使用除 Python 之外的 Try 的多用户输入

Question

我的任务是创建一个 python 程序，该程序将询问用户输入并计算每月还款额，这是我必须解决的公式：Formula。这样做并不太难，但是作为奖励，导师要求尝试使用户输入“牢不可破”，这意味着如果输入了预期之外的任何值，它不会破坏程序。我认为这可以通过 Try except 轻松完成，这就是我所做的，但是我相信我的代码可以以更简洁的方式编写，而不是对每个输入都使用 Try except，如下所示：

err = "Please enter a number only!!"
while True:
    try:
        A = int(input("How much were you loaned? "))
    except ValueError:
        print(err)
        continue
    else:
        break
while True:
    try:
        R = float(input("At what monthly rate interest? ")) / 100
    except ValueError:
        print(err)
        continue
    else:
        break
while True:
    try:
        N = int(input("And how many years is the loan for? "))
    except ValueError:
        print(err)
        continue
    else:
        break
RA = R * A
RN = 1 - (1 + R) ** -N
P = RA / RN
print("You will pay £", P, "yearly", "or, £", P / 12, "monthly") 

我觉得也许用户输入可以放入一个 For 循环，或者可能所有这些都放入一个 Try except 块？在考虑之后，我将所有输入放入一个 Try except 块中，但是一旦您未能正确输入用户输入的预期数据，它就会立即回到开头，而不是您所在的问题。这不是我想要的。看看：

err = "Please enter a number only!!"
while True:
    try:
        A = int(input("How much were you loaned? "))
        R = float(input("At what monthly rate interest? ")) / 100
        N = int(input("And how many years is the loan for? "))
    except ValueError:
        print(err)
        continue
    else:
        break
RA = R * A
RN = 1 - (1 + R) ** -N
P = RA / RN
print("You will pay £", P, "yearly", "or, £", P / 12, "monthly")

您认为我可以如何修改此代码以使其更简洁、更短且更高效，而无需对每个用户输入都进行尝试？

Answer 1

对其进行模块化，并使用一个通用函数来处理输入和循环，直到用户输入所要求的内容。

def get_num(prompt="Enter number: "):
    """
    (str) -> num
    Outputs a prompt and loops until a valid number is
    entered by the user, at which point that value is
    returned and the function terminates
    """
    while True:
        num = input(prompt)
        try:
            num = float(num)
            return num
        except:
            print("Must enter a valid number")

def do_calc():
    """(None) -> None
    Get user input, perform calculations and output results
    """
    A = int(get_num("How much were you loaned? "))
    R = get_num("At what monthly rate interest? ") / 100
    N = int(get_num("And how many years is the loan for? "))
    RA = R * A
    RN = 1 - (1 + R) ** -N
    P = RA / RN
    print(f"You will pay £{round(P, 2)} yearly or £{P/12:.2f} monthly")

do_calc()

使用 Python 3 中的 f-string 非常好用且优雅——只需将值或计算插入花括号内即可。根据评论中的建议，您可能希望使用显示的圆形函数或格式选项显示两位小数。

命名变量的更 Pythonic 方式是

• 金额而不是A
• 率而不是R
• num_years 而不是 N 等等。

Answer 2

提取函数中的重复过程，以及所有相关输入。
我在最终输出中添加了一些格式以避免过多的小数

def get_value(message, func, denominator = 1):
   try_again = True
   v = None
   while try_again:
      try:
         v = func(input(message))/denominator
      except ValueError:
         print("Please enter a number only!!")
      else:
         try_again= False   
   return v

A = get_value("How much were you loaned? ", int)
R = get_value("At what monthly rate interest? ", float, denominator=100.0)
N = get_value("And how many years is the loan for? ", int)

RA = R * A
RN = 1 - (1 + R) ** -N
P = RA / RN
print("You will pay £ {:.2f} yearly or £ {:.2f} monthly".format(P, P / 12))