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Python Comparison Solution

fahad lashari
fahad lashari
7,693 Points
Posted by fahad lashari
fahad lashari
7,693 Points

Did it in a weird way but it works lol

name = input("Please enter your name: ") number = input("Please enter a number: ")

str_number3 = str(int(number) / 3) str_number5 = str(int(number) / 5)

if int(str_number3[2]) <= 0 and int(str_number5[2]) <= 0: print(number, " is a FizzBuzz")

elif int(str_number3[2]) <= 0: print(number, " is a Fizz number")

elif int(str_number5[2]) <= 0: print(number, " is a Buzz number")

else: print(number, "is neither a Fizzy or Buzzy number")

Just wanted to see if I could this to work before watching the solution

5 Answers

Steven Parker
Steven Parker
229,695 Points
Steven Parker
Steven Parker
229,695 Points

I'm wondering how you tested this to get the impression that it worked — it won't actually work for most numbers.

But some hints about creating a workable strategy:

  • it's probably easier to work with numbers than strings
  • only one conversion will be needed on the input value
  • the "remainder" operator ("%") might be handy
fahad lashari
fahad lashari
7,693 Points
fahad lashari
fahad lashari
7,693 Points

The Operator % completely slipped my mind at the time I was doing this lol. I just wanted to try and get something together before watching the solution using only what I've learnt so far.

I do understand that it wont work with most numbers. Thanks for the tips I'll try and get another working version using those.

Regards

fahad lashari
fahad lashari
7,693 Points
fahad lashari
fahad lashari
7,693 Points 
#name = input("Please enter your name: ")
number = int(input("Please enter a number: "))

is_fizz = False
is_buzz = False

if number % 3 == 0 and number % 5 == 0:
  is_fizz = True
  is_buzz = True

elif number % 3 == 0 and number % 5 >= 1:
  is_fizz = True
  is_buzz = False

elif number % 3 >= 1 and number % 5 == 0:
  is_fizz = False
  is_buzz = True

elif number % 3 >= 1 and number % 5 >= 1:
  is_fizz = False
  is_buzz = False

if is_fizz and is_buzz:
  print(number, "is a FizzBuzz")

elif is_fizz and is_buzz != True:
  print(number, "is a Fizz")

elif is_fizz != True and is_buzz:
  print(number, "is a Buzz")

else:
  print(number, " is neither a Fizzy or Buzzy number")

I think this works a lot better.

Thanks for the help!

Steven Parker
Steven Parker
229,695 Points
Steven Parker
Steven Parker
229,695 Points

Much better! And you can simplify it a good deal from there by just assigning the booleans directly from the comparison. Then you don't need any conditional statements:

is_fizz = number % 3 == 0
is_buzz = number % 5 == 0
Aaron Wood
Aaron Wood
2,308 Points
Aaron Wood
Aaron Wood
2,308 Points 
name = input("Please enter your name: ")
number = int(input("Please enter a number: "))

if(number % 5 == 0):
    if(number % 3 == 0):
        print("{} is a fizzbuzz number!".format(number))
    else:
        print("{} is a buzz number!".format(number))
else:
    if(number % 3 == 0):
        print("{} is a fizz number!".format(number))
    else:
        print("{} is neither a fizzy or buzzy number.".format(number))
fahad lashari
fahad lashari
7,693 Points
fahad lashari
fahad lashari
7,693 Points

Thanks! Saw this in the video later. I further simplified it, I didn't know it was possible for calculations or any functionality to run whilst defining the variable. That makes everything so much more interesting lol. Thanks for the help! :)

Simplified code is as follows:

#name = input("Please enter your name: ")
number = int(input("Please enter a number: "))

is_fizz = number % 3 == 0
is_buzz = number % 5 == 0

if is_fizz and is_buzz:
  print(number, "is a FizzBuzz")

elif is_fizz and is_buzz != True:
  print(number, "is a Fizz")

elif is_fizz != True and is_buzz:
  print(number, "is a Buzz")

else:
  print(number, " is neither a Fizzy or Buzzy number")