PS 1 Solutions

Problem 1

Create vectors containing {3.1, 4.1, 5.1, 6.1, 7.1, 8.1} elements in three different ways using the following functions: c(), seq() with a by argument, and seq() with a length.out argument. Name them vector1, vector2, and vector3, respectively.

vector1 <- c(3.1, 4.1, 5.1, 6.1, 7.1, 8.1)

print(vector1)
#> [1] 3.1 4.1 5.1 6.1 7.1 8.1

vector2 <- seq(3.1, 8.1, by=1)

print(vector2)
#> [1] 3.1 4.1 5.1 6.1 7.1 8.1

vector3 <- seq(3.1, 8.1, length.out=6)

print(vector3)
#> [1] 3.1 4.1 5.1 6.1 7.1 8.1

Add three new elements to vector1: 10, 20, 30. 10 should be the first element in the vector, 20 should be the 3rd element in the vector given that 10 has been already added to it, and 30 should be the 6th element in the vector given that 10 and 20 have been already added to the vector.

vector1 <- append(vector1, 10, after=0)

vector1 <- append(vector1, 20, after=2)

vector1 <- append(vector1, 30, after=5)

print(vector1)
#> [1] 10.0  3.1 20.0  4.1  5.1 30.0  6.1  7.1  8.1

Create a new vector containing {2, 3, 6, 10, 15, 18, 22, 25, 27, 30, 31, 35, 42} elements. Name it vector4. Extract/Select elements that are greater than 5 and are divisible by 3 using a logical statement and create a new vector vector5 with these elements. Check whether element 27 is in vector5 using a logical operator.

vector4 <- c(2, 3, 6, 10, 15, 18, 22, 25, 27, 30, 31, 35, 42)

vector5 <- vector4[vector4>5 & vector4%%3==0]

print(vector5)
#> [1]  6 15 18 27 30 42

27 %in% vector5
#> [1] TRUE

Problem 2

Create a list with the following elements: [1, 50, 88], [“yesterday”, “today”, “tomorrow”], 22.5, [33.8, 42], “class_0001”. Name it list1.

list1 <- list(c(1, 50, 88), c("yesterday", "today", "tomorrow"), 22.5, c(33.8, 42), "class_0001")

print(list1)
#> [[1]]
#> [1]  1 50 88
#> 
#> [[2]]
#> [1] "yesterday" "today"     "tomorrow" 
#> 
#> [[3]]
#> [1] 22.5
#> 
#> [[4]]
#> [1] 33.8 42.0
#> 
#> [[5]]
#> [1] "class_0001"

Apply unlist() function to list1. What are the data type and data structure of the output? What do you think why you got this specific data type?

unlist(list1)
#>  [1] "1"          "50"         "88"         "yesterday" 
#>  [5] "today"      "tomorrow"   "22.5"       "33.8"      
#>  [9] "42"         "class_0001"

class(unlist(list1))
#> [1] "character"

str(unlist(list1))
#>  chr [1:10] "1" "50" "88" "yesterday" "today" ...

The unlisted list1 is a vector contains elements in “character” type. The unlist function returns vector in “character” type when the list consists of data in different types, such as c(1, 50, 88) and c("yesterday", "today", "tomorrow").

Create a new list (name it list2) by removing the third element of the first element of list1. Then extract/select the second and third elements of the second element of list2.

list2 <- list1

list2[[1]] <- list2[[1]][-3]

print(list2)
#> [[1]]
#> [1]  1 50
#> 
#> [[2]]
#> [1] "yesterday" "today"     "tomorrow" 
#> 
#> [[3]]
#> [1] 22.5
#> 
#> [[4]]
#> [1] 33.8 42.0
#> 
#> [[5]]
#> [1] "class_0001"

list2[[2]][c(2,3)]
#> [1] "today"    "tomorrow"

Create a new list with the following elements: 23, “new”, 45.7. Name it list3. Now create a new list (name it list4) by merging list2 and list3 with a c(). How many elements does list4 contain (use a built-in function to count). Extract the first element of list4 as a list and as a vector.

list3 <- list(23, "new", 45.7)

print(list3)
#> [[1]]
#> [1] 23
#> 
#> [[2]]
#> [1] "new"
#> 
#> [[3]]
#> [1] 45.7

