Python Napier 计算器问题

Question

所以，我已经为此工作了好几个小时，这是一项家庭作业，我只是不明白为什么代码没有完全执行。我提供了所有代码以查看在“assign2”函数之外是否有遗漏的内容。但是，我知道问题就在那里，并想找出问题所在。

我实际上是在尝试将最后生成的数字转换回代表 Napier arithmetic 的字母（即 a = 0、b = 1、c = 2...z = 25）并将它们放入一起放在我可以在主函数中打印的列表中。除了最后一部分之外，其他一切都有效，我试图找出原因。

def main():
  again = "y" 
  while again == "y" or again == "Y":
    var = checkalpha()
    num = assign(var) 
    print("The first number is: {}".format(num)) 
    var2 = checkalpha()
    num2 = assign(var2) 
    print("The second number is: {}".format(num2)) 
    arithmetic = getsign()  
    value = equation(num, num2, arithmetic) 
    newvar = assign2(value)  
    print("The result is {} or {}".format(value, newvar))  
    again = input("Would you like to repeat the program? Enter y for yes, n for no: ") 

def checkalpha():  
  num = input("Enter Napier number: ") 
  while not num.isalpha(): 
    print("Something is wrong. Try again.") 
    num = input("Enter Napier number: ")        
  return num  

def assign(char):
    value = 0
    for ch in char:
        value += 2 ** (ord(ch) - ord("a"))
    return value

def getsign():
operand = input("Enter the desired arithmetic operation: ")
while operand not in "+-*/":
    operand = input("Something is wrong. Try again. ")
return operand

def equation(num, num2, arithmetic):
  if arithmetic == "+":
    answer = num + num2
  elif arithmetic == "-":
    answer = num - num2
  elif arithmetic == "*":
    answer = num * num2
  elif arithmetic == "/":
    answer = num / num2
  else:
    input("Something is wrong. Try again. ")
  return answer

def assign2(n):
  new = []
  while n != 0:
    value = n%2
    x = n//2
    ch = chr(value + ord("a"))
    new.append(ch)
    n = x
  return new

main()

Answer 1

您的功能相当接近。问题在于ch = chr(value + ord("a"))。我们需要将位 position 编码为字母表中具有该位置的字母。如果该位置的位不为零，则将一个字母添加到列表中。在函数的最后，我们可以将字母列表连接成一个字符串。

这是您的函数的修复版本，其中包含一些测试代码，可验证它是否适用于 Location_arithmetic 上的维基百科文章中的示例

def assign2(n):
    new = []
    position = 0
    while n != 0:
        value = n % 2
        x = n // 2
        if value:
            ch = chr(position + ord("a"))
            new.append(ch)
        n = x
        position += 1
    return ''.join(new)

# test

data = [
    (87, 'abceg'),
    (3147, 'abdgkl'),
]

for n, napier_string in data:
    s = assign2(n)
    print(n, napier_string, s, napier_string == s)

输出

87 abceg abceg True
3147 abdgkl abdgkl True

---

这是该函数的 Pythonic 版本，名称更有意义。

def int_to_napier(n):
    new = []
    for position in range(26):
        if n == 0:
            break
        value, n = n % 2, n // 2
        if value:
            new.append(chr(position + ord("a")))
    return ''.join(new)

这是另一个通过循环包含小写字母的字符串来避免字符计算的方法。

from string import ascii_lowercase

def int_to_napier(n):
    new = []
    for ch in ascii_lowercase:
        if n == 0:
            break
        value, n = n % 2, n // 2
        if value:
            new.append(ch)
    return ''.join(new)