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STATS 380
SEMESTER ONE 2024
STATISTICS
Statistical Computing
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group,X17,X18,X19,X20,X21,X22 15-17,3.8,4 . 1,3 . 5,1 .4 e,1 . 1 e,1 . 0 e 18-24,20 . 2,19 . 9,16 . 2,11 .8,11 . 0,8 . 0 15-24,15 .3,14 . 9,12 .7,8 . 6,8 . 2,5 .8 25-34,22 .7,19,20 . 2,14 . 6,11 .4,10 . 1 35-44,17.3,19 . 9,14 . 1,12 .8,10 . 1,9 .8 45-54,16,16 . 2,17,13 . 0,11 .8,9 . 9 55-64,15,12 . 6,12 .8,12 . 5,12 .8,10 .7 65-74,7 . 6,7 . 9,7.3,6 . 2,6 .8,6 . 2 75+,2 . 2,4 . 1,3 . 9,2 . 6,3 .4,2 .4 |
Figure 1: A CSV file called “smoking-by-age . csv”.
Figure 1 shows the contents of a CSV file called “smoking-by-age . csv”.
The following code reads the CSV file into R and prints out the resulting data frame, smokingByAge. This data frame. will be used in several of the questions in this exam.
> smokingByAge <- read. csv(“smoking-by-age. csv”)
> smokingByAge
group X17 X18 X19 X20 X21 X22
1 15-17 3.8 4 . 1 3 . 5 1 .4 e 1 . 1 e 1 . 0 e
2 18-24 20 . 2 19 . 9 16 . 2 11 .8 11 . 0 8 . 0
3 15-24 15 .3 14 . 9 12 .7 8 . 6 8 . 2 5 .8
4 25-34 22 .7 19 . 0 20 . 2 14 . 6 11 .4 10 . 1
5 35-44 17.3 19 . 9 14 . 1 12 .8 10 . 1 9 .8
6 45-54 16 . 0 16 . 2 17 . 0 13 . 0 11 .8 9 . 9
7 55-64 15 . 0 12 . 6 12 .8 12 . 5 12 .8 10 .7
8 65-74 7 . 6 7 . 9 7.3 6 . 2 6 .8 6 . 2
9 75+ 2 . 2 4 . 1 3 . 9 2 . 6 3 .4 2 .4
1. [10 marks]
Write down what the output of the following code would be.
(a) [6 marks]
> sapply(smokingByAge, mode)
(b) [2 marks]
> dim(smokingByAge)
(c) [2 marks]
> colnames(smokingByAge)
The data frames shown below will be used in some of the remaining questions in this exam.
> maoriSmoking
group 2017 2018 2019 2020 2021 2022
16 Total Maori 33.4 33 .4 31 . 2 25 .7 22 .3 20 . 2
17 Maori men 29 .7 31 .4 27.3 25 . 6 23 . 0 20 .4
18 Maori women 36 .8 35 .4 35 . 0 25 .8 21 .7 20 . 2
> pacificSmoking
group 2017 2018 2019 2020 2021 2022
20 Total Pacific 23 . 1 24.7 22 . 5 19 . 9 18.8 10 .3
21 Pacific men 28 . 5 28 . 1 26 .8 20 .3 16 . 9 10 . 6
22 Pacific women 18 . 1 21 .8 19 . 0 19 . 5 20 .4 10 . 1
> asianSmoking
group 2017 2018 2019 2020 2021 2022
24 Total Asian 7 . 9 8 . 5 8 . 9 5 .8 3.3 3.7
25 Asian men 12 .8 14 . 1 14 . 0 9 . 0 4.7 6 . 1
26 Asian women 2 . 9 2 .3 2 . 5 1 . 5 1 .8 0 . 6
> euroOtherSmoking
group 2017 2018 2019 2020 2021 2022
28 Total European/Other 13 . 6 12 .7 11 .8 9 .4 9 . 2 7.7
29 European/Other men 14 . 9 14 . 2 11 . 9 10 .3 10 . 5 8 . 5
30 European/Other women 12 .3 11 . 2 11 .7 8 . 6 7 . 9 6 . 9
> test1
group
1 not this row
2 Total Row
> test2
group
1 not this row
2 or this row
3 Total Row
> test3
group
1 Total Row One
2 Total Row Two
2. [10 marks]
(a) [3 marks]
Write a function called findTotal(). The function should have a single argument, which is a data frame. It should use grepl() to search the group column of the data frame. for values that contain the text “Total”. The function should return a logical vector.
The findTotal() function should behave like this:
> findTotal(maoriSmoking)
[1] TRUE FALSE FALSE
> findTotal(pacificSmoking)
[1] TRUE FALSE FALSE
> findTotal(test1)
[1] FALSE TRUE
> findTotal(test2)
[1] FALSE FALSE TRUE
> findTotal(test3)
[1] TRUE TRUE
The values of the symbols maoriSmoking, pacificSmoking, test1, test2, and test3 can be found on page 3.
(b) [7 marks]
Write a function called getTotal(). The function should have a single argument, which is a data frame. It should use findTotal() to search the group column of the data frame. for values that contains the word “Total” and then subset() to extract those rows from the data frame. and thengsub() to remove the word “Total” from the group column of those rows. The function should return a data frame.
