中值的频率，而不关心它们在元组中的相对位置。因此，它也将匹配（375100100），即使元组的元素顺序相反。
如果您认为它太宽，您可以添加注释吗？我们的目标是找到
layer1
的元素，这些元素不包括
tuple1
作为子集。它不能处理项目的重复。。。集合是专门为不包含重复项而制作的。@Eldelshell是的，我知道。当需要在较大的元素中查找子集时，它们非常方便。但不能将它们用于重复项目的元素。因此，我正在寻找另一种比我的解决方案更优雅、性能更好的解决方案。太好了，我实际上不在乎订单，所以更好：）谢谢！我来试试这个！好的，它工作得很好，很好。然而，已经有很长一段时间了，我不知道它是如何工作的。什么是
super（）
<代码>其他
是一个命令吗？什么是
self.get（）
调用？计数器在哪里计数？我很想学习更多关于实现的知识，因为在这方面我有很多新东西！从我得到的信息来看，
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu将实现
中的
操作符，就像
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuq>对
所做的那样，但我实际上不理解在另一个
中声明一个元素
会执行什么操作。关于
\u。在这种情况下，该方法做两件事。（1） 确保
x
in
x in y
中的
x也是一个TupleCounter对象（如果不是，它会创建一个），并且（2）比较“self”中每个元素出现的次数与“other”中每个元素出现的次数<代码>计数器
，您正在子类化的类可能值得研究。我已经使用了
计数器
，我对它非常熟悉。这是子类化的想法，听起来很有趣。无论如何，我需要更多的研究，非常感谢！
needle = (100, 100, 375)
layer = [(100, 100, 100, 375), (20, 100, 100, 375), (20, 20, 100, 375), (20, 20, 20, 375), (20, 20, 30, 375), (20, 20, 40, 375), (20, 20, 50, 375), (20, 20, 60, 375), (20, 20, 70, 375), (20, 20, 80, 375), (20, 30, 100, 375), (20, 30, 30, 375), (20, 30, 40, 375), (20, 30, 50, 375), (20, 30, 60, 375), (20, 30, 70, 375), (20, 30, 80, 375), (20, 40, 100, 375), (20, 40, 40, 375), (20, 40, 50, 375), (20, 40, 60, 375), (20, 40, 70, 375), (20, 40, 80, 375), (20, 50, 100, 375), (20, 50, 50, 375), (20, 50, 60, 375), (20, 50, 70, 375), (20, 50, 80, 375), (20, 60, 100, 375), (20, 60, 60, 375), (20, 60, 70, 375), (20, 60, 80, 375), (20, 70, 100, 375), (20, 70, 70, 375), (20, 70, 80, 375), (20, 80, 100, 375), (20, 80, 80, 375), (30, 100, 100, 375), (30, 30, 100, 375), (30, 30, 30, 375), (30, 30, 40, 375), (30, 30, 50, 375), (30, 30, 60, 375), (30, 30, 70, 375), (30, 30, 80, 375), (30, 40, 100, 375), (30, 40, 40, 375), (30, 40, 50, 375), (30, 40, 60, 375), (30, 40, 70, 375), (30, 40, 80, 375), (30, 50, 100, 375), (30, 50, 50, 375), (30, 50, 60, 375), (30, 50, 70, 375), (30, 50, 80, 375), (30, 60, 100, 375), (30, 60, 60, 375), (30, 60, 70, 375), (30, 60, 80, 375), (30, 70, 100, 375), (30, 70, 70, 375), (30, 70, 80, 375), (30, 80, 100, 100), (30, 80, 100, 375), (30, 80, 80, 375), (40, 100, 100, 375), (40, 40, 100, 375), (40, 40, 40, 375), (40, 40, 50, 375), (40, 40, 60, 375), (40, 40, 70, 375), (40, 40, 80, 375), (40, 50, 100, 375), (40, 50, 50, 375), (40, 50, 60, 375), (40, 50, 70, 375), (40, 50, 80, 375), (40, 60, 100, 375), (40, 60, 60, 375), (40, 60, 70, 375), (40, 60, 80, 375), (40, 70, 100, 375), (40, 70, 70, 375), (40, 70, 80, 375), (40, 80, 100, 375), (40, 80, 80, 375), (50, 100, 100, 375), (50, 50, 100, 375), (50, 50, 50, 375), (50, 50, 60, 375), (50, 50, 70, 375), (50, 50, 80, 375), (50, 50, 80, 80), (50, 60, 100, 375), (50, 60, 60, 375), (50, 60, 70, 375), (50, 60, 80, 375), (50, 70, 100, 375), (50, 70, 70, 375), (50, 70, 70, 80), (50, 70, 80, 375), (50, 80, 100, 375), (50, 80, 80, 375), (50, 80, 80, 80), (60, 100, 100, 375), (60, 60, 100, 375), (60, 60, 60, 375), (60, 60, 70, 375), (60, 60, 80, 375), (60, 70, 100, 375), (60, 70, 70, 375), (60, 70, 80, 375), (60, 80, 100, 375), (60, 80, 80, 375), (60, 80, 80, 80), (70, 100, 100, 375), (70, 70, 100, 375), (70, 70, 70, 375), (70, 70, 80, 375), (70, 80, 100, 100), (70, 80, 100, 375), (70, 80, 80, 375), (80, 100, 100, 375), (80, 80, 100, 375), (80, 80, 80, 100), (80, 80, 80, 375)]

needle = TupleCounter(needle)
filtered = [t for t in layer if needle in TupleCounter(t)]
print(filtered)

[(100, 100, 100, 375), 
 (20, 100, 100, 375), 
 (30, 100, 100, 375), 
 (40, 100, 100, 375), 
 (50, 100, 100, 375), 
 (60, 100, 100, 375), 
 (70, 100, 100, 375), 
 (80, 100, 100, 375)
]