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Re:Fizz Buzz codegolf challenge in 15 languages by jaegerpicker on Friday December 14, 2012 @12:15PM (#4921949)
for i in range(1,101): print "FizzBuzz" if i %3 == 0 and i % 5 == 0 else "Fizz" if i % 3 == 0 else "Buzz" if i % 5 == 0 else str(i)

Scores 69 in python but the login/signup is broken :)

Re:Fizz Buzz codegolf challenge in 15 languages by glomph on Friday December 14, 2012 @02:11PM (#4922642)
OK I cheated and used google and found (60 chars):

  for i in range(1,101):print'FizzBuzz'[i*i%3*4:8--i**4%5]or i
Pastebin of my explanation: http://pastebin.com/raw.php?i=00Db30FF

EDIT: ugh, sometimes I hate markdown, moved post to pastebin.

Re:Fizz Buzz codegolf challenge in 15 languages by GregBuchholz on Friday December 14, 2012 @07:21PM (#4924049)
Awesomely ingenious use of math above. 59 if newlines count as 1.

    i=1
    while i<101:print'FizzBuzz'[i*i%3*4:8--i**4%5]or i;i+=1
Re:Fizz Buzz codegolf challenge in 15 languages by twiceaday on Sunday December 16, 2012 @10:48PM (#4930740)
Much simpler solution in 59 chars

  n=1
  while n<101:print'Fizz'*(n%3<1)+'Buzz'*(n%5<1)or n;n+=1
Re:Fizz Buzz codegolf challenge in 15 languages by twiceaday on Tuesday December 18, 2012 @07:26PM (#4940545)
I came up with a shorter way to write the first index!

  n%-3&4
unbelievably it works
Re:Fizz Buzz codegolf challenge in 15 languages by glomph on Monday December 24, 2012 @09:47AM (#4962938)
haha, clever!
Re:Fizz Buzz codegolf challenge in 15 languages by glomph on Saturday December 15, 2012 @07:49AM (#4925212)
Should have been empty string f[4:4]
Re:Fizz Buzz codegolf challenge in 15 languages by glomph on Friday December 14, 2012 @01:01PM (#4922224)
Some simple ways to cut down on characters:

    for i in range(1,101):print 'FizzBuzz' if not i%15 else 'Fizz' if not i%3  else 'Buzz' if not i%5 else i
Re:Fizz Buzz codegolf challenge in 15 languages by jaegerpicker on Friday December 14, 2012 @01:12PM (#4922277)
for i in range(1,101): print "FizzBuzz" if not i % 15 else "Fizz" if not i % 3 else "Buzz" if not i % 5 else str(i)

Even a little shorter :)

Re:Fizz Buzz codegolf challenge in 15 languages by adambard on Friday December 14, 2012 @01:34PM (#4922394)
And a bit shorter still :D. Down to 102 chars now.

    for i in range(1,101):print "Fizz" if not i%3 else '' + "Buzz" if not i%5 else str(i) if (i%3) else ''
Edit: Down to 82 characters now. I'm not sure if this would work on every Python, but it works on mine.

    for i in range(1,101):print(("Fizz"if i%3==0 else'')+("Buzz"if i%5==0 else''))or i
Re:Fizz Buzz codegolf challenge in 15 languages by c4p on Friday December 14, 2012 @02:56PM (#4922939)
You can drop the extra () which brings it down to 80.

  for i in range(1,101):print("Fizz"if i%3==0 else'')+("Buzz"if i%5==0 else'')or i
Re:Fizz Buzz codegolf challenge in 15 languages by GregBuchholz on Friday December 14, 2012 @05:00PM (#4923596)
74

    for i in range(1,101):print(''if i%3 else'Fizz')+(''if i%5 else'Buzz')or i
Re:Fizz Buzz codegolf challenge in 15 languages by y2bd on Friday December 14, 2012 @05:43PM (#4923771)
70 now, still same idea.

    for x in range(1,101):print['','Fizz'][x%3==0]+['','Buzz'][x%5==0]or x
Re:Fizz Buzz codegolf challenge in 15 languages by GregBuchholz on Friday December 14, 2012 @05:55PM (#4923818)
Excellent. Down to 68.

    for x in range(1,101):print['','Fizz'][x%3<1]+['','Buzz'][x%5<1]or x
Re:Fizz Buzz codegolf challenge in 15 languages by twiceaday on Sunday December 16, 2012 @10:43PM (#4930723)
My shortest Python solution is 59 characters. Two tricks, use while instead of range, and use string repetition instead of lists

  n=1
  while n<101:print'Fizz'*(n%3<1)+'Buzz'*(n%5<1)or n;n+=1
And here is a wacky one in 96 chars

  w=range(101)
  w[::3]=['Fizz']*34
  w[::5]=['Buzz']*21
  w[::15]=['FizzBuzz']*7
  for s in w[1:]:print s
Re:Fizz Buzz codegolf challenge in 15 languages by ajennings on Tuesday December 18, 2012 @01:04PM (#4938651)
You can save one more character with an exec loop

  n=1;exec"print'Fizz'*(n%3<1)+'Buzz'*(n%5<1)or n;n+=1;"*100
Re:Fizz Buzz codegolf challenge in 15 languages by jaegerpicker on Friday December 14, 2012 @02:23PM (#4922733)
Nice, The 82 character one works for me also, on Mac OS X python 2.7.3
Re:Fizz Buzz codegolf challenge in 15 languages by jaegerpicker on Friday December 14, 2012 @01:04PM (#4922238)
for i in range(1,101): print "FizzBuzz" if i % 15 == 0 else "Fizz" if i % 3 == 0 else "Buzz" if i % 5 == 0 else str(i)

derp, works better of course

Though I can't find out the score the page to test the code seems to be inaccessible if you are signed in. It just redirects you to the signed in page and the signout link is broken. Still it's a fun and cool idea

Re:Fizz Buzz codegolf challenge in 15 languages by neeleshs on Friday December 14, 2012 @01:16PM (#4922300)
for i in range(1,101):print i%15==0 and 'FizzBuzz' or i%3==0 and 'Fizz' or i%5==0 and 'Buzz' or i has a score of 102
Re:Fizz Buzz codegolf challenge in 15 languages by crodjer on Friday December 14, 2012 @02:00PM (#4922569)
Redirects issue fixed. Can you try it now?
Re:Fizz Buzz codegolf challenge in 15 languages by jaegerpicker on Friday December 14, 2012 @02:16PM (#4922683)
Works to login having trouble submitting however now
Re:Fizz Buzz codegolf challenge in 15 languages by e-dard on Friday December 14, 2012 @12:18PM (#4921964)
117 is my best Python solution yet, but I'm struggling to get higher than that!
Re:Fizz Buzz codegolf challenge in 15 languages by jaegerpicker on Friday December 14, 2012 @12:22PM (#4921996)
nvm it's just running realllllly slow...