在

elif(x in cube_nums for x in list2):
  print("Cubes are only present")

将elif改为if，并将else也改为if。

当我们使用一个elif时，它只会执行之前所有不满足的if和elif条件。

pattern_exist=False
if (x in square_nums for x in list2):
  pattern_exist=True
  print("Squares are present")

if(x in cube_nums for x in list2):
  pattern_exist=True
  print("Cubes are present")

if (!pattern_exist):
    print("No such pattern is present")

如果满足if条件，则只需终止，它不会运行elif块。您可以转换为if条件。
您可以看到我使用set()的另一种方法

• • 代码：-**

def list_oper(list1,list2):

    square_nums = set(map(lambda x: x ** 2, list1))
    cube_nums = set(map(lambda x: x ** 3, list1))
    #print(square_nums)
    #print(cube_nums)
    if not square_nums.difference(set(list2)) and not cube_nums.difference(set(list2)):
        return "Squares and Cubes both are present"
    elif not square_nums.difference(set(list2)):
        return "Squares are only present"
    elif not cube_nums.difference(set(list2)):
        return "Cubes are only present"
    else:
        return "No such pattern present"
    

list1 = list(map(int,input("Enter the list1: ").split()))
list2 = list(map(int,input("Enter the list2: ").split()))
    
print(list_oper(list1,list2))

测试用例1：当立方体和正方形都存在时

Enter the list1: 1 2 3
Enter the list2: 1 4 9 8 27
Squares and Cubes both are present

测试用例2：当存在正方形时

Enter the list1: 1 2 3
Enter the list2: 1 4 9
Squares are only present

测试用例3：当存在立方体时

Enter the list1: 1 2 3
Enter the list2: 1 4 8 27
Cubes are only present

测试用例4：当不满足任何条件时!!

Enter the list1: 1 2 4
Enter the list2: 1 4 9
No such pattern present

正确，1 ** 2 == 1如果您只想打印所有匹配的列表，

if all(x in square_nums for x in list2):
  print("Squares are only present")

elif语句。