检查列表是否包含 1、2 或 3

Question

The function below always returns false. I want to take a list as input and then find if it contains 1,2 and 3?

def arrayCheck(nums):
    # CODE GOES HERE
    if (1 in nums) and (2 in nums) and (3 in nums):
        return True
    else:
        return False

arr = input()  # takes a list as input
x = set(arr)  # coverts it into array
result = arrayCheck(x)
print(result)

I want to search the numbers 1,2 and 3 in a user defined list. So I converted the list into a set for having unique elements. Further I have used in method to search for the items. The problem is it returns false every time.

Answer 1

Your function works, you are inputting wrong data ( arrayCheck([1, 2, 3]) returns True ).

You can't take a list as input. Converting a string to a set will separate the characters to a set, which isn't what you want ( set("hello") == {'h', 'l', 'o', 'e'} ), so you have to input a list of ints.

Something like this:

inp = input().split(",")
inp = [int(i) for i in inp] # Convert all the values to ints
print(arrayCheck(inp))

and providing 1,2,3,4 as input will work.

By the way, you can return bools in functions:

def arrayCheck(nums):
    return (1 in nums) and (2 in nums) and (3 in nums)