Factor exercises

In the exercises below we cover the basics of factors. Before proceeding, first read chapter 4 of An Introduction to R, and the help pages for the cut, and table functions.

Answers to the exercises are available here.

Exercise 1
If x = c(1, 2, 3, 3, 5, 3, 2, 4, NA), what are the levels of factor(x)?
a. 1, 2, 3, 4, 5
b. NA
c. 1, 2, 3, 4, 5, NA

Exercise 2
Let x <- c(11, 22, 47, 47, 11, 47, 11). If an R expression factor(x, levels=c(11, 22, 47), ordered=TRUE) is executed, what will be the 4th element in the output?
a. 11
b. 22
c. 47

Exercise 3
If z <- c("p", "a" , "g", "t", "b"), then which of the following R expressions will replace the third element in z with "b".
a. factor(z[3]) <- "b"
b. levels(z[3]) <- "b"
c. z[3] <- "b"

Exercise 4
If z <- factor(c("p", "q", "p", "r", "q")) and levels of z are "p", "q" ,"r", write an R expression that will change the level "p" to "w" so that z is equal to: "w", "q" , "w", "r" , "q".

Exercise 5
If:
s1 <- factor(sample(letters, size=5, replace=TRUE)) and
s2 <- factor(sample(letters, size=5, replace=TRUE)),
write an R expression that will concatenate s1 and s2 in a single factor with 10 elements.

Exercise 6
Consider the iris data set in R. Write an R expression that will ‘cut’ the Sepal.Length variable and create the following factor with five levels.

(4.3, 5.02] (5.02, 5.74] (5.74, 6.46] (6.46, 7.18] (7.18, 7.9]
32 41 42 24 11


Exercise 7
Consider again the iris data set. Write an R expression that will generate a two-way frequency table with two rows and three colums. The rows should relate to Sepal.length (less than 5: TRUE or FALSE) and columns to Species, with the following output:

setosa versicolor virginica
FALSE 30 49 49
TRUE 20 1 1


Exercise 8
Consider the factor responses <- factor(c("Agree", "Agree", "Strongly Agree", "Disagree", "Agree")), with the following output:

[1] Agree Agree Strongly Agree Disagree Agree
Levels: Agree Disagree Strongly Agree

Later it was found that new a level "Strongly Disagree" exists. Write an R expression that will include "strongly disagree" as new level attribute of the factor and returns the following output:

[1] Agree Agree Strongly Agree Disagree Agree
Levels: Strongly Agree Agree Disagree Strongly Disagree


Exercise 9
Let x <- data.frame(q=c(2, 4, 6), p=c("a", "b", "c")). Write an R statement that will replace levels a, b, c with labels "fertiliser1", "fertliser2", "fertiliser3".

Exercise 10
If x <- factor(c("high", "low", "medium", "high", "high", "low", "medium")), write an R expression that will provide unique numeric values for various levels of x with the following output:

levels value
1 high 1
2 low 2
3 medium 3