Beginner FizzBuzz in Python

This is my first attempt at solving Fizzbuzz. I've always been pretty weak at algorithms, so I'm trying to improve. Specific feedback on performance would be appreciated.

def fizzBuzz(fizzNumber, buzzNumber):
  i = 0
  for i in range(1, 101):
      fizzBuzzValues = [i%fizzNumber, i%buzzNumber, i%(fizzNumber * buzzNumber)]
      if(0 in fizzBuzzValues):
          if(fizzBuzzValues[2] == 0):
              print("Fizzbuzz")
          elif(fizzBuzzValues[0] == 0):
              print("Fizz")
          elif(fizzBuzzValues[1] == 0):
              print("Buzz")
          else:
              break
      else:
          print(i)

fizzBuzz(3, 5)

Specific feedback on performance would be appreciated.

There are some things to improve performance-wise:

• on every single iteration, you are calculating all the 3 remainders, even though you need a single one (in the best case)

• the 0 in fizzBuzzValues check is not really needed (and the "in list" is O(n) in general, even though n=3 in this case, but you are doing that in a loop) - you can compare each of the remainders one by one

• if you can calculate something before the loop - always do it - the fizzNumber * buzzNumber can be defined before the loop

Modified code:

def fizz_buzz_new(fizz_number, buzz_number):
    fizz_buzz_number = fizz_number * buzz_number
    for i in range(1, 101):
        if i % fizz_buzz_number == 0:
            print("Fizzbuzz")
        elif i % fizz_number == 0:
            print("Fizz")
        elif i % buzz_number == 0:
            print("Buzz")
        else:
            print(i)

Some timeit tests (replaced print() with yield to avoid seeing impact of printing to stdout):

$ ipython3 -i test.py 

In [1]: %timeit for x in range(100): list(fizzBuzz(3, 5))
100 loops, best of 3: 3.52 ms per loop

In [2]: %timeit for x in range(100): list(fizz_buzz_new(3, 5))
100 loops, best of 3: 1.98 ms per loop

StackExchange Code Review Q#153730, answer score: 7