Exercises

In this practical exercise we are going to play around with the different types of elements in R.

  • Go to the course website and download the file “Practical B: template” (a Markdown file).

  • Save the template file in the project folder you created for this course, and, if necessary, open the R Project by clicking on the .Rproj file.

  • Open the file “Practical B: Exercises”. Make the exercises, if possible without looking at the answers in the file “Practical B: solutions”.

  • In any case; ask for help when you feel help is needed.

Exercise 1-5

  1. Make two vectors: one named vec1 with values 1 through 6 and one named vec2 with letters A through F.
vec1 <- c(1, 2, 3, 4, 5, 6)
vec2 <- c("A", "B", "C", "D", "E", "F")

To create a vector we used c(), which stands for ‘concatenation’. It is just a series of numbers or letters.

  1. Create two matrices, one from vec1 and one from vec2. The dimensions for both matrices are 3 rows by 2 columns. Find the function to create a matrix by typing ?matrix. Notice that when you start typing ?matrix in the code chunk, a pop-up window appears with information about the function.
?matrix
mat1 <- matrix(vec1, nrow = 3, ncol = 2)
mat2 <- matrix(vec2, nrow = 3, ncol = 2)

To create a matrix we used matrix(). For a matrix we need to specify the dimensions (in this case 3 rows and 2 columns) and the input (in this case vec1 or vec2) needs to match these dimensions.

  1. Inspect your vectors and matrices. Are all numerical?
vec1
## [1] 1 2 3 4 5 6
vec2
## [1] "A" "B" "C" "D" "E" "F"
mat1
##      [,1] [,2]
## [1,]    1    4
## [2,]    2    5
## [3,]    3    6
mat2
##      [,1] [,2]
## [1,] "A"  "D" 
## [2,] "B"  "E" 
## [3,] "C"  "F"

vec1 and mat1 contain numbers and vec2 and mat2 contain characters.

  1. Make a matrix from both vec1 and vec2 with 6 rows and 2 columns. Inspect this matrix.
mat3 <- matrix(c(vec1, vec2), 6, 2)
mat3
##      [,1] [,2]
## [1,] "1"  "A" 
## [2,] "2"  "B" 
## [3,] "3"  "C" 
## [4,] "4"  "D" 
## [5,] "5"  "E" 
## [6,] "6"  "F"

or

mat3b <- cbind(vec1, vec2)
is.matrix(mat3b)
## [1] TRUE
mat3b
##      vec1 vec2
## [1,] "1"  "A" 
## [2,] "2"  "B" 
## [3,] "3"  "C" 
## [4,] "4"  "D" 
## [5,] "5"  "E" 
## [6,] "6"  "F"

If one or more elements in the matrix represent characters, all other elements are also converted to characters. A matrix is just for either numeric or character elements. Notice that the second approach (the column bind approach from mat3b) returns a matrix where the column names are already set to the name of the bound objects.

To solve the problem of numbers represented as characters we can create a dataframe. A dataframe is essentially a matrix that allows for character elements. The use of a dataframe is often preferred over the use of a matrix in R, except for purposes where pure numerical calculations are done, such as in matrix algebra. However, most datasets do contain character information and a dataframe would normally be your preferred choice when working with your own collected datasets in R.

  1. Make a dataframe called dat3 where vec1 and vec2 are both columns. Name the columns V1 and V2, respectively. Use function data.frame().
dat3 <- data.frame(V1 = vec1, V2 = vec2)
dat3
##   V1 V2
## 1  1  A
## 2  2  B
## 3  3  C
## 4  4  D
## 5  5  E
## 6  6  F

Exercise 6-10

  1. A useful function to check the properties of any R object is the str() function. Try this function on dat3.

You can inspect the structure of a dataframe (or other R object) by using the str() function or by clicking on the object in the Environment tab in RStudio, which unfolds the properties of each element. See:

HTML5 Icon

Try both ways of inspecting the structure of dat3.

str(dat3)
## 'data.frame':    6 obs. of  2 variables:
##  $ V1: num  1 2 3 4 5 6
##  $ V2: chr  "A" "B" "C" "D" ...

