Python: Determine if an unsorted list is contained

Question

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

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

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.

EDIT:

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]

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]]

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

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

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

EDIT:

All lists have the same length.

Answer 1

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

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

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

Answer 2

This algorithm appears to be slightly faster:

l1 = [3, 4, 1, 2, 3]
l2 = [4, 2, 3, 3, 1]
same = True
for i in l1:
    if i not in l2:
        same = False
        break

For 1000000 loops, this takes 1.25399184227 sec on my computer, whilst

same = sorted(l1) == sorted(l2)

takes 1.9238319397 sec.

Answer 3

If you sort given list and then append it to myListOfLists you can use this:

if sorted(list1) in myListOfLists:

Answer 4

Use sets

def doSomething(myDictOfLists, 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
    list1Set = set(list1)
    if list1Set not in myDictOfLists:
        myDictOfLists[list1Set] = list1

    return myDictOfLists