用 Python 自动化无聊的东西 - Collat​​z 序列

Question

我对任何类型的编码都很陌生，目前使用的是 python 3.3。我已经成功地在 python 中准确地运行了 Collat​​z 序列：

while True: # The main game loop.
number = int(input('Enter number:\n'))  
def collatz(number):
    while number !=1:
        if number % 2==0: #even numbers
            number=number//2
            print(number)
        elif number % 2!=0: #odd numbers
            number=number*3+1
            print(number)
collatz(number)

但是，我不确定如何以及在何处添加 ValueError strong，因为当用户输入非整数时，如下所示：

except ValueError:
    print('Only integers accepted.')

我对 python 很陌生，所以如果有任何答案可以有一点解释，我会非常感激。谢谢

Answer 1

我假设您指的是Exception Handling，验证部分必须在开头完成。

    while True: # The main game loop.
        
        try:
            number = int(input('Enter number:\n'))
   
        except ValueError:
            print("Only integers accepted! Please try again ...")
        else:
            collatz(number)
#output:
#
#Enter number:
#abc
#Only integers accepted! Please try again ...
#Enter number:
#5
#16
#8
#4
#2
#1
#Enter number:

但程序会继续循环，需要终止条件。

Answer 2

def collatz():

    try:
        number = int(input("Enter number: "))

        while True:
            if number == 1 or number == 0:
                break

            elif number % 2 == 0:
                number = number // 2
                print(number)

            elif number % 2 == 1:
                number = 3 * number + 1
                print(number)

    except:
        print("Enter in a valid number")


collatz()

Answer 3

试试这个：

def collatz(number):
    if number % 2 == 0:
        print(number // 2)
        return number // 2
    else:
        print(3 * number + 1)
        return 3 * number + 1

while True:
    try:
        number = int(input("Enter a Number or type 0 to exit: "))
        if number == 0:
            break
        while True:
            if number != 1:
                number = collatz(number)
            else:
                break
    except ValueError:
        print("Enter an integer number")

Answer 4

number = None

while number != int():
    try:
        number = int(input("Enter number: "))
        break
    except:
        print("Enter a valid Number")

def collatz(number):
    while number != 1:
        if number % 2 == 0:
            number = number // 2
            print(number)
        else:
            number = (3 * number + 1)
            print(number)

collatz(number)

Answer 5

把它放在最顶端。参数约束应始终尽快发生，这样您就不会浪费时间运行您将出错的代码。

def progress(percentage):
    if percentage < 0 or percentage > 100:
        raise ValueError
    #  logic