General FizzBuzz: Difference between revisions
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ReeceGoding (talk | contribs) (→{{header|R}}: Shortened code and improved formatting.) |
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===... solution=== |
===... solution=== |
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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. |
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. |
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<lang r>genFizzBuzz<-function(n,...) |
<lang r>genFizzBuzz <- function(n, ...) |
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{ |
{ |
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args<-list(...) |
args <- list(...) |
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#R doesn't like vectors of mixed types, so c(3,"Fizz") is coerced to c("3","Fizz"). We must undo this. |
#R doesn't like vectors of mixed types, so c(3, "Fizz") is coerced to c("3", "Fizz"). We must undo this. |
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#Treating "[[" as if it is a function is a bit of R's magic. You can treat it like a function because it actually is one. |
#Treating "[[" as if it is a function is a bit of R's magic. You can treat it like a function because it actually is one. |
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factors<-as.integer(sapply(args, "[[", 1)) |
factors <- as.integer(sapply(args, "[[", 1)) |
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words<-sapply(args, "[[", 2) |
words <- sapply(args, "[[", 2) |
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sortedPermutation<-sort.list(factors)#Required by the task: We must go from least factor to greatest. |
sortedPermutation <- sort.list(factors)#Required by the task: We must go from least factor to greatest. |
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factors<-factors[sortedPermutation] |
factors <- factors[sortedPermutation] |
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words<-words[sortedPermutation] |
words <- words[sortedPermutation] |
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for(i in 1:n) |
for(i in 1:n) |
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{ |
{ |
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isFactor<-i %% factors == 0 |
isFactor <- i %% factors == 0 |
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if(any(isFactor)) |
print(if(any(isFactor)) paste0(words[isFactor], collapse = "") else i) |
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} |
} |
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invisible() |
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} |
} |
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genFizzBuzz(105,c(3,"Fizz"),c(5,"Buzz"),c(7,"Baxx")) |
genFizzBuzz(105, c(3, "Fizz"), c(5, "Buzz"), c(7, "Baxx")) |
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genFizzBuzz(105,c(5,"Buzz"),c(9,"Prax"),c(3,"Fizz"),c(7,"Baxx"))</lang> |
genFizzBuzz(105, c(5, "Buzz"), c(9, "Prax"), c(3, "Fizz"), c(7, "Baxx"))</lang> |
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===Names solution=== |
===Names solution=== |
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If we deviate from the task's example of how to input parameters and instead use R's names facilities to make our (number, name) pairs, we get a much cleaner solution. |
If we deviate from the task's example of how to input parameters and instead use R's names facilities to make our (number, name) pairs, we get a much cleaner solution. |
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<lang r>namedGenFizzBuzz<-function(n,namedNums) |
<lang r>namedGenFizzBuzz <- function(n, namedNums) |
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{ |
{ |
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factors<-sort(namedNums)#Required by the task: We must go from least factor to greatest. |
factors<-sort(namedNums)#Required by the task: We must go from least factor to greatest. |
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for(i in 1:n) |
for(i in 1:n) |
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{ |
{ |
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isFactor<-i %% factors == 0 |
isFactor <- i %% factors == 0 |
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if(any(isFactor)) |
print(if(any(isFactor)) paste0(names(factors)[isFactor], collapse = "") else i) |
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} |
} |
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invisible() |
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} |
} |
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namedNums<-c(Fizz=3,Buzz=5,Baxx=7)#Notice that we can name our inputs without a call |
namedNums <- c(Fizz=3, Buzz=5, Baxx=7)#Notice that we can name our inputs without a function call. |
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namedGenFizzBuzz(105,namedNums) |
namedGenFizzBuzz(105, namedNums) |
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shuffledNamedNums<-c(Buzz=5,Prax=9,Fizz=3,Baxx=7) |
shuffledNamedNums <- c(Buzz=5, Prax=9, Fizz=3, Baxx=7) |
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namedGenFizzBuzz(105,shuffledNamedNums)</lang> |
namedGenFizzBuzz(105, shuffledNamedNums)</lang> |
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=={{header|Racket}}== |
=={{header|Racket}}== |
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