Python conditional code not working properly

Question

I've just started to learn python and can't get this code to work properly.

def percentage():
    while True:
        try:
            x = int(input('First number:'))
            y = int(input('Second number:'))


            fraction = x/y
            percentage = fraction * 100
            final = "%d" % percentage

            return final

        except ZeroDivisionError:
            pass

        except ValueError:
            pass

if percentage() <= 1:
    print('E')

elif percentage() >= 99:
    print('F')

else:
    print(percentage(), "%")

The 'F' is printed only when I input x = 999 and y = 1000. Any other values just prompt the user again and again. The code should output 'E' when the percentage function returns <= 1, 'F' when it returns >= 99 and the actual percentage number when any other value is returned.

This fails with a TypeError as final a string that you then try to compare with 1. — user9706
– user9706, Commented Dec 30, 2022 at 19:12
Your function returns a str but then you try to compare it to a number. Instead, you probably want to just return percentage. After fixing that you will still not have the desired functionality because every time your code gets to a point where it reads percentage(), it will call the function again. Instead you want to save the result of calling percentage() in a variable by doing my_variable = percentage() and then comparing and printing just with my_variable — yagod
– yagod, Commented Dec 30, 2022 at 19:13

user9706 · Accepted Answer · 2022-12-30 19:18:26Z

4

I am getting a type error as final is a string which you then try to compare with an integer. The issue that you are asking about, however, is due to 3 calls of percentage(). Instead call it once and assign the return value a variable then run your tests against the variable:

def percentage():
    while True:
        try:
            x = int(input('First number:'))
            y = int(input('Second number:'))
            return 100 * x / y
        except ZeroDivisionError:
            print('x')
            pass
        except ValueError:
            print('y')
            pass


p = percentage() 
if p <= 1:
    print('E')
elif p >= 99:
    print('F')
else:
    print(p, "%")

answered Dec 30, 2022 at 19:18

user9706

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Comments

Derek O · Accepted Answer · 2022-12-30 19:31:22Z

I would restructure your program so that your percentage function returns the percentage as a float instead of a string, and deal with the string formatting in your print statements.

We should also set some variable equal to the output of your function using something like p = percentage(), and then use p in your conditionals so that we don't unintentionally run percentage() multiple times from within your conditionals.

def percentage():
    while True:
        try:
            x = int(input('First number:'))
            y = int(input('Second number:'))


            fraction = x/y
            percentage = fraction * 100
            # final = "%d" % percentage

            return percentage

        except ZeroDivisionError:
            pass

        except ValueError:
            pass

p = percentage()
if p < 1:
    print('E')

elif p >= 99:
    print('F')

else:
    print(f"{p:.2f}%")

Example program run:

First number:2
Second number:3
66.67%

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