代入这个:

def lists_equal(l1: list, l2: list) -> bool:
    """

    import collections
    compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
    ref:
        - https://stackoverflow.com/questions/9623114/check-if-two-unordered-lists-are-equal
        - https://stackoverflow.com/questions/7828867/how-to-efficiently-compare-two-unordered-lists-not-sets
    """
    compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
    set_comp = set(l1) == set(l2)  # removes duplicates, so returns true when not sometimes :(
    multiset_comp = compare(l1, l2)  # approximates multiset
    return set_comp and multiset_comp  #set_comp is gere in case the compare function doesn't work

您可以编写自己的函数来比较这些列表。

让我们得到两个列表。

list_1=['John', 'Doe'] 
list_2=['Doe','Joe']

首先，我们定义一个空字典，计数列表项并写入字典。

def count_list(list_items):
    empty_dict={}
    for list_item in list_items:
        list_item=list_item.strip()
        if list_item not in empty_dict:
            empty_dict[list_item]=1
        else:
            empty_dict[list_item]+=1
    return empty_dict


        

之后，我们将使用下面的函数比较两个列表。

def compare_list(list_1, list_2):
    if count_list(list_1)==count_list(list_2):
        return True
    return False
compare_list(list_1,list_2)

代入这个:

def lists_equal(l1: list, l2: list) -> bool:
    """

    import collections
    compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
    ref:
        - https://stackoverflow.com/questions/9623114/check-if-two-unordered-lists-are-equal
        - https://stackoverflow.com/questions/7828867/how-to-efficiently-compare-two-unordered-lists-not-sets
    """
    compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
    set_comp = set(l1) == set(l2)  # removes duplicates, so returns true when not sometimes :(
    multiset_comp = compare(l1, l2)  # approximates multiset
    return set_comp and multiset_comp  #set_comp is gere in case the compare function doesn't work

你可以对两者进行排序:

sorted(a) == sorted(b)

计数排序也可能更有效(但它要求对象是可哈希的)。

>>> from collections import Counter
>>> a = [1, 2, 3, 1, 2, 3]
>>> b = [3, 2, 1, 3, 2, 1]
>>> print (Counter(a) == Counter(b))
True

O(n): Counter()方法是最好的(如果你的对象是可哈希的):

def compare(s, t):
    return Counter(s) == Counter(t)

O(n log n): sorted()方法是次优方法(如果你的对象是可排序的):

def compare(s, t):
    return sorted(s) == sorted(t)

O(n * n):如果对象既不是可哈希的，也不是可排序的，你可以使用相等:

def compare(s, t):
    t = list(t)   # make a mutable copy
    try:
        for elem in s:
            t.remove(elem)
    except ValueError:
        return False
    return not t

最好的方法是对列表进行排序并进行比较。(使用Counter对不可哈希的对象无效。)这对于整数来说很简单:

sorted(a) == sorted(b)

对于任意对象，这就有点棘手了。如果您关心对象的身份，即相同的对象是否在两个列表中，您可以使用id()函数作为排序键。

sorted(a, key=id) == sorted(b, key==id)

(In Python 2.x you don't actually need the key= parameter, because you can compare any object to any object. The ordering is arbitrary but stable, so it works fine for this purpose; it doesn't matter what order the objects are in, only that the ordering is the same for both lists. In Python 3, though, comparing objects of different types is disallowed in many circumstances -- for example, you can't compare strings to integers -- so if you will have objects of various types, best to explicitly use the object's ID.)

另一方面，如果你想通过值比较列表中的对象，首先你需要定义“值”对于对象的意义。然后需要某种方法将其作为键提供(对于Python 3，作为一致类型)。一种适用于许多任意对象的潜在方法是根据它们的repr()进行排序。当然，这可能会浪费大量额外的时间和内存，为大型列表构建repr()字符串等等。

sorted(a, key=repr) == sorted(b, key==repr)

如果对象都是你自己的类型，你可以在它们上定义__lt__()，这样对象就知道如何将自己与其他对象进行比较。然后你就可以对它们进行排序，而不用担心key=参数。当然，你也可以定义__hash__()并使用Counter，这将更快。