1) prime number program

f<-function(x)
{
  flag=0
  if(x==1)
  {
    print(" 1 is nor prime nor composite")
  }else if(x==2)
    {
    print("2 is a prime number")
    }else
      { 
          m=x/2
          for(i in 2:m)
          {
            if(x%%i==0)
            {
              print(paste(x,"is not a prime number"))
              flag=1
              break
            }
          }
  
          if(flag==0)
          {
            print(paste(x,"is prime number"))
          }
      }
}
f(1)
## [1] " 1 is nor prime nor composite"
f(2)
## [1] "2 is a prime number"
f(3)
## [1] "3 is prime number"
f(4)
## [1] "4 is not a prime number"

2) Generate a multiplication table for numbers ranging from 1 to 10.

for(i in 1:10)
{
  print(paste("Multiplication table of ",  i))
  for(j in 1:10)
  {
    print(paste(i,"*",j,"=",i*j))
  }
}
## [1] "Multiplication table of  1"
## [1] "1 * 1 = 1"
## [1] "1 * 2 = 2"
## [1] "1 * 3 = 3"
## [1] "1 * 4 = 4"
## [1] "1 * 5 = 5"
## [1] "1 * 6 = 6"
## [1] "1 * 7 = 7"
## [1] "1 * 8 = 8"
## [1] "1 * 9 = 9"
## [1] "1 * 10 = 10"
## [1] "Multiplication table of  2"
## [1] "2 * 1 = 2"
## [1] "2 * 2 = 4"
## [1] "2 * 3 = 6"
## [1] "2 * 4 = 8"
## [1] "2 * 5 = 10"
## [1] "2 * 6 = 12"
## [1] "2 * 7 = 14"
## [1] "2 * 8 = 16"
## [1] "2 * 9 = 18"
## [1] "2 * 10 = 20"
## [1] "Multiplication table of  3"
## [1] "3 * 1 = 3"
## [1] "3 * 2 = 6"
## [1] "3 * 3 = 9"
## [1] "3 * 4 = 12"
## [1] "3 * 5 = 15"
## [1] "3 * 6 = 18"
## [1] "3 * 7 = 21"
## [1] "3 * 8 = 24"
## [1] "3 * 9 = 27"
## [1] "3 * 10 = 30"
## [1] "Multiplication table of  4"
## [1] "4 * 1 = 4"
## [1] "4 * 2 = 8"
## [1] "4 * 3 = 12"
## [1] "4 * 4 = 16"
## [1] "4 * 5 = 20"
## [1] "4 * 6 = 24"
## [1] "4 * 7 = 28"
## [1] "4 * 8 = 32"
## [1] "4 * 9 = 36"
## [1] "4 * 10 = 40"
## [1] "Multiplication table of  5"
## [1] "5 * 1 = 5"
## [1] "5 * 2 = 10"
## [1] "5 * 3 = 15"
## [1] "5 * 4 = 20"
## [1] "5 * 5 = 25"
## [1] "5 * 6 = 30"
## [1] "5 * 7 = 35"
## [1] "5 * 8 = 40"
## [1] "5 * 9 = 45"
## [1] "5 * 10 = 50"
## [1] "Multiplication table of  6"
## [1] "6 * 1 = 6"
## [1] "6 * 2 = 12"
## [1] "6 * 3 = 18"
## [1] "6 * 4 = 24"
## [1] "6 * 5 = 30"
## [1] "6 * 6 = 36"
## [1] "6 * 7 = 42"
## [1] "6 * 8 = 48"
## [1] "6 * 9 = 54"
## [1] "6 * 10 = 60"
## [1] "Multiplication table of  7"
## [1] "7 * 1 = 7"
## [1] "7 * 2 = 14"
## [1] "7 * 3 = 21"
## [1] "7 * 4 = 28"
## [1] "7 * 5 = 35"
## [1] "7 * 6 = 42"
## [1] "7 * 7 = 49"
## [1] "7 * 8 = 56"
## [1] "7 * 9 = 63"
## [1] "7 * 10 = 70"
## [1] "Multiplication table of  8"
## [1] "8 * 1 = 8"
## [1] "8 * 2 = 16"
## [1] "8 * 3 = 24"
## [1] "8 * 4 = 32"
## [1] "8 * 5 = 40"
## [1] "8 * 6 = 48"
## [1] "8 * 7 = 56"
## [1] "8 * 8 = 64"
## [1] "8 * 9 = 72"
## [1] "8 * 10 = 80"
## [1] "Multiplication table of  9"
## [1] "9 * 1 = 9"
## [1] "9 * 2 = 18"
## [1] "9 * 3 = 27"
## [1] "9 * 4 = 36"
## [1] "9 * 5 = 45"
## [1] "9 * 6 = 54"
## [1] "9 * 7 = 63"
## [1] "9 * 8 = 72"
## [1] "9 * 9 = 81"
## [1] "9 * 10 = 90"
## [1] "Multiplication table of  10"
## [1] "10 * 1 = 10"
## [1] "10 * 2 = 20"
## [1] "10 * 3 = 30"
## [1] "10 * 4 = 40"
## [1] "10 * 5 = 50"
## [1] "10 * 6 = 60"
## [1] "10 * 7 = 70"
## [1] "10 * 8 = 80"
## [1] "10 * 9 = 90"
## [1] "10 * 10 = 100"

