R で解く FizzBuzz 問題

Nov 18, 20114 likes2,082 views

The document presents four different functions (fb1, fb2, fb3, fbcpp) for solving the FizzBuzz problem and benchmarks their performance on different sized test cases. fb3 is the fastest approach, using modulo operations and ifelse statements to vectorize the logic. The document then explores compiling fb1-fb3 to C++ using Rcpp to further improve performance, with fbcpp showing the fastest results.

> fizzbuzz <- function(n) {
# Write your code here
}

> fizzbuzz(20)
[1] "1" "2" "Fizz"
[4] "4" "Buzz" "Fizz"
[7] "7" "8" "Fizz"
[10] "Buzz" "11" "Fizz"
[13] "13" "14" "FizzBuzz"
[16] "16" "17" "Fizz"
[19] "19" "Buzz"

fb1 <- function(n) {
fb <- character(n)
for (i in 1:n) {
if (i %% 3 == 0 && i %% 5 == 0)
fb[i] <- "FizzBuzz"
else if (i %% 3 == 0)
fb[i] <- "Fizz"
else if (i %% 5 == 0)
fb[i] <- "Buzz"
else
fb[i] <- i
}
fb
}

fb2 <- function(n) {
sapply(1:n, function(i) {
if (i %% 3 == 0 && i %% 5 == 0)
"FizzBuzz"
else if (i %% 3 == 0)
"Fizz"
else if (i %% 5 == 0)
"Buzz"
else
i
})
}

fb3 <- function(n) {
isFizz <- rep(c(F,F,T), length=n)
isBuzz <- rep(c(F,F,F,F,T), length=n)
isNumber <- !isFizz & !isBuzz

fizz <- ifelse(isFizz, "Fizz", "")
buzz <- ifelse(isBuzz, "Buzz", "")
number <- ifelse(isNumber, 1:n, "")

paste(fizz, buzz, number, sep="")
}

> N <- 100
> benchmark(fb1(N), fb2(N), fb3(N),
order="relative",
replications=10000)

test elapsed relative
3 fb3(N) 4.172 1.000000
2 fb2(N) 6.669 1.598514
1 fb1(N) 7.779 1.864573

> N <- 10000
> benchmark(fb1(N), fb2(N), fb3(N),
order="relative",
replications=100)

test elapsed relative
3 fb3(N) 2.907 1.000000
2 fb2(N) 7.823 2.691090
1 fb1(N) 8.200 2.820777

> N <- 1000000
> benchmark(fb1(N), fb2(N), fb3(N),
order="relative",
replications=5)

test elapsed relative
3 fb3(N) 28.287 1.000000
1 fb1(N) 46.198 1.633188
2 fb2(N) 52.223 1.846184

library(compiler) # Installed by default
library(inline) # Needs manual installation

fbcpp <- cxxfunction(
signature(ns="integer"),
body=src, # Shown in the next slide
plugin="Rcpp")

fb1.c <- cmpfun(fb1)
fb2.c <- cmpfun(fb2)
fb3.c <- cmpfun(fb3)
fbcpp.c <- cmpfun(fbcpp)

int n = Rcpp::as<int>(ns);
Rcpp::CharacterVector fb(n);
for (int index = 0; index < n; index++) {
int i = index + 1;
if (i % 3 == 0 && i % 5 == 0)
fb[index] = "FizzBuzz";
else if (i % 3 == 0)
fb[index] = "Fizz";
else if (i % 5 == 0)
fb[index] = "Buzz";
else {
char s[11]; // Integer is 10 or less digit
sprintf(s, "%d", i);
fb[index] = s;
}
}
return(fb);

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R で解く FizzBuzz 問題

26. > fizzbuzz <- function(n) { # Write your code here } > fizzbuzz(20) [1] "1" "2" "Fizz" [4] "4" "Buzz" "Fizz" [7] "7" "8" "Fizz" [10] "Buzz" "11" "Fizz" [13] "13" "14" "FizzBuzz" [16] "16" "17" "Fizz" [19] "19" "Buzz"

28. fb1 <- function(n) { fb <- character(n) for (i in 1:n) { if (i %% 3 == 0 && i %% 5 == 0) fb[i] <- "FizzBuzz" else if (i %% 3 == 0) fb[i] <- "Fizz" else if (i %% 5 == 0) fb[i] <- "Buzz" else fb[i] <- i } fb }

30. fb2 <- function(n) { sapply(1:n, function(i) { if (i %% 3 == 0 && i %% 5 == 0) "FizzBuzz" else if (i %% 3 == 0) "Fizz" else if (i %% 5 == 0) "Buzz" else i }) }

35. fb3 <- function(n) { isFizz <- rep(c(F,F,T), length=n) isBuzz <- rep(c(F,F,F,F,T), length=n) isNumber <- !isFizz & !isBuzz fizz <- ifelse(isFizz, "Fizz", "") buzz <- ifelse(isBuzz, "Buzz", "") number <- ifelse(isNumber, 1:n, "") paste(fizz, buzz, number, sep="") }

40. > N <- 100 > benchmark(fb1(N), fb2(N), fb3(N), order="relative", replications=10000) test elapsed relative 3 fb3(N) 4.172 1.000000 2 fb2(N) 6.669 1.598514 1 fb1(N) 7.779 1.864573

41. > N <- 10000 > benchmark(fb1(N), fb2(N), fb3(N), order="relative", replications=100) test elapsed relative 3 fb3(N) 2.907 1.000000 2 fb2(N) 7.823 2.691090 1 fb1(N) 8.200 2.820777

42. > N <- 1000000 > benchmark(fb1(N), fb2(N), fb3(N), order="relative", replications=5) test elapsed relative 3 fb3(N) 28.287 1.000000 1 fb1(N) 46.198 1.633188 2 fb2(N) 52.223 1.846184

47. library(compiler) # Installed by default library(inline) # Needs manual installation fbcpp <- cxxfunction( signature(ns="integer"), body=src, # Shown in the next slide plugin="Rcpp") fb1.c <- cmpfun(fb1) fb2.c <- cmpfun(fb2) fb3.c <- cmpfun(fb3) fbcpp.c <- cmpfun(fbcpp)

48. int n = Rcpp::as<int>(ns); Rcpp::CharacterVector fb(n); for (int index = 0; index < n; index++) { int i = index + 1; if (i % 3 == 0 && i % 5 == 0) fb[index] = "FizzBuzz"; else if (i % 3 == 0) fb[index] = "Fizz"; else if (i % 5 == 0) fb[index] = "Buzz"; else { char s[11]; // Integer is 10 or less digit sprintf(s, "%d", i); fb[index] = s; } } return(fb);

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R で解く FizzBuzz 問題

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