Lecture notes and solutions to the exercises of session 2 in the spring semester 2023

#S2 - Lecture notes (SpSe23)

 Lecture notes and solutions to the exercises of session 2 in the spring semester 2023

S2-ExerciseSolutions-I.R

# Session 2 - Exercise solutions I

# Execute the following mathematical computations via the console:

# a)
5 + 12

# b)
(2*3)**2

# c)
2*5.8

# d)
(8**2 + 5**4)/3 
((8**2) + (5**4))/3 

S2-ExerciseSolutions-II.R

# Session 2 - Exercise solutions II

a_ <- 2+3
b_ <- (5*a_)/2
c_ <- (b_ + 1)**2
d_ <- sqrt(c_)
d_

S2-ExerciseSolutions-III.R

# Session 2 - Exercise solutions III

# Define a vector with the elements -2, 2, 4, 6, 9 and NA
t_vec <- c(-2, 2, 4, 6, 9, NA)

# Apply the following functions and understand what they are doing:

median(t_vec)
median(t_vec, na.rm = TRUE)
# Computes the median

is.na(t_vec)
# Gives TRUE/FALSE for each element of t_vec, indicating of the
#  element is missing or not

anyNA(t_vec)
# Indicates whether there is any missing element

sum(t_vec, na.rm = TRUE)
# Sums up all elements

var(t_vec, na.rm = TRUE)
# Computes the sample variance (check out help(var), if in doubt)

# More info:
# From the help function you may know that 
#  var() computes sample variances
var_vec <- c(1, 6, 7, 9, 12)
sample_var <- var(var_vec)
# To get the population variance:
# Determine length of data
n_elements <- length(sample_var)
pop_var <- var(sample_var)*(n_elements - 1)/n_elements
# See, e.g., here: https://stackoverflow.com/a/37733398

sample_var
pop_var

S2-ExerciseSolutions-IV.R

# Session 2 - Exercise solutions IV

#' Normalize a vector into 0-1 range
#' 
#' This function takes a vector and normalizes its elements into the range 
#'  between 0 and 1. It does not yet handle special cases, such as vectors with
#'  missing values, but we learn how to deal with this later on.
#' @param x_vector A vector without missing values
#' @returns A vector with the same number of elements as `x_vector`, but 
#'  normalized into the 0-1 range
norm_function <- function(x_vector){
  min_value <- min(x_vector)
  max_value <- max(x_vector)
  result <- (x_vector - min_value) / (max_value - min_value)
  return(result)
}

# Example application:
vec_used <- c(1, 2, 3, 4, 5, 6, -2)
vec_used_normed <- norm_function(vec_used)
vec_used_normed

# Note that all values defined within the function are not available after
#  calling it:
max_value

S2-LectureNotes.R

# Lecture notes session 2

# 1. R as a calculator-----------------
2 + 3 # Addition
6 - 8 # Substraction
5/3 # Division
4*2 # Multiplication
5**2 # Exponentiation
(3-2)*2 # Using brackets

# 2. Assignments ----------------------

assign("intermediate_result", 4-2)
intermediate_result

final_result <- intermediate_result*5
TRUE <- 5**2 # Gives an error
rm(final_result) # Removes name-object association

# 'One object - many names' is possible:
name_1 <- 66**2
name_2 <- 66**2
name_2 <- name_1

# One name can only be used for at most one object:
name_1 <- 66**2*4
name_1 <- 66**2*4*5 # Overwrites previous assignment
# Digression: you can concatenate objects and refer to them with one name:
name_1_2 <- c(name_1, name_2)

# Be aware between the distinction of names and objects:

# Gives an error:
assign(a, 5) 

# Assigns the name 'name_of' to 5:
a <- "name_of" 
assign(a, 5)

# Assigns the name 'a' to 5:
assign("a", 5)

# 3. Defining functions ----------------------

# Solution for the short practice on slide 27:
function1 <- function(x1, x2){
  eq_1 <- x1**2 + 2*x1*x2 + x2**2
  return(eq_1)
}

# Example: writing a function to normalize the vector
# For a more detailed solution see file S2-ExerciseSolutions-IV.R

x_vec <- c(2, 4, 8)

sigma <- function(vector_1){
  min_vec_1 <- min(vector_1)
  max_vec_2 <- max(vector_1)
  step1 <- vector_1 - min_vec_1
  step2 <- max_vec_2 - min_vec_1
  step3 <- step1/step2
  return(step3)
}