如何在python3的int和float中获取用户的输入[关闭]

Question

如何在 python3 的 int 和 float 中获取用户的输入？

显然编程了一个计算器，但我希望用户的输入是 int 和 float，而不仅仅是 int 或 float，而是两者。这基本上就是我所关心的，将 int 更改为 int 和 float。可悲的是，Java 使用双精度。

def calculator():
    def add(x, y):
        return x + y

    def sub(x, y):
        return x - y

    def mul(x, y):
        return x * y

    def div(x, y):
        return x / y

    def power(x, y):
        return x ** y

    def modulus(x, y):
        return x % y

    print(""" -- Select Operation --
    1. Addition
    2. Subtraction
    3. Multiplication
    4. Division
    5. Power
    6. Modulus
    """)

    choice = input("Enter 1 or 2 or 3 or 4 or 5 or 6 from the Select Operation >> ")
    while choice not in ('1', '2', '3', '4', '5', '6'):
        choice = input("Invalid Input! Please Enter 1 or 2 or 3 or 4 or 5 or 6 from the Select Operation >> ")
    print("\n")
    print("============================================================")
    print("Gathering data...")
    # How to make user input in both int and float not int alone?
    num1 = int(input("Enter First number >> "))
    num2 = int(input("Enter Second number >> "))

    if choice == '1':
        print(num1, "+", num2, "=", add(num1, num2))
    elif choice == '2':
        print(num1, "-", num2, "=", sub(num1, num2))
    elif choice == '3':
        print(num1, "*", num2, "=", mul(num1, num2))
    elif choice == '4':
        print(num1, "/", num2, "=", div(num1, num2))
    elif choice == '5':
        print(num1, "^", num2, "=", power(num1, num2))
    elif choice == '6':
        print(num1, "%", num2, "=", modulus(num1, num2))
    repeat_cal()

def repeat_cal():
    print("Do you want to Select Operation again")
    choose_again = input("Enter Y for YES or N for NO>> ")
    if choose_again.upper() == 'Y':
        calculator()
    elif choose_again.upper() == 'N':
        print("-----------------Good Bye!---------------------------")
        exit()
    else:
        repeat_cal()

calculator()

Answer 1

我建议您只将输入作为浮点数。这样，如果用户输入一个浮点数，你就可以支持它。

如果用户输入一个int，它也将被支持。如果出于某种原因，您想检查输入是否为 int，则只需检查该数字是否在小数点后为 0。

Answer 2

使用字符串构造函数将输入作为十进制。十进制以避免浮点精度损失。 Decimal 可以表示任意精度，包括整数，并且总是返回一个计算器预期的值。

def calculator():
    def add(x, y):
        return x + y

    def sub(x, y):
        return x - y

    def mul(x, y):
        return x * y

    def div(x, y):
        return x / y

    def power(x, y):
        return x ** y

    def modulus(x, y):
        return x % y

    print(""" -- Select Operation --
    1. Addition
    2. Subtraction
    3. Multiplication
    4. Division
    5. Power
    6. Modulus
    """)
    from decimal import Decimal
    choice = input("Enter 1 or 2 or 3 or 4 or 5 or 6 from the Select Operation >> ")
    while choice not in ('1', '2', '3', '4', '5', '6'):
        choice = input("Invalid Input! Please Enter 1 or 2 or 3 or 4 or 5 or 6 from the Select Operation >> ")
    print("\n")
    print("============================================================")
    print("Gathering data...")
    # How to make user input in both int and float not int alone?
    num1 = Decimal(input("Enter First number >> "))
    num2 = Decimal(input("Enter Second number >> "))

    if choice == '1':
        print(num1, "+", num2, "=", str(add(num1, num2)))
    elif choice == '2':
        print(num1, "-", num2, "=", str(sub(num1, num2)))
    elif choice == '3':
        print(num1, "*", num2, "=", str(mul(num1, num2)))
    elif choice == '4':
        print(num1, "/", num2, "=", str(div(num1, num2)))
    elif choice == '5':
        print(num1, "^", num2, "=", str(power(num1, num2)))
    elif choice == '6':
        print(num1, "%", num2, "=", str(modulus(num1, num2)))
    repeat_cal()

def repeat_cal():
    print("Do you want to Select Operation again")
    choose_again = input("Enter Y for YES or N for NO>> ")
    if choose_again.upper() == 'Y':
        calculator()
    elif choose_again.upper() == 'N':
        print("-----------------Good Bye!---------------------------")
        exit()
    else:
        repeat_cal()

calculator()