3) Create an R script that calculates the square root of a given integer vector x of length one, if the value contained in x is negative it should return NA.

g<-function(x)
{
    if(x>0)
  {
    print(sqrt(x))
  }else
    {
    print("NA")
  }
}

let’s pass a possitive value

g(5)
## [1] 2.236068

Now, let’s pass a negative value

g(-5)
## [1] "NA"

4) Write a function which takes a vector from 1 to 100. Make a for-loop which runs through the whole vector. Multiply the elements which are smaller than 5 and larger than 90 with 10 and the other elements with 0.1., so that you can feed it any vector you like (as argument). Use a for-loop in the function to do the computation with each element.

ho<-function(n)
{
  for(i in n)
  {
    if(i<5 | i>90)
    {
      print(i*10)
    }else
    {
      print(i*0.1)
    }
  }
 
}
ho(1:100)
## [1] 10
## [1] 20
## [1] 30
## [1] 40
## [1] 0.5
## [1] 0.6
## [1] 0.7
## [1] 0.8
## [1] 0.9
## [1] 1
## [1] 1.1
## [1] 1.2
## [1] 1.3
## [1] 1.4
## [1] 1.5
## [1] 1.6
## [1] 1.7
## [1] 1.8
## [1] 1.9
## [1] 2
## [1] 2.1
## [1] 2.2
## [1] 2.3
## [1] 2.4
## [1] 2.5
## [1] 2.6
## [1] 2.7
## [1] 2.8
## [1] 2.9
## [1] 3
## [1] 3.1
## [1] 3.2
## [1] 3.3
## [1] 3.4
## [1] 3.5
## [1] 3.6
## [1] 3.7
## [1] 3.8
## [1] 3.9
## [1] 4
## [1] 4.1
## [1] 4.2
## [1] 4.3
## [1] 4.4
## [1] 4.5
## [1] 4.6
## [1] 4.7
## [1] 4.8
## [1] 4.9
## [1] 5
## [1] 5.1
## [1] 5.2
## [1] 5.3
## [1] 5.4
## [1] 5.5
## [1] 5.6
## [1] 5.7
## [1] 5.8
## [1] 5.9
## [1] 6
## [1] 6.1
## [1] 6.2
## [1] 6.3
## [1] 6.4
## [1] 6.5
## [1] 6.6
## [1] 6.7
## [1] 6.8
## [1] 6.9
## [1] 7
## [1] 7.1
## [1] 7.2
## [1] 7.3
## [1] 7.4
## [1] 7.5
## [1] 7.6
## [1] 7.7
## [1] 7.8
## [1] 7.9
## [1] 8
## [1] 8.1
## [1] 8.2
## [1] 8.3
## [1] 8.4
## [1] 8.5
## [1] 8.6
## [1] 8.7
## [1] 8.8
## [1] 8.9
## [1] 9
## [1] 910
## [1] 920
## [1] 930
## [1] 940
## [1] 950
## [1] 960
## [1] 970
## [1] 980
## [1] 990
## [1] 1000

