13 ms·

fizz = cycle ["","","Fizz"] buzz = cycle ["","","","","Buzz"] fizzbuzz = zipWith (++) fizz buzz fb = zipWith (\n cs -> if cs == "" then show n else

fizz = cycle ["","","Fizz"] buzz = cycle ["","","","","Buzz"] fizzbuzz = zipWith (++) fizz buzz fb = zipWith (\n cs -> if cs == "" then show n else cs) [1..] fizzbuzz mapM_ putStrLn fb

GyrosOfWar 13y ago
Very cool. This really makes me want to get better at Haskell. I know enough of it to understand what's going on here, but not enough to be able to come up with something like this.
waps 13y ago
I love this answer. It's like "yes I know and hate fizzbuzz, also you don't know crap", but in code. Poetry. In python: from itertools import * fizz = cycle(["", "", "fizz"]) buzz = cycle([""] * 4 + ["buzz"]) fizzbuzz = izip(fizz, buzz, xrange(1,100)) print "\n".join(map(lambda (x,y,z): x+y if x or y else str(z), fizzbuzz)) (bugs found by marekmroz fixed) Can you do [1,2,3] * 4 in haskell as well ?
- marekmroz 13y ago
  Funny thing is, as soon as I saw OP's solution I fired up IPython to do the same thing in Python. :) Admittedly mine was not as elegant as I haven't thought of using izip with 3 iterables. Still, I don't think that lambda and islice is really needed... xrange(100) guarantees a finite number of iterations already. from itertools import * fizzer = cycle(['Fizz','','']) buzzer = cycle(['Buzz','','','','']) fizzbuzzer = izip(xrange(100), fizzer, buzzer) for f in fizzbuzzer: print f[1] + f[2] if f[1] or f[2] else f[0] EDIT: the output from your version does not look right. "Fizz" and "Buzz" are on the wrong index positions. 0 1 fizz 3 buzz fizz 6 7 fizz buzz ...
  - waps 13y ago
    heh you're right. Now I feel so dumb. Off by one error. I needed to use xrange(1, 100) instead of just xrange. You're also correct islice is not needed.
- gamegoblin 13y ago
  I more gray-bearded haskell wizard than I might have a better solution, but to me the direct haskell equivalent of python's [1,2,3]*4 is concat $ replicate 4 [1,2,3]

fizz = cycle ["","","Fizz"] buzz = cycle ["","","","","Buzz"] fizzbuzz = zipWith (++) fizz buzz fb = zipWith (\n cs -> if cs == "" then show n else cs) [1..] fizzbuzz mapM_ putStrLn fb

GyrosOfWar 13y ago

Very cool. This really makes me want to get better at Haskell. I know enough of it to understand what's going on here, but not enough to be able to come up with something like this.

waps 13y ago

I love this answer. It's like "yes I know and hate fizzbuzz, also you don't know crap", but in code. Poetry. In python: from itertools import * fizz = cycle(["", "", "fizz"]) buzz = cycle([""] * 4 + ["buzz"]) fizzbuzz = izip(fizz, buzz, xrange(1,100)) print "\n".join(map(lambda (x,y,z): x+y if x or y else str(z), fizzbuzz)) (bugs found by marekmroz fixed) Can you do [1,2,3] * 4 in haskell as well ?

marekmroz 13y ago
Funny thing is, as soon as I saw OP's solution I fired up IPython to do the same thing in Python. :) Admittedly mine was not as elegant as I haven't thought of using izip with 3 iterables. Still, I don't think that lambda and islice is really needed... xrange(100) guarantees a finite number of iterations already. from itertools import * fizzer = cycle(['Fizz','','']) buzzer = cycle(['Buzz','','','','']) fizzbuzzer = izip(xrange(100), fizzer, buzzer) for f in fizzbuzzer: print f[1] + f[2] if f[1] or f[2] else f[0] EDIT: the output from your version does not look right. "Fizz" and "Buzz" are on the wrong index positions. 0 1 fizz 3 buzz fizz 6 7 fizz buzz ...
- waps 13y ago
  heh you're right. Now I feel so dumb. Off by one error. I needed to use xrange(1, 100) instead of just xrange. You're also correct islice is not needed.
gamegoblin 13y ago
I more gray-bearded haskell wizard than I might have a better solution, but to me the direct haskell equivalent of python's [1,2,3]*4 is concat $ replicate 4 [1,2,3]