Question

我正在使用Python为我的最终项目编写一个简单的计算器，而我在验证用户输入的值时遇到的问题是浮点数据类型。我想这样做，如果值是一个字符串类型，它将打印“值必须是一个整数或十进制 - 请输入一个有效的数字”，然后循环它以询问用户输入，直到用户给出有效进入。我试过了，但我卡住了。所以这是我目前的代码：

keepProgramRunning = True

print ("Welcome to the Calculator Application!")
good = True
while keepProgramRunning:

    print ("1: Addition")

    print ("2: Subtraction")

    print ("3: Multiplication")

    print ("4: Division")

    print ("5: Quit Application")


    choice = input("Please choose what you would like to do: ")

    if choice == "1":
        n1 = float(input ("Enter your first number: "))
        n2 = float(input ("Enter your second number: "))
        print ("Your result is: ", n1 + n2)
    elif choice == "2":
        n1 = float(input ("Enter your first number: "))
        n2 = float(input ("Enter your second number: "))
        print ("Your result is: ", n1 - n2)
    elif choice == "3":
        n1 = float(input ("Enter your first number: "))
        n2 = float(input ("Enter your second number: "))
        print ("Your result is: ", n1 * n2)
    elif choice == "4":
        n1 = float(input ("Enter your first number: "))
        n2 = float(input ("Enter your second number: "))
        try:
            print ("Your result is: ", n1 / n2)
        except:
            if n2 == 0:
                print ("Zero Division Error - Enter Valid Number")
                while good:
                    n2 = float(input ("Enter your second number: "))
                    if n2!=0:
                        good =False
                        print ("Your result is: ", n1 / n2)
    elif choice == "5":
        print ("Thank you for using the calculator. Goodbye!")
        keepProgramRunning = False
    else:
        print ("Please choose a valid option.")

Answer 1

假设您在这里使用Python 3.x，每一行：

n1 = float(input ("Enter your first number: "))

...如果给出了无法转换为浮点数的内容，则会引发ValueError。

因此，不是验证然后转换，只是尝试转换，让转换器成为自己的验证器。

例如，而不是：

n1 = float(input ("Enter your first number: "))
n2 = float(input ("Enter your second number: "))
print ("Your result is: ", n1 + n2)

......你可以这样做：

while True:
    try:
        n1 = float(input ("Enter your first number: "))
        n2 = float(input ("Enter your second number: "))
    except ValueError:
        print("When I ask for a number, give me a number. Come on!")
    else:
        print ("Your result is: ", n1 + n2)
        break

如果你想分别检查每个值，只需在try上做两个较小的循环而不是一个较大的循环。

不是将此代码复制和粘贴6次，而是将其重构为函数会更好。像这样：

def get_two_floats():
    while True:
        try:
            n1 = float(input ("Enter your first number: "))
            n2 = float(input ("Enter your second number: "))
        except ValueError:
            print("When I ask for a number, give me a number. Come on!")
        else:
            return n1, n2

或者，如果您想分别验证每一个：

def get_float():
    while True:
        try:
            return float(input ("Enter your second number: "))
        except ValueError:
            print("When I ask for a number, give me a number. Come on!")

def get_two_floats();
    return get_float(), get_float()

然后你可以这样做：

if choice == "1":
    n1, n2 = get_two_floats()
    print ("Your result is: ", n1 + n2)
elif choice == "2":
    n1, n2 = get_two_floats()
    print ("Your result is: ", n1 - n2)
# etc.

作为旁注：要将除法除以零，而不是处理所有异常，然后根据输入找出导致错误的原因，只需处理ZeroDivisionError。（一般情况下，除非您要使用except:，重新sys.exc_info()，或类似的东西，否则裸raise是一个坏主意。使用except SpecificException:是几乎总是更好。或者更常见的是，except SpecificException as e:，因此您可以使用e执行某些操作，例如错误消息中的print。）

Answer 2

# get original input
n1 = raw_input("enter your number: ")

while not (n1.isdigit()):
# check of n1 is a digit, if not get valid entry
    n1 = raw_input ("enter a valid number: ")

num1 = float(n1) # convert string to float



n2 = raw_input("enter number: ")
while not (n2.isdigit()):
    n2 = raw_input("enter a valid number: ")

num2 = float(n2)

Answer 3

while True:
        try:
          *Your Code*
except ValueError:
        print("Please enter a number:")
        else:
        break

验证浮点数据类型python

3 个答案: