Python > 检查一个 int 是否介于另外两个 int 之间

问：

有问题的功能

def IsInRange(lr,hr,num):
    return lr < num < hr

稍后，我们要求输入。

lr = floaterror('Provide your low range number, please: ')
hr = floaterror('Provide your high range number, please: ')
num = floaterror('Provide number to range find, please: ')

作业要求我们说它是否在范围内。所以，我做到了：

if IsInRange(lr, hr, num) == False:
    print(num, "is not in range.")
elif IsInRange(lr, hr, num) == True:
    print(num, "is in range.")

它有效。但是，让我们说，

lr = 1
hr = 1
num = 1

它说 1 不在范围内......我想创建一个例外，如果发生这种情况，它会打印“您的检查值不能等于您的范围”而不是“#不在范围内”。但我正在为此苦苦挣扎。我无法更改每个赋值返回的 IsInRange 函数。我确信这是其中的一部分，也是我编写的完整代码。我把它包括在下面。

def add(num1,num2):
    return num1+num2
def sub(num1,num2):
    return num1-num2
def mlt(num1,num2):
    return num1*num2
def dvd(num1,num2):
    return num1/num2
def IsInRange(lr,hr,num):
    return lr < num < hr

end = True
print("Basic Math Function and In-Range Checker")
while end == True:
    user = input('Please press Enter to continue with a simply mathtastical time, or "q" to quit. ')
    if user == 'q':
        print('You chose to avoid the math! Exiting!')
        break
    else:
        keep_on_rolling = True
        while keep_on_rolling:
        #1st exception to catch invalid entries, that is, not integers.
        #You'll be prompted to enter a correct value.
            def floaterror(one):
                while True:
                    try:
                        return float(input(one))
                    except ValueError:
                        print("Numbers only, please. Try again, with digits!")
        #End 1st catch
            num1 = floaterror('Give us your first number and press enter, please: ')
            num2 = floaterror('Give us your second number and press enter, please: ')
            lr = floaterror('Provide your low range number, please: ')
            hr = floaterror('Provide your high range number, please: ')
            num = floaterror('Provide number to range find, please: ')
            print('The result of', num1, 'added with', num2, 'is', add(num1,num2))
            print('The result of', num1, 'subtracting', num2, 'is', sub(num1,num2))
            print('The result of', num1, 'multiplied by', num2, 'is', mlt(num1,num2))
        #2nd error catch, tryig to destroy the universe by dividing by zero.
            try:
                print('The result of', num1, 'divided by', num2, 'is', (dvd(num1,num2)))
            except ZeroDivisionError:
                print("Uff da. We don't divide by zero in this house.")
        #End 2nd catch.
            else:
                if IsInRange(lr, hr, num) == False:
                    print(num, "is not in range.")
                elif IsInRange(lr, hr, num) == True:
                    print(num, "is in range.")
            break

我没有保留任何失败的代码。 但是我尝试在 else 下添加一个异常：你不能这样做，我尝试了 if lr == num == hr 来打印我想要的答案。我尝试在有问题的函数出现的地方添加一个异常。我尝试在输入后添加异常。但是不断出现语法错误，或者如果没有错误，它似乎根本没有出现。我还尝试定义一个使用输入的新函数，但仍然一无所获。