验证浮点数据类型python

Question

我正在用 Python 为我的最终项目编写一个简单的计算器，但我无法验证用户输入的值是否为浮点数据类型。我想这样做，如果值是字符串类型，它将打印“值必须是整数或小数 - 请输入有效数字”，然后将其循环回询问用户输入，直到用户给出有效的条目。我试过了，但我卡住了。所以这是我到目前为止的代码：

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

假设您在这里使用的是 Python 3.x，以下每一行：

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

... 如果给定无法转换为浮点数的东西，将引发ValueError。

所以，与其先验证再转换，不如尝试转换，让转换器成为自己的验证器。

例如，而不是这个：

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

---

如果您想分别检查每个值，只需在 try 上执行两个较小的循环，而不是一个大循环。

---

与其将这段代码复制粘贴 6 次，不如将其重构为一个函数。像这样的：

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项：

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()

那么你可以这样做：

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.

---

附带说明：要捕获除以零，而不是处理所有异常，然后尝试根据输入找出导致错误的原因，只需处理 ZeroDivisionError。 （一般来说，一个裸露的except: 是个坏主意，除非你打算使用sys.exc_info()、re-raise-ing 或类似的东西。使用except SpecificException: 几乎总是更好。或者，更频繁，except SpecificException as e:，所以你可以用e做一些事情，比如print在错误消息中。）

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