# merge

list4 <- c(list2, list3)

print(list4)
#> [[1]]
#> [1]  1 50
#> 
#> [[2]]
#> [1] "yesterday" "today"     "tomorrow" 
#> 
#> [[3]]
#> [1] 22.5
#> 
#> [[4]]
#> [1] 33.8 42.0
#> 
#> [[5]]
#> [1] "class_0001"
#> 
#> [[6]]
#> [1] 23
#> 
#> [[7]]
#> [1] "new"
#> 
#> [[8]]
#> [1] 45.7

length(list4)
#> [1] 8

# Extract as a list

list(list4[[1]])
#> [[1]]
#> [1]  1 50

# Extract as a vector

list4[[1]]
#> [1]  1 50

Problem 3

Create a data frame (name it df1) with the following variables:
- Name - {“James”, “Linda”, “Stacy”, “Mary”, “Tom”, “Anna”, “Bob”, “Jeniffer”, “Lucas”, “Amy”, “Jim”}
- Major - {“Math”, “Math”, “Genetics”, “Statistics”, “Accounting”, “Art”, “Music”, “Business”, “Finance”, “Finance”, “Math”}
- Grad_Year - {2023, 2025, 2025, 2024, 2026, 2024, 2025, 2025, 2023, 2026, 2024}
- GPA - {3.9, 3.75, 4.0, 4.0, 3.4, 3.9, 3.3, 3.8, 3.55, 4.0, 3.6}

Name <- c("James", "Linda", "Stacy", "Mary", "Tom", "Anna", "Bob", "Jeniffer", "Lucas", "Amy", "Jim")

Major <- c("Math", "Math", "Genetics", "Statistics", "Accounting", "Art", "Music", "Business", "Finance", "Finance", "Math")

Grad_Year <- c(2023, 2025, 2025, 2024, 2026, 2024, 2025, 2025, 2023, 2026, 2024)

GPA <- c(3.9, 3.75, 4.0, 4.0, 3.4, 3.9, 3.3, 3.8, 3.55, 4.0, 3.6)

df1 <- data.frame(Name, Major, Grad_Year, GPA)

print(df1)
#>        Name      Major Grad_Year  GPA
#> 1     James       Math      2023 3.90
#> 2     Linda       Math      2025 3.75
#> 3     Stacy   Genetics      2025 4.00
#> 4      Mary Statistics      2024 4.00
#> 5       Tom Accounting      2026 3.40
#> 6      Anna        Art      2024 3.90
#> 7       Bob      Music      2025 3.30
#> 8  Jeniffer   Business      2025 3.80
#> 9     Lucas    Finance      2023 3.55
#> 10      Amy    Finance      2026 4.00
#> 11      Jim       Math      2024 3.60

After you created df1, capitalize its column names. Get a 6-number summary of GPA column. What is a data type of GRAD_YEAR column? Convert it into the correct data structure. Display the frequency of its elements.

colnames(df1) <- c("NAME", "MAJOR", "GRAD_YEAR", "GPA")

summary(df1$GPA)
#>    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
#>   3.300   3.575   3.800   3.745   3.950   4.000

class(df1$GRAD_YEAR)
#> [1] "numeric"

df1$GRAD_YEAR <- as.factor(df1$GRAD_YEAR)

summary(df1$GRAD_YEAR)
#> 2023 2024 2025 2026 
#>    2    3    4    2

GRAD_YEAR column should be factor type, since it represents meaningful year number, not simply numbers.

Use a simple ifelse statement to add a new column NEXT_YEAR to the df1. This would be a boolean column, indicating TRUE if a student is graduating next year (2024) and FALSE if not (Check GRAD_YEAR column to see what year a student plans to graduate).

df1$NEXT_YEAR <- ifelse(df1$GRAD_YEAR==2024, TRUE, FALSE)

print(df1)
#>        NAME      MAJOR GRAD_YEAR  GPA NEXT_YEAR
#> 1     James       Math      2023 3.90     FALSE
#> 2     Linda       Math      2025 3.75     FALSE
#> 3     Stacy   Genetics      2025 4.00     FALSE
#> 4      Mary Statistics      2024 4.00      TRUE
#> 5       Tom Accounting      2026 3.40     FALSE
#> 6      Anna        Art      2024 3.90      TRUE
#> 7       Bob      Music      2025 3.30     FALSE
#> 8  Jeniffer   Business      2025 3.80     FALSE
#> 9     Lucas    Finance      2023 3.55     FALSE
#> 10      Amy    Finance      2026 4.00     FALSE
#> 11      Jim       Math      2024 3.60      TRUE

Extract/Select all students in the df1 who is majoring in Math and is not going to graduate this year (2023).

df1[df1$MAJOR=="Math" & df1$GRAD_YEAR!="2023", ]
#>     NAME MAJOR GRAD_YEAR  GPA NEXT_YEAR
#> 2  Linda  Math      2025 3.75     FALSE
#> 11   Jim  Math      2024 3.60      TRUE

Problem 4

Create a matrix with 5 columns filled by columns with the following elements: 10, 3, 6, 23, -5, -4, 13, 17, 5, 6, -7, -10, 13, 39, 20, 2, 1, 9, 11, -22, 23, -15, -3, 6, 12. Name it matrix1.

