General FizzBuzz: Difference between revisions
→{{header|R}}: Changed variable name, refactored if statement, and added new solution.
ReeceGoding (talk | contribs) (Added R.) |
ReeceGoding (talk | contribs) (→{{header|R}}: Changed variable name, refactored if statement, and added new solution.) |
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=={{header|R}}==
===... solution===
The task asks that we assume 3 factors for the sake of simplicity. However, R makes the k factors case not much more complicated, so we will do that. The only major downside is that checking for malformed user input becomes so difficult that we will not bother.
<lang r>genFizzBuzz<-function(n,...)
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for(i in 1:n)
{
if(
}
invisible()
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genFizzBuzz(105,c(3,"Fizz"),c(5,"Buzz"),c(7,"Baxx"))
genFizzBuzz(105,c(5,"Buzz"),c(9,"Prax"),c(3,"Fizz"),c(7,"Baxx"))</lang>
===Names solution===
If we deviate from the task's example of how to input parameters and instead use R's names function to make our (number, name) pairs, we get a much cleaner solution.
<lang r>namedGenFizzBuzz<-function(n,namedNums)
{
factors<-sort(namedNums)#Required by the task: We must go from least factor to greatest.
for(i in 1:n)
{
isFactor<-i %% factors == 0
if(any(isFactor)){print(paste0(names(factors)[isFactor],collapse = ""))}else{print(i)}
}
invisible()
}
namedNums<-c(3,5,7); names(namedNums)<-c("Fizz","Buzz","Baxx")
namedGenFizzBuzz(105,namedNums)
shuffledNamedNums<-c(5,9,3,7); names(shuffledNamedNums)<-c("Buzz","Prax","Fizz","Baxx")
namedGenFizzBuzz(105,shuffledNamedNums)</lang>
=={{header|Racket}}==
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