5) Create a character vector with length of number-of-rows-of-iris-dataset, such that,each element gets a character value – “greater than 5. if the corresponding ‘Sepal.Length’ > 5, else it should get “lesser than 5. Create a logic for the same problem statement using apply() function

Let’s first look how the iris dataset look like.

head(iris)
##   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1          5.1         3.5          1.4         0.2  setosa
## 2          4.9         3.0          1.4         0.2  setosa
## 3          4.7         3.2          1.3         0.2  setosa
## 4          4.6         3.1          1.5         0.2  setosa
## 5          5.0         3.6          1.4         0.2  setosa
## 6          5.4         3.9          1.7         0.4  setosa

let’s look at the structure of iris dataset.

str(iris)
## 'data.frame':    150 obs. of  5 variables:
##  $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
##  $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
##  $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
##  $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
##  $ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...

Solve the problem using ifelse

a<-ifelse(iris$Sepal.Length>5,"greater than 5","lesser than 5")
a
##   [1] "greater than 5" "lesser than 5"  "lesser than 5"  "lesser than 5" 
##   [5] "lesser than 5"  "greater than 5" "lesser than 5"  "lesser than 5" 
##   [9] "lesser than 5"  "lesser than 5"  "greater than 5" "lesser than 5" 
##  [13] "lesser than 5"  "lesser than 5"  "greater than 5" "greater than 5"
##  [17] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [21] "greater than 5" "greater than 5" "lesser than 5"  "greater than 5"
##  [25] "lesser than 5"  "lesser than 5"  "lesser than 5"  "greater than 5"
##  [29] "greater than 5" "lesser than 5"  "lesser than 5"  "greater than 5"
##  [33] "greater than 5" "greater than 5" "lesser than 5"  "lesser than 5" 
##  [37] "greater than 5" "lesser than 5"  "lesser than 5"  "greater than 5"
##  [41] "lesser than 5"  "lesser than 5"  "lesser than 5"  "lesser than 5" 
##  [45] "greater than 5" "lesser than 5"  "greater than 5" "lesser than 5" 
##  [49] "greater than 5" "lesser than 5"  "greater than 5" "greater than 5"
##  [53] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [57] "greater than 5" "lesser than 5"  "greater than 5" "greater than 5"
##  [61] "lesser than 5"  "greater than 5" "greater than 5" "greater than 5"
##  [65] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [69] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [73] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [77] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [81] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [85] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [89] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [93] "greater than 5" "lesser than 5"  "greater than 5" "greater than 5"
##  [97] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [101] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [105] "greater than 5" "greater than 5" "lesser than 5"  "greater than 5"
## [109] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [113] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [117] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [121] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [125] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [129] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [133] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [137] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [141] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [145] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [149] "greater than 5" "greater than 5"

Solve the problem using apply() function. * First create a user-defined function. * Then use apply function

j<-function(sepl_length)
{
  ifelse(iris$Sepal.Length>5,"greater than 5","lesser than 5")
}

Now use apply function to use the above function on the dataset.

