Python：返回 None 而不是 False

Question

The below code is returning None instead of False . Please help me understand the reason.

def same_first_last(nums):
    for i in nums:
        if nums[0] == nums[-1] and len(nums) >= 1:
            return True
        else:
            return False


print(same_first_last([]))

Any help is appreaciated !!!

Answer 1

There are two scenarios based on your question: -

1. Size of nums > 0:

   If nums list is non-empty, then it will either satisfy your if condition ie nums[0] == nums[-1] and len(nums) >= 1 in which case it will return True or else it will return False

2. Size of nums == 0

   If this is the case then code doesn't even go inside the for loop in which case It will not be able to go inside the if else block, that's why python returns None by default, so your code is also returning None.

   To avoid this you can change the code as below:- ( You don't even need to go inside the for loop as your if else conditions are static ie not changing with the iterator i from the for loop of your question.

def same_first_last(nums):
    return nums and nums[0] == nums[-1]

print(same_first_last([]))

here in the above return nums and nums[0] == nums[-1] , it a join of two conditions if nums and if nums[0] == nums[-1]

a) if nums is checking if the list or the string you are iterating is empty or not. If empty it return False else True

b) if nums[0] == nums[-1] : - It's checking for the condition from your question (checking if the front element and the last element are equal or not)

When both the above conditions met, then only it will return True or else it will return False

Answer 2

Since you are calling your method with an empty list, you are never entering the for-loop in your method and therefor you can never reach your return-statement. when your method is called with an empty List, your for-loop is skipped completely. You can change your method like this:

def same_first_last(nums): 
    for i in nums: 
        if nums[0] == nums[-1] and len(nums) >= 1: 
            return True
        else:
            return False
    return <enter default value here>

But you dont even need the loop, so you can simplify your code even a little bit more:

def same_first_last(nums): 
    if len(nums) > 0: 
        return nums[0] == nums[-1]
    else:
        return False

Answer 3

You are not entering the loop, so the function effectively has no return value. This can be verified by printing something inside the loop:

def same_first_last(nums):
    for i in nums:
        print("in the loop")
        if nums[0] == nums[-1] and len(nums) >= 1:
            return True
        else:
            return False


print(same_first_last([]))

You can actually make this code work by swapping the conditions in the if() statement and removing the for loop like so:

def same_first_last(nums):
    return len(nums) >= 1 and nums[0] == nums[-1]

print(same_first_last([]))