python不能返回变量但可以打印

所以我遇到了一个奇怪的问题。

def findFourPlus(itemCount, seq, goal): 
    goalDifference = float("inf") 
    closestPartial = [] 
    hello = subset_sum(itemCount, seq, goal, goalDifference, closestPartial, partial=[]) 
    return hello #doesn't return value from subset_sum() 
 
def subset_sum(itemCount, seq, goal, goalDifference, closestPartial, partial): 
    s = sum(partial) 
 
    # check if the partial sum is equals to target 
    if(len(partial) == itemCount): 
        if s == goal: 
                print("FOUND YAA") 
                return partial #right now doesn't return anything. I intend for it to break out of this function as soon as it finds one pair of solution.  
 
 
        else: 
            if( abs(goal - s) < goalDifference): 
                #print(abs(goal-s), goalDifference, closestPartial) 
                goalDifference = abs(goal - s) 
                closestPartial[:] = partial 
                #print(abs(goal-s), goalDifference, closestPartial) 
                return closestPartial 
 
    for i in range(len(seq)): 
        n = seq[i] 
        remaining = seq[i+1:] 
        print(subset_sum(itemCount, remaining, goal, goalDifference, closestPartial, partial + [n])) 

在 subset_sum() 函数中，我可以打印出 partial 变量，它会返回给我正确的值

>>> findFourPlus(3, [1,2,3,4,5,6,7,8,9,10], 20) 
FOUND YAA 
[1, 9, 10] 
FOUND YAA 
[2, 8, 10] 
FOUND YAA 
[3, 7, 10] 
FOUND YAA 
[3, 8, 9] 
FOUND YAA 
[4, 6, 10] 
FOUND YAA 
[4, 7, 9] 
FOUND YAA 
[5, 6, 9] 
FOUND YAA 
[5, 7, 8] 

但是，如果我将 print(partial) 行更改为 return partial，它仍然会打印出 FOUNDYAA 行，因为它已打印，但不会返回部分

>>> findFourPlus(3, [1,2,3,4,5,6,7,8,9,10], 20) 
FOUND YAA 
FOUND YAA 
FOUND YAA 
FOUND YAA 
FOUND YAA 
FOUND YAA 
FOUND YAA 
FOUND YAA 

在编写 findFourPlus 和 subset_sum 之前，我编写了其他函数，当我尝试返回某些内容时，这些函数运行良好。我在这里做错了什么？

编辑:我在这里似乎让读者感到困惑。我最终要做的是通过调用 findFourPlus() 函数将整数集(列表)存储在一个变量中。假设我想看看列表 [1,2,3,4,5] 中的任意两个加起来是否等于 5，我会调用 findFourPlus(2, [1,2,3,4 ,5], 5)，它有望返回给我 [1, 4]。有了它，我可以通过调用将答案 [1,4] 存储到一个变量中，比如答案

answer = findFourPlus(2, [1,2,3,4,5], 5) // if I call answer, it'll return to me [1,4] 

一个例子:

>>> a = findFourPlus(2, [1,2,3,4,5], 6) # prints out everything, which is good for test cases 
[1, 2] 
[1, 3] 
[1, 4] 
FOUND YAA 
[1, 5] 
None 
[2, 3] 
FOUND YAA 
[2, 4] 
[2, 5] 
None 
[3, 4] 
[3, 5] 
None 
[4, 5] 
None 
None 
>>> a # doesn't return anything when called 
>>>  

打印都是为了确保我的逻辑是正确的。很抱歉造成混淆，但希望这能澄清问题!

请您参考如下方法:

当你执行一个函数时，它不会自动将它的返回值转储到输出中，所以在你的 for 循环中，

for i in range(len(seq)): 
    n = seq[i] 
    remaining = seq[i+1:] 
    print(subset_sum(itemCount, remaining, goal, goalDifference, closestPartial, partial + [n]))