Exercises Week 1

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Basic R operations, data frames, functions and packages

Exercise 1. Assign the numeric 3 to an object called a.

Solution Ex. 1

a <- 3

Exercise 2. What are the differences between these operations?

a <- 3

a = 3

3 -> a

3 = a

Try to explain your results.

Solution Ex. 2

Using the assign operator <- gives the direction of the assignment. This means that a <- 3 is equivalent to a = 3. However the assign oerator <- can be reversed and thus gives the direction of assignment ->. Using the equal sign as a assignment oeperator is restricted to one direction.

Exercise 3. Combine the numeric values 3, 5, 6, 7 in a vector.

Solution Ex. 3

my_vector <- c(3, 5, 6, 7)

Exercise 4. Print the second value from your vector.

Solution Ex. 4

my_vector[2]

Exercise 5. Change the second value of your vector to 23.

Solution Ex. 5

my_vector[2] <- 23

Exercise 6. A parameter can be regarded as a constant in a given situation but can take different values in other situations. The benefit of working with parameters is that we can use the name of the parameter to describe for example mathematical operations.

Write code that corresponds to the following equation:

\[a = \frac{beta}{alpha}\]

given that \(beta = 3\) and \(alpha = 6\).

Solution Ex. 6

beta <- 3
alpha <- 6

a <- beta/alpha

Exercise 7. What is the problem with the following operation:

vector_a <- c(3, 5, 2, 7, 9)

vector_b <- c(3, 3)

vector_a * vector_b

And how is this correct?

vector_c <- c(3, 5, 2, 7)

vector_d <- c(3, 2)

vector_c * vector_d

How do R calculate the second situation?

Solution Ex. 7

Read the warning meassage in from situation 1. What does that mean?

Solution 7a

Vector can only be multiplied if they have corresponding length or the longer object is a multiple of the shorter. In the second situation the shorter is a multiple of the longer object. The operations are \(3*5,~ 2*5,~ 3*2,~ 2*7\).

Exercise 8. A data frame is basically multiple vectors of data stacked in columns. Create a data frame by hand that corresponds to the table below.

ID	Age	Sex	Performance	Body_mass
id1	24	M	250	82
id2	25	F	180	65
id3	22	F	187	55
id4	26	F	210	59
id5	25	M	205	75
id6	29	M	199	89

Solution 8

A data frame can be created using the data.frame()-function, name the columns in the data frame and enter values using the c() function.

Solution 8

my_df <- data.frame(ID = c("id1", "id2", "id3", "id4", "id5", "id6"), 
                    Age = c(24, 25, 22, 26, 25, 29), 
                    Sex = c("M", "F", "F", "F", "M", "M"),
                    Performance = c(250, 180, 187, 210, 205, 199), 
                    Body_mass = c(82, 65, 55, 59, 75, 89))


my_df

Write code to: 1. return the first row in the data frame 2. create a new variable calculated as Performance per Body mass 3. print the the new variable from the data frame

Solution 9.1

my_df[1, ] # read as [row, column]

Solution 9.2

my_df$p_bm <- my_df$Performance / my_df$Body_mass

# An alternative is

my_df$p_bm <- my_df[,4] / my_df[,5]

Solution 9.3

my_df$p_bm

Exercise 10. We can use functions to calculate mean, standard deviation, minimum and so on. Calculate the mean and standard deviation of Performance and Body mass. Store the results in a new data frame.

Solution 10

data.frame(variable = c("Peromance", "Body_mass"), 
           m = c(mean(my_df$Performance), mean(my_df$Body_mass)), 
           s = c(sd(my_df$Performance), sd(my_df$Body_mass)))

Exercise 11. We will be using several packages during the course. Install the follwing packages using the install.package() function:

Package
dplyr
tidyr
readxl
ggplot2

Solution 10

install.packages(c("dplyr", "tidyr", "readxl", "ggplot2"))