Python：确定未排序的列表是否包含在“列表的列表”中，无论元素的顺序如何

我有一个与这个问题类似的问题：确定 2 个列表是否具有相同的元素，无论顺序如何？ https://stackoverflow.com/questions/8866652/determine-if-2-lists-have-the-same-elements-regardless-of-order

确定未排序列表的最佳/最快方法是什么list1包含在“列表的列表”中myListOfLists，无论元素的顺序如何list1？我的尝试包含在函数中doSomething(...)我多次称之为：

def doSomething(myListOfLists, otherInputs):

    list1 = []
    ...  # do something here with `otherInputs' 
    ...  # which gives `list1' some values

    # now only append `list1' to `myListOfLists' if it doesn't already exist
    # and if it does exist, remove it

    removeFromList = False
    for myList in myListOfLists:
        if sorted(list1) == sorted(myList):
            removeFromList = True
            break

    if removeFromList:
        myListOfLists.remove(list1)
    else:
        myListOfLists.append(list1)

    return myListOfLists

问题是我需要运行该函数doSomething(...)大约1.0e5倍。作为myListOfLists每次调用都会变得更大doSomething(...)这变得非常耗时。

EDIT:

对任务的一些澄清。让我在这里举一个所需输出的示例：

a = []
doSomething(a, [1,2,3])
>> a = [1,2,3]

Because [1,2,3]不在a，它被附加到a.

doSomething(a, [3,4,5])
>> a = [[1,2,3], [3,4,5]]

Because [3,4,5]不在a，它被附加到a.

doSomething(a, [1,2,3])
>>[3,4,5]

Because [1,2,3] is in a，它被从a.

EDIT:

所有列表都具有相同的长度。

您可以在此处使用集合：

def doSomething(myListOfLists, otherInputs):
    s = set(otherInputs)           #create set from otherInputs
    for item in myListOfLists: 
        #remove the common items between `s` and current sublist from `s`.
        s -= s.intersection(item) 
        #if `s` is empty, means all items found. Return True
        if not s:                  
            return True
    return not bool(s)
... 
>>> doSomething([[1, 2, 7],[6, 5, 4], [10, 9, 10]], [7, 6, 8])
False
>>> doSomething([[1, 2, 7],[6, 5, 4], [10, 8, 10]], [7, 6, 8])
True

更新1：任何子列表包含完全相同的项目otherInputs.

def doSomething(myListOfLists, otherInputs):
    s = set(otherInputs)
    return any(set(item) == s for item in myListOfLists)
... 
>>> doSomething([[6, 8, 7],[6, 5, 4], [10, 8, 10]], [7, 6, 8])
True
>>> doSomething([[1, 2, 7],[6, 5, 4], [10, 8, 10]], [7, 6, 8])
False

更新2： otherInputs是任何子列表的子集：

def doSomething(myListOfLists, otherInputs):
    s = set(otherInputs)
    return any(s.issubset(item) for item in myListOfLists)
... 
>>> doSomething([[6, 8, 7],[6, 5, 4], [10, 8, 10]], [7, 6, 8])
True
>>> doSomething([[6, 8, 7, 10],[6, 5, 4], [10, 8, 10]], [7, 6, 8])
True
>>> doSomething([[1, 2, 7],[6, 5, 4], [10, 8, 10]], [7, 6, 8])
False