for i in range(1, 101): print (i if i%3 else "Fizz") if i%5 else "Buzz" if i%15 ... | Hacker News

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danbmil99 on April 10, 2014 | parent | context | favorite | on: Unpythonic Python

for i in range(1, 101): print (i if i%3 else "Fizz") if i%5 else "Buzz" if i%15 else "FizzBuzz"

[edit: proper place https://news.ycombinator.com/item?id=7564988]

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