使用 Python 实现 collat​​z 函数

Question

我目前无法完成“自动化无聊的东西”中的这个挑战：

我的代码是：

def collatz(number):
    global seqNum
    if (seqNum % 2 == 0):
        return seqNum // 2
    elif (seqNum % 2 == 1):
        return 3 * seqNum + 1


print('What number would you like to use?')
seqNum = input()
number = int(seqNum)
i = number

while i > 1:
    collatz(seqNum)
    print(number)

我收到了这个错误：

"Traceback (most recent call last):
  File "C:/Users/Administrative/AppData/Local/Programs/Python/Python36-32/collatzSeq.py", line 15, in <module>
    collatz(seqNum)
  File "C:/Users/Administrative/AppData/Local/Programs/Python/Python36-32/collatzSeq.py", line 3, in collatz
    if (seqNum % 2 == 0):
TypeError: not all arguments converted during string formatting"

我知道我在编写代码的方式上做错了，但我不明白它到底是什么。非常感谢任何和所有帮助！

我也在使用 python 3。

Answer 1

1. 您正在对字符串进行算术运算，而不是整数。

2. 不需要global 变量。将参数传递给函数，并让它相应地返回一个值。

---

def collatz(number):
    if (number % 2 == 0):
        return number // 2

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

print('What number would you like to use?')

i = int(input())
while i > 1:     
    i = collatz(i)
    print(i)

Answer 2

这里有几个问题，但导致您的异常的一个是您在函数中使用了seqNum，这是input() 返回的内容。 input() 返回一个字符串（至少在 Python 3 上）。而strings the % 是“格式化运算符”，它也解释了异常消息，其中谈到了“字符串格式化”。

你可以这样写（使用number而不是seqNum）：

def collatz(number):  
    # you pass the number to the function and you return, so no need for global        
    if number % 2 == 0:       # python doesn't need parenthesis for "if"s
        return number // 2
    else:                     # it can only be even OR odd so no need to calculate the modulo again
        return 3 * number + 1

# You can put the question as argument for "input" instead of printing it
seqNum = input('What number would you like to use?')  
number = int(seqNum)

while number > 1 :
    number = collatz(number)   # assign the result of the function to "number"
    print(number)

Answer 3

seqNum 是一个字符串。

>>> "3" % 2
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: not all arguments converted during string formatting
>>>

Answer 4

# This could be the simplest solution
def collatz(number):
    while True:
        if number <= 1:
            break
        elif number % 2 == 0:
            number = number // 2
            print(number)
        elif number % 2 == 1:
            number = 3 * number + 1
            print (number)
try:
    print('Enter a number \n')
    number = int(input())
    collatz(number)
except ValueError:
    print('Invalid value, Enter a number.')

Answer 5

我认为对于像我这样的初学者来说这是最容易理解的：

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

try:
    collatz(number=int(input("Enter the number: \n")))
except ValueError:
    print('Error: Invalid argument \nPlease enter a number')

Answer 6

看来您应该将i 传递给您的函数，而不是seqNum。

并且在您的函数中删除所有对seqNum 的引用并改用number。

Answer 7

您好，我是编码新手，也在看这个练习。如果这里有帮助的话，我使用 1 x 函数 + 2 x while 循环所采用的方法。我还注意到程序没有很好地处理零值作为输入：

# This program runs the Collatz sequence - Automate book, Chapter 3 practice project
# It includes the 'Input Validaton' additional exercise
# It also inlcudes a further test for zero value input as this makes collatz non-terminating

def collatz(number):
    #test even
    if number % 2 == 0:
        return number // 2
    #or implicit it is odd
    else:
        return 3 * number + 1

# Get the user input and validate - loop continues until non-zero integer entered
while True:
    try:
        print('Enter a non-zero number')
        number = int(input())
        if number == 0:
            continue
        else:
            break
    except ValueError:
        print('Error: You must enter and integer')

# Iterate over the input number until it == 1              
while number != 1:
    # return value assigned to global var
    number = collatz(number) 
    # output the result of collatz to screen
    print(number)