Trying to turn fizzbuzz into a function in python 3

Question

I have only just started to learn python as my first language and whilst i worked out the code for fizzbuzz, i cannot for the life of me get it to do the items below. I also want it to print horizontally instead of vertically. Any help would be great (heads spinning). Create a function which does this. For example

fizzbuzz(20) 

would print 1,2,fizz,4,buzz,fizz,7,8,fizz,buzz,11,fizz,13,14,fizzbuzz,16,17,fizz,19,buzz

def fizzbuzz(n):

    for x in range (101):
        if x%3==0 and x%5==0:
            print("fizz buzz")
        elif x%3==0:
            print('fizz')
        elif x%5==0:
            print('buzz')
        else:
            print (x)    

def main():
  print(fizzbuzz(20))

Answer 1

Collect the items into a list. Then print the list at the end of function. You can join the items together with a comma in between using ', '.join(...).

def fizzbuzz(n):
    result = []
    for x in range(1, n+1):
        if x % 3 == 0 and x % 5 == 0:
            result.append("fizz buzz")
        elif x % 3 == 0:
            result.append('fizz')
        elif x % 5 == 0:
            result.append('buzz')
        else:
            result.append(str(x))
    return result

def main():
    print(', '.join(fizzbuzz(20)))

main()

---

It's good to also know that print(..., end=', ') would print "horizontally" with a comma and space at the end, and this would almost solve your problem except that the very last item would also have a comma and space at the end, which is not what you desire.

---

It's usually a good idea to (1) separate printing from computing, (2) make functions reusable. You often want to compute more frequently than print. In the future you may wish to pass the computation on to some other function before you print. So functions that immediate print are not as useful. So I recommend taking the print statements out of fizzbuzz.

You could return the string ', '.join(result), but how useful would that be? The list result might be (in the future) more useful for further processing, so I chose to return result. The printing I left for the main function.

Answer 2

Slightly more elegant

def fizzbuzz(n):
    for x in range(1,n+1):
        if not x % 15:
            yield 'fizz buzz'
        elif not x % 3:
            yield 'fizz'
        elif not x % 5:
            yield 'buzz'
        else:
            yield x

if __name__ == "__main__":
    print ','.join(fizzbuzz(20))

Answer 3

Yet another one with list comprehension and Python3

def is_mod_zero(num, *div):
    return not [d for d in div if num % d is not 0]

def fizz_buzz(num):
    if is_mod_zero(num, 3, 5):
        return 'FizzBuzz'
    if is_mod_zero(num, 3):
        return 'Fizz'
    if is_mod_zero(num, 5):
        return 'Buzz'
    return num

if __name__ == "__main__":
    for i in range(100):
        print(fizz_buzz(i))

Answer 4

def fizzbuzz(numbers, fizz, buzz):
    x = ['Fizzbuzz' if x % fizz == 0 and x % buzz == 0 else 'Fizz' if x % fizz == 0 else 'Buzz' if x % buzz == 0 else x for x in numbers]
    return x

Answer 5

def fizzBuzz(n):
    for x in range(1, n + 1):
        if x % 3 == 0 and x % 5 == 0:
            print("fizzbuzz")
        elif x % 3 == 0:
            print("fizz")
        elif x % 5 == 0:
            print("buzz")
        else:
            print(x)
print(fizzBuzz(50))  

Answer 6

i made a fizzbuzz that works for any number and words, so you can do fizzbuzzfuzz if you wanted. i made it in 6 lines (7 if you count the line that runs through the fizzfuzz)

def f(n:int,m:list,w:list):
    s=''
    for x in m:
        if n%x==0:    s+=w[m.index(x)] 
    if s=='':  s=n
    return s
for i in range(1, 100 +1):  print(f(i,[3,5],['Fizz','Buzz']))

Answer 7

You can add end=', ' to print in order to print on the same line, with whatever character you want separating the values.

class FizzBuzz:
    @staticmethod
    def fizz_buzz(n):
        if n % 15 == 0:
            return 'FizzBuzz'
        elif n % 3 == 0:
            return 'Fizz'
        elif n % 5 == 0:
            return 'Buzz'
        else:
            return str(n)

    def __str__(self, rng):
        [print(self.fizz_buzz(n), end=', ') for n in range(1, rng + 1)]

But it leaves , at the end. Instead:

    def __str__(self, rng):
        print(', '.join(self.fizz_buzz(n) for n in range(1, rng + 1)))

Answer 8

Here's a fun one (obviously not the best for readability, but still kinda fun to think about list comprehensions):

def fizzBuzz():
    print(", ".join(["FizzBuzz" if x%15==0 else "Fizz" if x%3 == 0 else "Buzz" if x%5==0 else str(x) for x in range(1,101)]))

Don't forget, 3 and 5 are coprime! So you can check x % 15 instead of (x % 3 and x % 5) for "FizzBuzz".

