Different implementations of the FizzBuzz problem using R Language

The FizzBuzz is a well-known problem in R Language. ‘Fizz’ refers to numbers that are multiples of three, ‘Buzz’ refers to numbers that are multiples of five, and ‘FizzBuzz’ refers to numbers that are multiples of both three and five.

Method 1: implementations of the FizzBuzz using Classical Approach

This is a classical approach to solving the FizzBuzz problem, We start by running the for loop for every number that exists in the sequence range. We have created a sequence using the ‘:’ operator. The ‘:’ operator takes the step size as 1. Now inside the for loop we have if, else if, and else conditional statements. We will then check the divisibility using the modulus operator (%%) of the number.

If the number is divisible by 15 we output ‘FizzBuzz’, else if we check that if a number is divisible by 3, if the number is divisible by 3 then we output ‘Fizz’, further we check if the divisible by 5, if the number is divisible by 5, we then output ‘Buzz’. If none of these conditions are satisfied the else conditional statement gets executed, where the number is directly outputted as it is.

R

for (number in 1:100) {
  if (number %% 15 == 0){
    cat('FizzBuzz\n')
  }
  else if (number %% 3 == 0){
    cat('Fizz\n')
  }
  else if (number %% 5 == 0){
    cat('Buzz\n')
  }
  else{
    cat(number)
    cat('\n')
  }
}

Output:

1 2 Fizz  4 Buzz  Fizz  7 8 Fizz  Buzz  11 Fizz  13 14 Fizz  Buzz  16 17 Fizz  19 Buzz  Fizz  22 23 Fizz  Buzz  26 Fizz  28 29 Fizz  Buzz  31 32 Fizz  34 Buzz  Fizz  37 38 Fizz  Buzz  41 Fizz  43 44 Fizz  Buzz  46 47 Fizz  49 Buzz  Fizz  52 53 Fizz  Buzz  56 Fizz  58 59 Fizz  Buzz  61 62 Fizz  64 Buzz  Fizz  67 68 Fizz  Buzz  71 Fizz  73 74 Fizz  Buzz  76 77 Fizz  79 Buzz  Fizz  82 83 Fizz  Buzz  86 Fizz  88 89 Fizz  Buzz  91 92 Fizz  94 Buzz  Fizz  97 98 Fizz  Buzz 

Method 2: Efficient solution than the Classical using loop

In this solution, we try to reduce the number of conditional statements. Here, we are using paste0 because we want to concatenate without a separator. Let us assume the number is 9. We check the following condition ‘9%%3==0’. The output of the following condition will be True. So the statements inside the if condition will be executed. We are concatenating all elements that are o which is an empty string, and a new string “Fizz”. We then store all elements back in o. We finally display the result for each number.

R

for (number in 1:100) {
  o <- ''
  if(number%%3==0){
    o<-paste0(o,"Fizz")
  }
  if(number%%5==0){
    o<-paste0(o,"Buzz")
  }
  if(o==''){
    o<-number
  }
  cat(o)
  cat('\n')
}

Output:

1 2 Fizz  4 Buzz  Fizz  7 8 Fizz  Buzz  11 Fizz  13 14 Fizz  Buzz  16 17 Fizz  19 Buzz  Fizz  22 23 Fizz  Buzz  26 Fizz  28 29 Fizz  Buzz  31 32 Fizz  34 Buzz  Fizz  37 38 Fizz  Buzz  41 Fizz  43 44 Fizz  Buzz  46 47 Fizz  49 Buzz  Fizz  52 53 Fizz  Buzz  56 Fizz  58 59 Fizz  Buzz  61 62 Fizz  64 Buzz  Fizz  67 68 Fizz  Buzz  71 Fizz  73 74 Fizz  Buzz  76 77 Fizz  79 Buzz  Fizz  82 83 Fizz  Buzz  86 Fizz  88 89 Fizz  Buzz  91 92 Fizz  94 Buzz  Fizz  97 98 Fizz  Buzz 

Method 3: Implementations of the FizzBuzz using the function

In this solution, we try to eliminate them for loop and instead utilized a function sapply(). The sapply() function takes a vector as the input and returns the result of the function on the sequence that we have passed as an input. Then we have the code of a function that had a quite similar functioning as explained in solution 2.

R

fizz_buzz <- function(number){
  o <- ''
  if(number%%3==0){
    o<-paste0(o,"Fizz")
  }
  if(number%%5==0){
    o<-paste0(o,"Buzz")
  }
  if(o==''){
    o<-number
  }
  cat(o)
  cat('\n')
}
fb<-sapply(1:100, fizz_buzz)[0]

Output:

1 2 Fizz  4 Buzz  Fizz  7 8 Fizz  Buzz  11 Fizz  13 14 Fizz  Buzz  16 17 Fizz  19 Buzz  Fizz  22 23 Fizz  Buzz  26 Fizz  28 29 Fizz  Buzz  31 32 Fizz  34 Buzz  Fizz  37 38 Fizz  Buzz  41 Fizz  43 44 Fizz  Buzz  46 47 Fizz  49 Buzz  Fizz  52 53 Fizz  Buzz  56 Fizz  58 59 Fizz  Buzz  61 62 Fizz  64 Buzz  Fizz  67 68 Fizz  Buzz  71 Fizz  73 74 Fizz  Buzz  76 77 Fizz  79 Buzz  Fizz  82 83 Fizz  Buzz  86 Fizz  88 89 Fizz  Buzz  91 92 Fizz  94 Buzz  Fizz  97 98 Fizz  Buzz 

Method 4: Implementations of the FizzBuzz using package ‘fizzbuzzR’ 

Here we are utilizing a function namely ‘fizzbuzz’ in the ‘fizzbuzzR’ package. It takes the following arguments:

• start: This is the starting value of the loop
• end: The loop will end at this value
• mod1: If we specify mod1 as 3, we are then saying that we will replace all the divisors of 3 with ‘Fizz’
• mod2: If we specify mod2 as 5, we are then saying that we will replace all the divisors of 5 with ‘Buzz’

R

install.packages("fizzbuzzR")
library(fizzbuzzR)
fizzbuzz(start = 1, end = 100, mod1 = 3, mod2 = 5)

Output:

1 2 Fizz  4 Buzz  Fizz  7 8 Fizz  Buzz  11 Fizz  13 14 Fizz  Buzz  16 17 Fizz  19 Buzz  Fizz  22 23 Fizz  Buzz  26 Fizz  28 29 Fizz  Buzz  31 32 Fizz  34 Buzz  Fizz  37 38 Fizz  Buzz  41 Fizz  43 44 Fizz  Buzz  46 47 Fizz  49 Buzz  Fizz  52 53 Fizz  Buzz  56 Fizz  58 59 Fizz  Buzz  61 62 Fizz  64 Buzz  Fizz  67 68 Fizz  Buzz  71 Fizz  73 74 Fizz  Buzz  76 77 Fizz  79 Buzz  Fizz  82 83 Fizz  Buzz  86 Fizz  88 89 Fizz  Buzz  91 92 Fizz  94 Buzz  Fizz  97 98 Fizz  Buzz