python - 如何在 python 中重复查找每个 T 数字的数字总和的数字总和，直到它成为一个数字？

Question

这是我编写的代码，用于从重复次数的数字中获取数字总和，直到总和低于 10:

T = int(input())
for i in range(T):
    N = int(input())
    def P():
            M = [int(d) for d in str(N)]
            N = sum(M)
            if N<10:
                    print(N)
            else :
                    return P()
    P()

在运行这段代码时，它给我一个错误，如:

 Traceback (most recent call last): 
 File"C:/Users/AdityaShrivastava/AppData/Roaming/Python/Python36/Scripts/tes 
 ting.py", line 11, in <module>
 P()
 File "C:/Users/Aditya 
 Shrivastava/AppData/Roaming/Python/Python36/Scripts/testing.py", line 5, in 
 P
 M = [int(d) for d in str(N)]
 UnboundLocalError: local variable 'N' referenced before assignment

Answer 1

您正在使用递归来解决这个问题。简单地使用循环更有效:

def gimmeNumber(text):
    """"Helper: Asks for input until valid integer number is inputted. Retuns the number"""
    while True:
        T = input(text).strip()
        if T and T.isdigit():
            T = int(T)
            break
        print("Thats not a number ...")
    return T

def sumDigits(number):
    return sum(int(x) for x in str(number))

T = gimmeNumber("How many numbers? ")
for _ in range(T):
    s = 0
    N = gimmeNumber("Give me a number: ")
    # calculate the cross-sum
    s = sumDigits(N)
    while s > 9: # repeat while greater then 9  
        s = sumDigits(s)

    print(s)

输入:4，然后是 999,888,333,111

输出:

9
6
9
3

---

正如@Arne 建议的那样，将 gimmeNumber(text) 改为使用 try/except 更适合 python Ask forgiveness not permission心态，我同意。

不过，上面的变体也可以工作，而且对于初学者来说更容易理解。这是 try/except 一个:

def gimmeNumber(text):
    """"Helper: Asks for input until valid integer number is inputted. Retuns the number"""
    while True:
        try:
            T = int(input(text).strip())
            break
        except ValueError as e:
            print("Thats not a number ...")
    return T

有关输入验证的更多信息，我建议阅读 Asking the user for input until they give a valid response 的答案。 .