Python:确定“列表列表"中是否包含未排序的列表，而与元素的顺序无关 [英] Python: Determine if an unsorted list is contained in a 'list of lists', regardless of the order to the elements

问题描述

我对此问题有一个类似的问题:

I have a similar question to this question: Determine if 2 lists have the same elements, regardless of order?

确定列表列表" myListOfLists中是否包含未排序列表list1的最佳/最快方法是什么，而不管list1中元素的顺序如何?我的尝试被包装在函数doSomething(...)中，该函数多次调用:

What is the best/quickest way to determine whether an unsorted list list1 is contained in a 'list of lists' myListOfLists, regardless of the order to the elements in list1? My attempt is wrapped up in the function doSomething(...) which I call many times:

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

此问题是我需要运行功能doSomething(...)大约1.0e5次.随着myListOfLists在每次调用doSomething(...)时变大，这变得非常耗时.

The problem with this is that I need to run the function doSomething(...) approximately 1.0e5 times. As myListOfLists gets bigger with every call of doSomething(...) this becomes massively time consuming.

对任务进行了一些澄清.让我在此处给出所需输出的示例:

Some clarification of the task. Let me give an example of the desired output here:

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

由于[1,2,3]不在a中，因此将其附加到a.

Because [1,2,3] is not in a, it is appended to a.

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

由于[3,4,5]不在a中，因此将其附加到a.

Because [3,4,5] is not in a, it is appended to a.

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

由于a中的[1,2,3] 是，因此已将其从a中删除.

Because [1,2,3] is in a, it is removed from a.

所有列表的长度相同.

推荐答案

您可以在此处使用设置:

You can use sets here:

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::任何子列表都包含与otherInputs完全相同的项目.

Update 1: Any Sublist contains exactly same items as 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:

Update 2: otherInputs is a subset of any of the sublist:

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

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