另一种解决方案是使用十字路口。

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

set(one).intersection(set(two)) == set(one)

集合的交点包含集合1

(OR)

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

set(one) & (set(two)) == set(one)

大多数解决方案都认为列表没有副本。如果你的列表确实有重复，你可以试试这个:

def isSubList(subList,mlist):
    uniqueElements=set(subList)
    for e in uniqueElements:
        if subList.count(e) > mlist.count(e):
            return False     
    # It is sublist
    return True

它确保子列表的元素永远不会与list不同，或者公共元素的数量更多。

lst=[1,2,2,3,4]
sl1=[2,2,3]
sl2=[2,2,2]
sl3=[2,5]

print(isSubList(sl1,lst)) # True
print(isSubList(sl2,lst)) # False
print(isSubList(sl3,lst)) # False

如果你在问一个列表是否“包含”在另一个列表中，那么:

>>>if listA in listB: return True

如果你想知道listA中的每个元素在listB中是否有相等数量的匹配元素，试试:

all(True if listA.count(item) <= listB.count(item) else False for item in listA)

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

all(x in two for x in one)

解释:生成器通过遍历列表1检查该项是否在列表2中来创建布尔值。all()如果每一项都为真则返回True，否则返回False。

还有一个好处是，所有元素都在缺失元素的第一个实例上返回False，而不必处理每个元素。

如果我迟到了，请原谅。；）

为了检查集合A是否是集合B的子集，Python有a.is子集(B)和A <= B。它只对集合起作用，而且工作得很好，但内部实现的复杂性是未知的。参考:https://docs.python.org/2/library/sets.html #集对象

我提出了一种算法来检查列表A是否是列表B的子集。

To reduce complexity of finding subset, I find it appropriate to sort both lists first before comparing elements to qualify for subset. It helped me to break the loop when value of element of second list B[j] is greater than value of element of first list A[i]. last_index_j is used to start loop over list B where it last left off. It helps avoid starting comparisons from the start of list B (which is, as you might guess unnecessary, to start list B from index 0 in subsequent iterations.) Complexity will be O(n ln n) each for sorting both lists and O(n) for checking for subset. O(n ln n) + O(n ln n) + O(n) = O(n ln n). Code has lots of print statements to see what's going on at each iteration of the loop. These are meant for understanding only.

检查一个列表是否是另一个列表的子集

is_subset = True;

A = [9, 3, 11, 1, 7, 2];
B = [11, 4, 6, 2, 15, 1, 9, 8, 5, 3];

print(A, B);

# skip checking if list A has elements more than list B
if len(A) > len(B):
    is_subset = False;
else:
    # complexity of sorting using quicksort or merge sort: O(n ln n)
    # use best sorting algorithm available to minimize complexity
    A.sort();
    B.sort();

    print(A, B);

    # complexity: O(n^2)
    # for a in A:
    #   if a not in B:
    #       is_subset = False;
    #       break;

    # complexity: O(n)
    is_found = False;
    last_index_j = 0;

    for i in range(len(A)):
        for j in range(last_index_j, len(B)):
            is_found = False;

            print("i=" + str(i) + ", j=" + str(j) + ", " + str(A[i]) + "==" + str(B[j]) + "?");

            if B[j] <= A[i]:
                if A[i] == B[j]:
                    is_found = True;
                last_index_j = j;
            else:
                is_found = False;
                break;

            if is_found:
                print("Found: " + str(A[i]));
                last_index_j = last_index_j + 1;
                break;
            else:
                print("Not found: " + str(A[i]));

        if is_found == False:
            is_subset = False;
            break;

print("subset") if is_subset else print("not subset");

输出

[9, 3, 11, 1, 7, 2] [11, 4, 6, 2, 15, 1, 9, 8, 5, 3]
[1, 2, 3, 7, 9, 11] [1, 2, 3, 4, 5, 6, 8, 9, 11, 15]
i=0, j=0, 1==1?
Found: 1
i=1, j=1, 2==1?
Not found: 2
i=1, j=2, 2==2?
Found: 2
i=2, j=3, 3==3?
Found: 3
i=3, j=4, 7==4?
Not found: 7
i=3, j=5, 7==5?
Not found: 7
i=3, j=6, 7==6?
Not found: 7
i=3, j=7, 7==8?
not subset

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

set(x in two for x in one) == set([True])

如果list1在list 2中:

(x in two for x in one)生成一个True列表。 当我们做一个集合(x in two for x in one)只有一个元素(True)。