Inspecting the structure of your data is vital, as you probably have imported your data from some other source. If we, at a later stage, start analyzing our data without the correct measurement level, we may run into problems. One problem that often occurs is that categorical variables (factors in R) are not coded as such.

  1. Select 1) the third row, 2) the second column and 3) the intersection of these two in the dataframe dat3 that you have created in Question 4.
dat3[3, ] #3rd row
##   V1 V2
## 3  3  C
dat3[, 2] #2nd column
## [1] "A" "B" "C" "D" "E" "F"
dat3$V2   #also 2nd column
## [1] "A" "B" "C" "D" "E" "F"
dat3[3, 2] #intersection
## [1] "C"

The [3, 2] index is very useful in ‘R’. The first number (before the comma) represents the row and the second number (after the comma) represents the column. For a vector there are no two dimensions and only one dimension can be called. For example, vec1[3] would yield 3. Try it.

Columns can also be called by the $ sign, but only if a name has been assigned. With dataframes assigning names happens automatically.

  1. Imagine that the first variable V1 in our dataframe dat3 is not coded correctly, but actually represents grouping information about cities. Convert the variable to a factor and add the labels Utrecht, New York, London, Singapore, Rome and Cape Town.
dat3$V1 <- factor(dat3$V1, labels = c("Utrecht", "New York", "London", "Singapore", "Rome", "Capetown"))
dat3
##          V1 V2
## 1   Utrecht  A
## 2  New York  B
## 3    London  C
## 4 Singapore  D
## 5      Rome  E
## 6  Capetown  F

You can verify the changes with str() or by inspecting the object dat3 in the RStudio Environment tab.

  1. Open the workspace boys.RData.

There are two ways to go about opening workspaces that are available on the internet. You either need to download the boys.RData file from the course page and put it in the project folder. Then run the below code

load("boys.RData")

or double-click the boys.RData file on your machine (right-click and open with RStudio if it does not open by default in RStudio, but in R).

Alternatively, you can import workspaces directly from the internet by running and loading the connection

con <- url("https://www.gerkovink.com/fundamentals/data/boys.RData")
load(con)

In the above code we store the connection in object con and then load the connection with load(con).

  1. Most packages have datasets included. Since we have not learned to load packages yet, you are presented with such a data set in a workspace. Open the boys dataset (it is from package mice, by the way) by typing boys in the console, and subsequently by using the function View().

The output is not displayed here as the data set is simply too large.

Using View() is preferred for inspecting datasets that are large. View() opens the dataset in a spreadsheet-like window (conform MS Excel, or SPSS). If you View() your own datasets, you can not edit the datasets’ contents.

Exercise 11-15

  1. Find out the dimensions of the boys data set and inspect the first and final 6 cases in the data set.

To do it numerically, find out what the dimensions of the boys dataset are.

dim(boys)
## [1] 748   9

There are 748 cases on 9 variables. To select the first and last six cases, use

boys[1:6, ]
##      age  hgt   wgt   bmi   hc  gen  phb tv   reg
## 3  0.035 50.1 3.650 14.54 33.7 <NA> <NA> NA south
## 4  0.038 53.5 3.370 11.77 35.0 <NA> <NA> NA south
## 18 0.057 50.0 3.140 12.56 35.2 <NA> <NA> NA south
## 23 0.060 54.5 4.270 14.37 36.7 <NA> <NA> NA south
## 28 0.062 57.5 5.030 15.21 37.3 <NA> <NA> NA south
## 36 0.068 55.5 4.655 15.11 37.0 <NA> <NA> NA south
boys[743:748, ]
##         age   hgt  wgt   bmi   hc  gen  phb tv   reg
## 7410 20.372 188.7 59.8 16.79 55.2 <NA> <NA> NA  west
## 7418 20.429 181.1 67.2 20.48 56.6 <NA> <NA> NA north
## 7444 20.761 189.1 88.0 24.60   NA <NA> <NA> NA  west
## 7447 20.780 193.5 75.4 20.13   NA <NA> <NA> NA  west
## 7451 20.813 189.0 78.0 21.83 59.9 <NA> <NA> NA north
## 7475 21.177 181.8 76.5 23.14   NA <NA> <NA> NA  east

