关于Python的一个函数

Question

I'm a absolute beginner, and when I made a function to count the number of even ints on a given list, it didn't went as expected, and I can't see where I'm doing it wrong.

nums = [2,2,4,4,5,6,7,8,9]

def count_even(nums):
    for number in nums:
        num = 0
        if number % 2 == 0:
            num += number
            return num
        else:
            continue

The output is:

count_even(nums)
2

It stops on nums[1] for some obscure reason. Or it just prints the first "2" and adds it, and I don't know how to fix it, yet.

Answer 1

You have three problems here.

• You're setting num = 0 every time through the loop, instead of just once at the start. So you're only going to get the very last count that you do.

• You're doing num += number , instead of num += 1 , so instead of counting the even numbers, you're adding them.

• You're doing a return num as soon as the first even number is found, instead of only at the end of the function, after the loop. (And that also means that if there are no even numbers, instead of returning 0 , you return None ).

While we're at it, you don't need else: continue , because continuing is already what happens by default when you fall off the end of a loop.

---

Anyway, this means it's not stopping at nums[1] , it's stopping at nums[0] —but it's adding 2 instead of 1 there, which makes things confusing. (It's always fun when bugs interact like that. Even more fun when they happen to exactly cancel out for your test case, like if you did nums = [6,2,4,4,5,6,7,8,9] and got back 6 and thought everything was working…)

---

So:

def count_even(nums):
    num = 0
    for number in nums:
        if number % 2 == 0:
            num += 1
    return num

Answer 2

Your function stops on nums[1] because the return keyword exits the function and returns num (which is set to 0 + number ). If you want to print out the sum of all the even numbers in nums you could move the return statement to the end and take the num = 0 expression out of the for loop (because it will reset to 0 after each iteration), like this:

def count_even(nums):
    num = 0
    for number in nums:
        if number % 2 == 0:
            num += number
        else:
            continue
    return num

Also, your else statement is redundant here, so you could simply remove it, leaving you with the following:

def count_even(nums):
    num = 0
    for number in nums:
        if number % 2 == 0:
            num += number
    return num

And you could simplify that further to:

def count_even(nums):
    evens = [number for number in nums if number % 2 == 0]
    return sum(evens)

Answer 3

Another solution:

def count_even(nums):
    return len([i for i in nums if i % 2 == 0])

Answer 4

There are a couple of problems in your code:

• Your num needs be defined outside the loop, so you can count every even number found. If you define it inside the loop, it resets to 0 every iteration.
• When you find an even number, you increment it towards the counter num . You need to instead increment num by 1.
• You return immediately when a even number is found, this means the function exits on the first even number. You instead want to return the total num at the end.
• You have an unnecessary continue in the else block. If a uneven number is found, simply ignore it.

Which these suggestions, you could implement your code like this:

nums = [2,2,4,4,5,6,7,8,9]

def count_even(nums):
    num = 0

    for number in nums:
        if number % 2 == 0:
            num += 1

    return num

print(count_even(nums))
# 6