matrix1 <- matrix(c(10, 3, 6, 23, -5, -4, 13, 17, 5, 6, -7, -10, 13, 39, 20, 2, 1, 9, 11, -22, 23, -15, -3, 6, 12), ncol=5)

print(matrix1)
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]   10   -4   -7    2   23
#> [2,]    3   13  -10    1  -15
#> [3,]    6   17   13    9   -3
#> [4,]   23    5   39   11    6
#> [5,]   -5    6   20  -22   12

Use a For loop to replace elements of matrix1 as follows: if an element has a negative value, replace it with 0, if an element is greater than or equal to 10, replace it with 10, and if an element is greater than or equal to 0 and less than 10, replace it with 5.

for (i in 1:dim(matrix1)[1]) {
  
  for (j in 1:dim(matrix1)[2]) {
    
    if (matrix1[i,j] < 0) {
      
      matrix1[i, j] = 0
      
    }
    
    else if (matrix1[i,j] >= 10) {
      
      matrix1[i,j] = 10
      
    }
    
    else {
      
      matrix1[i,j] = 5
    }
  }
}

print(matrix1)
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]   10    0    0    5   10
#> [2,]    5   10    0    5    0
#> [3,]    5   10   10    5    0
#> [4,]   10    5   10   10    5
#> [5,]    0    5   10    0   10

Create an empty vector vector6. Use a While loop to populate vector6 as follows: populate vector6 with odd numbers that are greater than 0 and less than 30 skipping odd numbers that are divisible by 3.

vector6 <- c()

i <- 0

while (i < 30) {
  
  i <- i + 1
  
  if (i%%2==0 || i%%3==0) {
    
    next
    
  }
  
  vector6 <- c(vector6, i)
  
}

print(vector6)
#>  [1]  1  5  7 11 13 17 19 23 25 29

Problem 5

Create a function (name it fun1) that will take a numeric vector as input and will extract elements of the input vector that are greater than 10. Do it in two ways: using for loop combined with conditional statement, and without it (using logical and extraction operators). Test your function with the following vector: [1, 4, 45, 23, 7, 9, 12, 15, 33]

vector <- c(1, 4, 45, 23, 7, 9, 12, 15, 33)

fun1 <- function(vec) {
  
  result <- c()
  
  for (i in 1:length(vec)) {
    
    if (vec[i] > 10) {
      
      result <- c(result, vec[i])
      
    }
  }
  
  return(result)
}

fun1(vector)
#> [1] 45 23 12 15 33

fun1 <- function(vec) {
  
  return(vec[vec>10])
  
}

fun1(vector)
#> [1] 45 23 12 15 33

Create a function (name it fun2) that will take two matrices as inputs (mat1 and mat2) and will produce another matrix that has its elements populated as follows: if an element in mat1 is greater than the corresponding element in mat2, then the corresponding element in the output matrix would be "Greater than"; if an element in mat1 is less than the corresponding element in mat2, then the corresponding element in the output matrix would be "Less than"; and if an element in mat1 is equal to the corresponding element in mat2, then the corresponding element in the output matrix would be "Equal to".

For example, if mat1[1, 1] = 10 and mat2[1, 1] = 5, then the output matrix will have "Greater than" element at [1, 1] position. Test your function with the following matrices: [1, 4, 9, 5, 2, 7, 4, 3, 10], 3 rows and filled by row; [2, 3, 10, 9, 1, 4, 4, 3, 9], 3 rows and filled by row.

mat1 <- matrix(c(1, 4, 9, 5, 2, 7, 4, 3, 10), nrow=3, byrow=TRUE)

mat2 <- matrix(c(2, 3, 10, 9, 1, 4, 4, 3, 9), nrow=3, byrow=TRUE)

fun2 <- function(mat1, mat2) {
  
  mat3 <- matrix(nrow=dim(mat1)[1], ncol=dim(mat1)[2])
  
  for (i in 1:dim(mat1)[1]) {
    
    for (j in 1:dim(mat1)[2]) {
      
      if (mat1[i,j]>mat2[i,j]) {
        
        mat3[i,j] <- "Greater than"
        
      }
      
      else if (mat1[i,j]<mat2[i,j]) {
        
        mat3[i,j] <- "Less than"
      }
      
      else {
        
        mat3[i,j] <- "Equal to"
      }
    }
  }
  
  return(mat3)
}

fun2(mat1, mat2)
#>      [,1]        [,2]           [,3]          
#> [1,] "Less than" "Greater than" "Less than"   
#> [2,] "Less than" "Greater than" "Greater than"
#> [3,] "Equal to"  "Equal to"     "Greater than"

Problem Set 1

Module 5