or, more efficiently:

head(boys)
##      age  hgt   wgt   bmi   hc  gen  phb tv   reg
## 3  0.035 50.1 3.650 14.54 33.7 <NA> <NA> NA south
## 4  0.038 53.5 3.370 11.77 35.0 <NA> <NA> NA south
## 18 0.057 50.0 3.140 12.56 35.2 <NA> <NA> NA south
## 23 0.060 54.5 4.270 14.37 36.7 <NA> <NA> NA south
## 28 0.062 57.5 5.030 15.21 37.3 <NA> <NA> NA south
## 36 0.068 55.5 4.655 15.11 37.0 <NA> <NA> NA south
tail(boys)
##         age   hgt  wgt   bmi   hc  gen  phb tv   reg
## 7410 20.372 188.7 59.8 16.79 55.2 <NA> <NA> NA  west
## 7418 20.429 181.1 67.2 20.48 56.6 <NA> <NA> NA north
## 7444 20.761 189.1 88.0 24.60   NA <NA> <NA> NA  west
## 7447 20.780 193.5 75.4 20.13   NA <NA> <NA> NA  west
## 7451 20.813 189.0 78.0 21.83 59.9 <NA> <NA> NA north
## 7475 21.177 181.8 76.5 23.14   NA <NA> <NA> NA  east

The functions head() and tail() are very useful functions. For example, from looking at both functions we can observe that the data are very likely sorted based on age.

  1. Suppose you want to plot the height versus the weight of the boys. You would like to have the weight variable (wgt) on the x-axis and the height variable (hgt) on the y-axis. How can you achieve such a plot? Tip: use the args() function.

A. plot(boys$wgt, boys$hgt)

B. plot(boys$hgt, boys$wgt)

C. plot(x=boys$wgt, y=boys$hgt)

D. plot(y=boys$hgt, x=boys$wgt)

There are three correct answers: A, C and D.

# look at the arguments structure in the function
args(plot)
## function (x, y, ...) 
## NULL
# this reveals that the function plots the first value on the x-axis and the second value on the y-axis.

Make the plot using the correct code.

plot(x=boys$wgt, y=boys$hgt)

  1. It seems that the boys data are sorted based on age. Verify this.

To verify if the data are indeed sorted, we can run the following command to test the complement of that statement. Remember that we can always search the help for functions: e.g. we could have searched here for ?sort and we would quickly have ended up at function is.unsorted() as it tests whether an object is not sorted.

is.unsorted(boys$age)
## [1] FALSE

which returns FALSE, indicating that boys’ age is indeed sorted (we asked if it was unsorted!). To directly test if it is sorted, we could have used

!is.unsorted(boys$age)
## [1] TRUE

which tests if data data are not unsorted. In other words the values TRUE and FALSE under is.unsorted() turn into FALSE and TRUE under !is.unsorted(), respectively.