Answer 9

    a=1
    while a<101:print((a%3<1)*'Fizz'+(a%5<1)*'Buzz'or a);a+=1 

it is verticle but very short %=mod it works as, if a%3>1 it will be the same as 1fizz +0buzz if a%3 and a %5>1 then it prints fizz+buzz else it prints a

Answer 10

def fizzbuzz(num):
    if x % 3 == 0 and x % 5 == 0:
    return "Fizz Buzz"
    elif x % 3 == 0:
        return "Fizz"
    elif x % 5 == 0:
        return "Buzz"
    else:
        return x    

MAXNUM = 100
for x in range (MAXNUM):
    print fizzbuzz(x)

Answer 11

Shorter yet:

for n in range(100):
    print("Fizz"*(not n % 3) + "Buzz"*(not n % 5) or n)

Yes, but why?

To understand this, let's look at the parts separately.

"Fizz"*(not n % 3)

In Python, we can "multiply" strings, so "a"*3 would result in "aaa". You can also multiply a string with a boolean: "a" * True is "a", whereas "a" * False is an empty string, "". That's what's happening to our "Fizz" here. When n % 3 == 0 (ie. n is 3, 6, 9, ...), then not n % 3 will be the same as not 0, which is True. Conversely, when n is 1, 2, 4, 5, 7, ... then n % 3 will be either 1 or 2, and not n % 3 will be false. In Other words, whenever n is divisible by 3, the term "Fizz"*(not n % 3) will multiply the string "Fizz" by True, and when it's not, it will multiply by False, resulting in an empty string.

The same logic applies to the next part, "Buzz"*(not n % 5). It'll give us an empty string when n is not divisible by 5, and the string "Buzz" when it is.

Now we're adding those two things together:

"Fizz"*(not n % 3) + "Buzz"*(not n % 5)

When n is neither divisible by 3 nor 5, this will be adding two empty strings together (ie. "" + ""), which will of course give us another empty string. In that case, the whole print statement reads print("" or n). Since an empty string is False-y, it will print our number n. If n is divisible by 3 (but not 5), this would be print("Fizz" or n), and since "Fizz" is Truthy, it will just print that and omit the number.

Bonus points

Or, if you really want to impress your interviewer,

for n in range(100):
    print("FizzBuzz"[n%-3&4:12&8-(n%-5&4)] or n)

Answer 12

one more :) just for fun :)

get_index = lambda i: bool(i%3)+2*bool(i%5) if i !=0 else 3
afb = lambda x: ('fizzbuzz','buzz', 'fizz', str(x))[get_index(x)]
fizzbuzz = lambda inpt: print(','.join([ afb(i) for i in inpt ]))
fizzbuzz(range(101))

'0,1,2,fizz,4,buzz,fizz,7,8,fizz,buzz,11,fizz,13,14,fizzbuzz,16,17,fizz,19,buzz,fizz,22,23,fizz,buzz,26,fizz,28,29,fizzbuzz,31,32,fizz,34,buzz,fizz,37,38,fizz,buzz,41,fizz,43,44,fizzbuzz,46,47,fizz,49,buzz,fizz,52,53,fizz,buzz,56,fizz,58,59,fizzbuzz,61,62,fizz,64,buzz,fizz,67,68,fizz,buzz,71,fizz,73,74,fizzbuzz,76,77,fizz,79,buzz,fizz,82,83,fizz,buzz,86,fizz,88,89,fizzbuzz,91,92,fizz,94,buzz,fizz,97,98,fizz,buzz'

Answer 13

I'm a novice coder so some of the answers I did not understand or it did not seems to directly apply to my problem. This answer incorporates Fizz_Buzz as a variable and the range of x is determined by the user. I looked at the above solutions and came up with this:

def Fizz_Buzz(x): 
     for x in range (0,x):      
        if x % 3 == 0 and x % 5 == 0:           
            print('FizzBuzz')       
        elif x % 3 == 0:            
            print('Fizz')       
        elif x % 5 == 0:            
            print ('Buzz')      
        else:           
            print (x)

Answer 14

I've found the following works well (Python 3):

def fizzbuzz():
for i in range(1,101):
    print("Fizz"*(i%3==0)+"Buzz"*(i%5==0) or i)

print (fizzbuzz())