FizzBuzz without an if

Apparently using an if to solve FizzBuzz is a sign that you’re not a great programmer… Well, that’s how I would have done it :-(

I wanted to see what the map alternative looked like, and I found this python example

print(*map(lambda i: 'Fizz'*(not i%3)+'Buzz'*(not i%5) or i, range(1,101)),sep='\n')

What’s interesting to me is that there’s an or in there - if neither of the modulo operations match, then the resulting string is empty, and that can be used in an or (failing the left-hand side). Whoa! That doesn’t work in R because characters can’t be used in logical operations

"" | 2
Error in "" | 2 : 
  operations are possible only for numeric, logical or complex types

The rest I can more or less reproduce in R, using strrep in place of the infix * used for ‘repeat string’

sapply(1:20, 
       \(x) paste0(
         strrep("Fizz", x %% 3 == 0), 
         strrep("Buzz", x %% 5 == 0), 
         strrep(x, x %% 3 != 0 && x %% 5 != 0)
       )
)
 [1] "1"        "2"        "Fizz"     "4"        "Buzz"     "Fizz"    
 [7] "7"        "8"        "Fizz"     "Buzz"     "11"       "Fizz"    
[13] "13"       "14"       "FizzBuzz" "16"       "17"       "Fizz"    
[19] "19"       "Buzz"    

This also requires the additional comparison for when the number is a divisor of neither number, so kudos to python for working nicer in that case.

Is there an even better way to solve this without an if?