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Python Comparison Challenge

Posted by Brandyn Burgett
.a{fill-rule:evenodd;}techdegree
Brandyn Burgett
Python Development Techdegree Student 219 Points

Any way to make cleaner code?

It works for the most part, I was having a hard time with displaying an error message if someone entered a name into the number question. i used .isdigit() at first, but i couldn't find a way to have 2 else statements next to each other so I just completely removed it. I started to get a feeling that this wasn't important for where we're at right now in the course. I'm not sure if I followed the directions properly, but it works. Any way to make this code a little cleaner and more concise?

name = input("Please enter your name: ")
number = input("Please enter a number: ")
true_number = int(number) 
print(name)
print(number)
if true_number % 3 == 0 and true_number % 5 == 0:
    is_fizzbuzz = True
    print("is a FizzBuzz Number")
elif true_number % 3 == 0 and true_number % 5 != 0:
    is_fizz = True  
    print("is a Fizz Number")
elif true_number % 3 != 0 and true_number % 5 == 0:
    is_buzz = True   
    print("is a Buzz Number")
else: 
    print("This is neither a Fizzy or a Buzzy Number")

Any Help is appreciated

2 Answers

Steven Parker
Steven Parker
229,786 Points
Steven Parker
Steven Parker
229,786 Points

You can directly assign the results of a comparison to a variable, and avoid re-testing conditions previously eliminated by other tests:

is_fizz = true_number % 3 == 0
is_buzz = true_number % 5 == 0
if is_fizz and is_buzz:
    print("is a FizzBuzz Number")
elif is_fizz:
    print("is a Fizz Number")
elif is_buzz:
    print("is a Buzz Number")
else: 
    print("This is neither a Fizzy or a Buzzy Number")
Brandyn Burgett
.a{fill-rule:evenodd;}techdegree
Brandyn Burgett
Python Development Techdegree Student 219 Points

Thanks, it really helped me clean up my code

Arnon Pilz
Arnon Pilz
198 Points
Arnon Pilz
Arnon Pilz
198 Points

Hey Brandin, trying to understand my mistakes... what's the purpose of "_" = True?