  1. Inspect the boys dataset with str(). Use one or more functions to find distributional summary information (at least information about the minimum, the maximum, the mean and the median) for all of the variables. Give the standard deviation for age and bmi. Tip: make use of the help (?) and help search (??) functionality in R.
str(boys)
## 'data.frame':    748 obs. of  9 variables:
##  $ age: num  0.035 0.038 0.057 0.06 0.062 0.068 0.068 0.071 0.071 0.073 ...
##  $ hgt: num  50.1 53.5 50 54.5 57.5 55.5 52.5 53 55.1 54.5 ...
##  $ wgt: num  3.65 3.37 3.14 4.27 5.03 ...
##  $ bmi: num  14.5 11.8 12.6 14.4 15.2 ...
##  $ hc : num  33.7 35 35.2 36.7 37.3 37 34.9 35.8 36.8 38 ...
##  $ gen: Ord.factor w/ 5 levels "G1"<"G2"<"G3"<..: NA NA NA NA NA NA NA NA NA NA ...
##  $ phb: Ord.factor w/ 6 levels "P1"<"P2"<"P3"<..: NA NA NA NA NA NA NA NA NA NA ...
##  $ tv : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ reg: Factor w/ 5 levels "north","east",..: 4 4 4 4 4 4 4 3 3 2 ...
summary(boys) #summary info
##       age              hgt              wgt              bmi       
##  Min.   : 0.035   Min.   : 50.00   Min.   :  3.14   Min.   :11.77  
##  1st Qu.: 1.581   1st Qu.: 84.88   1st Qu.: 11.70   1st Qu.:15.90  
##  Median :10.505   Median :147.30   Median : 34.65   Median :17.45  
##  Mean   : 9.159   Mean   :132.15   Mean   : 37.15   Mean   :18.07  
##  3rd Qu.:15.267   3rd Qu.:175.22   3rd Qu.: 59.58   3rd Qu.:19.53  
##  Max.   :21.177   Max.   :198.00   Max.   :117.40   Max.   :31.74  
##                   NA's   :20       NA's   :4        NA's   :21     
##        hc          gen        phb            tv           reg     
##  Min.   :33.70   G1  : 56   P1  : 63   Min.   : 1.00   north: 81  
##  1st Qu.:48.12   G2  : 50   P2  : 40   1st Qu.: 4.00   east :161  
##  Median :53.00   G3  : 22   P3  : 19   Median :12.00   west :239  
##  Mean   :51.51   G4  : 42   P4  : 32   Mean   :11.89   south:191  
##  3rd Qu.:56.00   G5  : 75   P5  : 50   3rd Qu.:20.00   city : 73  
##  Max.   :65.00   NA's:503   P6  : 41   Max.   :25.00   NA's :  3  
##  NA's   :46                 NA's:503   NA's   :522
sd(boys$age) #standard deviation for age
## [1] 6.894052
sd(boys$bmi, na.rm = TRUE) #standard deviation for bmi
## [1] 3.053421

Note that bmi contains 21 missing values, e.g. by looking at the summary information. Therefor we need to use na.rm = T to calculate the standard deviation on the observed cases only.

  1. Select all boys that are 20 years or older. How many are there?

The logical operators (TRUE vs FALSE) are a very powerful tool in R. For example, we can just select the rows (respondents) in the data that are older than 20 by putting the logical operator within the row index of the dataset:

boys2 <- boys[boys$age >= 20, ]
nrow(boys2)
## [1] 12

or, alternatively using subset(),

boys2.1 <- subset(boys, age >= 20)
nrow(boys2.1)
## [1] 12

Exercise 16-17

  1. Select all boys that are older than 19, but younger than 19.5. How many are there?
boys3 <- boys[boys$age > 19 & boys$age < 19.5, ]
nrow(boys3)
## [1] 18

or, alternatively,

boys3.2 <- subset(boys, age > 19 & age < 19.5)
nrow(boys3.2)
## [1] 18
  1. What is the mean age of boys younger than 15 years of age that do not live in region north?
mean(boys$age[boys$age < 15 & boys$reg != "north" ], na.rm = TRUE)
## [1] 6.044461

or, alternatively,

mean(subset(boys, age < 15 & reg != "north")$age, na.rm=TRUE)
## [1] 6.044461

The mean age is 6.0444609 years

Today we have learned to use R at its basics. This offers tremendous flexibility, but may also be inefficient when our aim is some complex analysis, data operation of data manipulation. Doing advanced operations in basic R may require lots and lots of code. Tomorrow we will start using packages that allow us to do complicated operations with just a few lines of code.

As you start using R in your own research, you will find yourself in need of packages that are not part of the default R installation. The beauty of R is that its functionality is community-driven. People can add packages to CRAN that other people can use and improve. Chances are that a function and/or package has been already developed for the analysis or operation you plan to carry out. If not, you are of course welcome to fill the gap by submitting your own package.

End of practical