Ultra-Beginner Python FizzBuzz ... Am I missing something?

Question

I just started programming in Python this morning, and it is (more or less) my first programming language. I've done a bit of programming before, but never really did much except for "Hello World" in a few languages. I searched around for some Python FizzBuzz solutions, and they all seem significantly more complicated then mine, so I think I must be missing something, even though it works correctly. Could you guys point out any errors I've made, or things I can improve?

count = 0
while (count < 101):
    if (count % 5) == 0 and (count % 3) == 0:
        print "FizzBuzz"
        count = count +1
    elif (count % 3) == 0:
        print "Fizz"
        count = count + 1
    elif (count % 5) == 0:
        print "Buzz"
        count = count +1
    else:
        print count
        count = count + 1

Answer 1

Lose the useless brackets

This:

while (count < 101):

can just be:

while count < 101:

Increment out of the ifs

Wouldn't be easier to do:

count = 0
while count < 101:
    if count % 5 == 0 and count % 3 == 0:
        print "FizzBuzz"
    elif count % 3 == 0:
        print "Fizz"
    elif count % 5 == 0:
        print "Buzz"
    else:
        print count

    count = count + 1    # this will get executed every loop

A for loop will be better

for num in xrange(1,101):
    if num % 5 == 0 and num % 3 == 0:
        print "FizzBuzz"
    elif num % 3 == 0:
        print "Fizz"
    elif num % 5 == 0:
        print "Buzz"
    else:
        print num

I've also renamed count to num since it doesn't count much, is just a number between 1 and 100.

Let's use only one print

Why do 4 different print, when what is really changing is the printed message?

for num in xrange(1,101):
    if num % 5 == 0 and num % 3 == 0:
        msg = "FizzBuzz"
    elif num % 3 == 0:
        msg = "Fizz"
    elif num % 5 == 0:
        msg = "Buzz"
    else:
        msg = str(num)
    print msg

Light bulb!

"FizzBuzz" is the same of "Fizz" + "Buzz".

Let's try this one:

for num in xrange(1,101):
    msg = ''
    if num % 3 == 0:
        msg += 'Fizz'
    if num % 5 == 0:       # no more elif
        msg += 'Buzz'
    if not msg:      # check if msg is an empty string
        msg += str(num)
    print msg

Copy and paste this last piece of code here to see what it does.

Python is a very flexible and powerful language, so I'm sure there could be other hundred and one different possible solutions to this problem :)

Edit: Improve more

There's still something "quite not right" with these lines:

if not msg:
    msg += str(num)

IMHO it would be better to do:

for num in xrange(1,101):
    msg = ''
    if num % 3 == 0:
        msg += 'Fizz'
    if num % 5 == 0:
        msg += 'Buzz'
    print msg or num

There! Now with:

print msg or num

is clear that num is the default value to be printed.

Answer 2

Your code looks fine. It certainly works. The only thing I'd say about it is that you are repeating the incrementation in all if blocks. You could just move that out of and after them and you'll achieve the same thing.

if (count % 5) == 0 and (count % 3) == 0:
    print "FizzBuzz"
elif (count % 3) == 0:
    print "Fizz"
elif (count % 5) == 0:
    print "Buzz"
else:
    print count
count = count + 1

You could also condense the if conditions a bit if you recognize that values of 0 are considered False and non-zero is True. You're testing if they're equal to 0 so you could just write it as

if not(count % 5) and not(count % 3):
    print "FizzBuzz"

I would however not do your loop like that using a while loop but instead use a for loop. Then you wouldn't even need the increment statement there.

for count in range(0, 101):
    if not(count % 5) and not(count % 3):
        print "FizzBuzz"
    elif not(count % 3):
        print "Fizz"
    elif not(count % 5):
        print "Buzz"
    else:
        print count