Exercises for Recap Session 1

Possible solutions

Author

Prof. Dr. Claudius Gräbner-Radkowitsch

Published

18 04 2024

Modified

18 04 2024

Exercise 1: Basic object types I

  1. Create a vector containing the numbers 2, 5, 2.4 and 11.
ex1_vec <- c(2, 5, 2.4, 11)
  1. Replace the second element with 5.9.
ex1_vec[2] <- 5.9
ex1_vec
[1]  2.0  5.9  2.4 11.0
  1. Add the elements 3 and 1 to the beginning, and the elements "8.0" and "9.2" to the end of the vector.
va_1 <- c(3, 1)
va_2 <- c("8.0", "9.2")
ex1_vec_extended <- c(va_1, ex1_vec, va_2)
ex1_vec_extended
[1] "3"   "1"   "2"   "5.9" "2.4" "11"  "8.0" "9.2"
  1. Create a vector with the numbers from -8 to 9 (step size: 0.5)
ex1_vec_4 <- seq(-8, 9, by = 0.5)
ex1_vec_4
 [1] -8.0 -7.5 -7.0 -6.5 -6.0 -5.5 -5.0 -4.5 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0
[16] -0.5  0.0  0.5  1.0  1.5  2.0  2.5  3.0  3.5  4.0  4.5  5.0  5.5  6.0  6.5
[31]  7.0  7.5  8.0  8.5  9.0
  1. Compute the square root of each element of the first vector using vectorisation.
sqrt(ex1_vec_4)
Warning in sqrt(ex1_vec_4): NaNs produced
 [1]       NaN       NaN       NaN       NaN       NaN       NaN       NaN
 [8]       NaN       NaN       NaN       NaN       NaN       NaN       NaN
[15]       NaN       NaN 0.0000000 0.7071068 1.0000000 1.2247449 1.4142136
[22] 1.5811388 1.7320508 1.8708287 2.0000000 2.1213203 2.2360680 2.3452079
[29] 2.4494897 2.5495098 2.6457513 2.7386128 2.8284271 2.9154759 3.0000000
  1. Create a character vector containing then strings "Number_1" to "Number_5". Use suitable helper functions to create this vector quickly.
ex1_char_vec <- paste0("Number_", seq(1, 5))
ex1_char_vec
[1] "Number_1" "Number_2" "Number_3" "Number_4" "Number_5"

Exercise 2: Basic object types II

Consider the following vector:

ex_2_vec <- c(1.9, "2", FALSE)
  1. What is the type of this vector? Why?
typeof(ex_2_vec)
[1] "character"

Atomic vectors only contain objects of the same type, and there is a hierarchy. Elements that themselves are of a type lower in the hierarchy are coerced to the same type as the object highest in the hierarchy. The hierarchy is as as follows:

  1. character
  2. double
  3. integer
  4. logical

Therefore, the type of ex_2_vec is character. The underlying reason is that you can, for instance, always transform a double value into a character but not vice versa.

  1. What happens if you coerce this vector into type integer? Why?
as.integer(ex_2_vec)
Warning: NAs introduced by coercion
[1]  1  2 NA

Because integer is lower in the hierarchy than character, the transformation is not straightforward. By coincidence, the first two elements can actually be coerced into integers (albeit maybe not with the expected result), but there is no way you can transform the logical value FALSE into an integer, which is why a missing value is produced.

  1. What does sum(is.na(x)) tell you about a vector x? What is happening here?
x <- c(1,2,3,NA,NA,8)

First, is.na(x) creates a vector with logical values indicating whether a value of the original vector is missing (i.e. NA):

is.na(x)
[1] FALSE FALSE FALSE  TRUE  TRUE FALSE

Then, sum() computes the sum over this vecor of boolean values:

sum(is.na(x))
[1] 2

Here, TRUE counts as one and FALSE as zero, so sum() gives the number of cases in which is.na(x) has evaluated to TRUE:

  1. Is it a good idea to use as.integer() on double characters to round them to the next integer? Why (not)? What other ways are there to do the rounding?