b<-apply(iris[,c(-3:-5)],2,j) # Note that apply is for matrix not vector so, i am passing 2 columns
b[,"Sepal.Length"]
##   [1] "greater than 5" "lesser than 5"  "lesser than 5"  "lesser than 5" 
##   [5] "lesser than 5"  "greater than 5" "lesser than 5"  "lesser than 5" 
##   [9] "lesser than 5"  "lesser than 5"  "greater than 5" "lesser than 5" 
##  [13] "lesser than 5"  "lesser than 5"  "greater than 5" "greater than 5"
##  [17] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [21] "greater than 5" "greater than 5" "lesser than 5"  "greater than 5"
##  [25] "lesser than 5"  "lesser than 5"  "lesser than 5"  "greater than 5"
##  [29] "greater than 5" "lesser than 5"  "lesser than 5"  "greater than 5"
##  [33] "greater than 5" "greater than 5" "lesser than 5"  "lesser than 5" 
##  [37] "greater than 5" "lesser than 5"  "lesser than 5"  "greater than 5"
##  [41] "lesser than 5"  "lesser than 5"  "lesser than 5"  "lesser than 5" 
##  [45] "greater than 5" "lesser than 5"  "greater than 5" "lesser than 5" 
##  [49] "greater than 5" "lesser than 5"  "greater than 5" "greater than 5"
##  [53] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [57] "greater than 5" "lesser than 5"  "greater than 5" "greater than 5"
##  [61] "lesser than 5"  "greater than 5" "greater than 5" "greater than 5"
##  [65] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [69] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [73] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [77] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [81] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [85] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [89] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
##  [93] "greater than 5" "lesser than 5"  "greater than 5" "greater than 5"
##  [97] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [101] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [105] "greater than 5" "greater than 5" "lesser than 5"  "greater than 5"
## [109] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [113] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [117] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [121] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [125] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [129] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [133] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [137] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [141] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [145] "greater than 5" "greater than 5" "greater than 5" "greater than 5"
## [149] "greater than 5" "greater than 5"

6) Write a program to print fibonacchi series upto a range specified.

fib_series<-function(x)
{
  n1=0
  n2=1
  if(x<0)
  {
    print("The number should not be less than zero")
  }else if(x==1)
  {
    print(x)
  }else
  {
    print(n1)
    print(n2)
    for(i in 3:x)
    {
      n3=n1+n2
      print(n3)
      n1=n2
      n2=n3
    }
  }
}

Now, we can specify a range.

fib_series(8)
## [1] 0
## [1] 1
## [1] 1
## [1] 2
## [1] 3
## [1] 5
## [1] 8
## [1] 13

7) Find Sum of Series 1²+2²+3²+…..+n².

fun<-function(x)
{
 sum_ser=0
 for(i in 1:x)
 {
   sum_ser<-sum_ser+(i^2)
 }
 print(sum_ser)
}
fun(5)
## [1] 55

8) Find Sum of Series 1²+2²+3²+…..+n² using the Recursive function.

fun_rec<-function(y)
{
  if(y==0)
  {
    return(0)
  }else{
    return((y*y)+fun_rec(y-1))
  }
}
fun_rec(5)
## [1] 55

9) Consider the vector 1:K, where K is a positive integer. Write an R command that determines how many elements in the vector are exactly divisible by 3.

v<-1:100 # Let k value be 100
v
##   [1]   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17
##  [18]  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34
##  [35]  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51
##  [52]  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68
##  [69]  69  70  71  72  73  74  75  76  77  78  79  80  81  82  83  84  85
##  [86]  86  87  88  89  90  91  92  93  94  95  96  97  98  99 100
j<-ifelse(v%%3==0,TRUE,FALSE) # If the number is divisible, then true else false
v[j] # Displays only those numbers whose index is true
##  [1]  3  6  9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69
## [24] 72 75 78 81 84 87 90 93 96 99
length(v[j]) # verify its length
## [1] 33

Find the sum of natural numbers using the Recursive function.

fun_rec<-function(y)
{
  if(y==0)
  {
    return(0)
  }else{
    return(y+fun_rec(y-1))
  }
}
fun_rec(5)
## [1] 15