No, because as.integer() is not acutally rounding numbers (as, for example, as.integer(2.1) would make you think), but only removing the decimal part of the number:

as.integer(2.9) # you might expect 2...
[1] 2

Better use round():

round(2.9)
[1] 3

Exercise 3: Define a function

Create functions that take a vector as input and returns:

  1. The last value.
get_last_val <- function(x){
  last_val <- x[length(x)]
  return(last_val)
}
  1. Every element except the last value and any missing values.
get_beginning <- function(x){
  beginning <- x[-length(x)] # Removes last value
  na_positions <- which(is.na(beginning)) # Get positions of NA values
  beginning_nonas <- beginning[-na_positions] # Removes these values
  return(beginning_nonas)
}
  1. Only even numbers.

Hint: Use the operation x %% y to get the remainder from diving x by y, the so called ‘modulo y’. For even numbers, the modulo 2 is zero.

get_even <- function(x){
  modulo_2s <- x%%2 # Module 2 is zero for even numbers only
  even_nbs <- x[modulo_2s==0] # Keep only those for which modulo 2 is zero
  na_positions <- which(is.na(even_nbs)) # Get positions of NA values
  even_nbs_nonas <- even_nbs[-na_positions] # Removes these values
  return(even_nbs_nonas)
}

Apply your function to the following example vector:

ex_3_vec <- c(1, -8, 99, 3, NA, 4, -0.5, 50)
get_last_val(ex_3_vec)
[1] 50
get_beginning(ex_3_vec)
[1]  1.0 -8.0 99.0  3.0  4.0 -0.5
get_even(ex_3_vec)
[1] -8  4 50

Exercise 4: Lists

  1. Create a list that contains three elements called 'a', 'b' and 'c'. The first element should correspond to a double vector with the elements 1.5, -2.9 and 99. The second element should correspond to a character vector with the elments 'Hello', '3', and 'EUF'. The third element should contain three times the entry FALSE.
ex_4_list <- list(
  'a' = c(1.5, -2.9, 99),
  'b' = c('Hello', "'3'", 'EUF'),
  'c' = rep(FALSE, 3)
)
  1. Transform this list into a data.frame and a tibble. Then apply str() to get information about the respective structure. How do the results differ?
ex_4_df <- as.data.frame(ex_4_list)
ex_4_tb <- tibble::as_tibble(ex_4_list)
str(ex_4_list)
List of 3
 $ a: num [1:3] 1.5 -2.9 99
 $ b: chr [1:3] "Hello" "'3'" "EUF"
 $ c: logi [1:3] FALSE FALSE FALSE
str(ex_4_df)
'data.frame':   3 obs. of  3 variables:
 $ a: num  1.5 -2.9 99
 $ b: chr  "Hello" "'3'" "EUF"
 $ c: logi  FALSE FALSE FALSE
str(ex_4_tb)
tibble [3 × 3] (S3: tbl_df/tbl/data.frame)
 $ a: num [1:3] 1.5 -2.9 99
 $ b: chr [1:3] "Hello" "'3'" "EUF"
 $ c: logi [1:3] FALSE FALSE FALSE

str() only differs with regard to the first line describing the type.

Exercise 5: Data frames and the study semester distribution at EUF

The package DataScienceExercises contains a data set called EUFstudentsemesters, which contains information about the distribution of study semesters of enrolled students at the EUF in 2021. You can shortcut the data set as follows:

euf_semesters <- DataScienceExercises::EUFstudentsemesters
  1. What happens if you extract the column with study semesters as a vector and transform it into a double?
unique(euf_semesters[["Semester"]])
[1] "6"           "4"           "2"           "8"           "9 or higher"
[6] "7"           "5"           "3"           "1"          
semesters <- as.double(euf_semesters[["Semester"]])
Warning: NAs introduced by coercion
unique(semesters)
[1]  6  4  2  8 NA  7  5  3  1

We see that the previous entry "9 or higher" has been transformed into NA.

  1. What is the average study semester of those students being in their 8th or earlier semester?
mean(semesters, na.rm = TRUE)
[1] 4.177026
  1. How many students are in their 9th or higher study semester?
sum(euf_semesters$Semester=="9 or higher")
[1] 469
  1. What does typeof(euf_semesters) return and why?
typeof(euf_semesters)
[1] "list"

It returns list, because while euf_semesters is a tibble, typeof() always gives the underlying basic object type. For tibbles, this is list.