The getTotal() function should behave like this:
> getTotal(maoriSmoking)
group 2017 2018 2019 2020 2021 2022
16 Maori 33.4 33 .4 31 . 2 25 .7 22 .3 20 . 2
> getTotal(pacificSmoking)
group 2017 2018 2019 2020 2021 2022
20 Pacific 23 . 1 24.7 22 . 5 19 . 9 18.8 10 .3
> getTotal(test1)
group
2 Row
> getTotal(test2)
group
3 Row
> getTotal(test3)
group
1 Row One
2 Row Two
The values of the symbols maoriSmoking, pacificSmoking, test1, test2, and test3 can be found on page 3.
The following code creates a list called tables and prints out the list. This list will be used in some of the remaining questions in this exam.
> tables <- list(maoriSmoking,
+ pacificSmoking,
+ asianSmoking,
+ euroOtherSmoking)
> tables
[[1]]
group 2017 2018 2019 2020 2021 2022
16 Total Maori 33.4 33 .4 31 . 2 25 .7 22 .3 20 . 2
17 Maori men 29 .7 31 .4 27.3 25 . 6 23 . 0 20 .4
18 Maori women 36 .8 35 .4 35 . 0 25 .8 21 .7 20 . 2
[[2]]
group 2017 2018 2019 2020 2021 2022
20 Total Pacific 23 . 1 24.7 22 . 5 19 . 9 18.8 10 .3
21 Pacific men 28 . 5 28 . 1 26 .8 20 .3 16 . 9 10 . 6
22 Pacific women 18 . 1 21 .8 19 . 0 19 . 5 20 .4 10 . 1
[[3]]
group 2017 2018 2019 2020 2021 2022
24 Total Asian 7 . 9 8 . 5 8 . 9 5 .8 3.3 3.7
25 Asian men 12 .8 14 . 1 14 . 0 9 . 0 4.7 6 . 1
26 Asian women 2 . 9 2 .3 2 . 5 1 . 5 1 .8 0 . 6
[[4]]
group 2017 2018 2019 2020 2021 2022
28 Total European/Other 13 . 6 12 .7 11 .8 9 .4 9 . 2 7.7
29 European/Other men 14 . 9 14 . 2 11 . 9 10 .3 10 . 5 8 . 5
30 European/Other women 12 .3 11 . 2 11 .7 8 . 6 7 . 9 6 . 9
3. [10 marks]
(a) [2 marks]
Write R code that uses the list tables (from page 6) and the functions lapply() and getTotal() (from page 5) to create a list of data frames that just contain “Total” rows and assign the result to the symbol totalTables.
The list totalTables should look like this:
> totalTables
[[1]]
group 2017 2018 2019 2020 2021 2022
16 Maori 33.4 33 .4 31 . 2 25 .7 22 .3 20 . 2
[[2]]
group 2017 2018 2019 2020 2021 2022
20 Pacific 23 . 1 24.7 22 . 5 19 . 9 18.8 10 .3
[[3]]
group 2017 2018 2019 2020 2021 2022
24 Asian 7 . 9 8 . 5 8 . 9 5 .8 3.3 3.7
[[4]]
group 2017 2018 2019 2020 2021 2022
28 European/Other 13 . 6 12 .7 11 .8 9 .4 9 . 2 7.7
(b) [3 marks]
Write R code that uses the functions do. call() and rbind() to com- bine the list of data frames, totalTables, into a single data frame. called ethnicSmoking.
The data frame. ethnicSmoking should look like this:
> ethnicSmoking
group 2017 2018 2019 2020 2021 2022
16 Maori 33.4 33 .4 31 . 2 25 .7 22 .3 20 . 2
20 Pacific 23 . 1 24.7 22 . 5 19 . 9 18.8 10 .3
24 Asian 7 . 9 8 . 5 8 . 9 5 .8 3.3 3.7
28 European/Other 13 . 6 12 .7 11 .8 9 .4 9 . 2 7.7
(c) [5 marks]
Write down what the output of the following code would be.
> groups <- split(ethnicSmoking[1:2, ], ethnicSmoking$group[1:2])
> groups
> ranges <- lapply(groups, function(df) range(df[-1]))
> ranges
> do. call(rbind, ranges)
The following code creates a plot (Figure 2) that shows the prevalence of smoking overtime for different ethnic groups.
The code makes use of the data in the ethnicSmoking data frame. (from page 7). The code and plot will be used in some of the remaining questions in this exam.
> par(mar=margins) > plot. new()
> plot. window(xlim, ylim) > axis(1, at=years)
> axis(right) > box()
> for (i in 1:numGroups) {
+ y <- ethnicSmoking[i, -1]
+ lines(years, y)
+ points(years, y, pch=pch) + }
> mtext(ethnicSmoking$group, side=left, at=ethnicSmoking$”2017″,
+ adj=adj, line=.5, las=horizontal)
Figure 2: A plot of smoking prevalence for different ethnic groups. The grey border around the outside is not drawn by R; it is just there to show the extent of the “page” that R is drawing on.
4. [10 marks]
The code shown on page 8 makes use of some symbols that have not yet been assigned a value.
Write R code that assigns values to each of the following symbols:
(i) margins
(ii) xlim
(iii) ylim
(iv) years
(v) right
(vi) numGroups
(vii) pch
(viii) left
(ix) adj
(x) horizontal
HINTS:
• The left margin of the plot has been made wider to leave room for the ethnic group labels.
• The y-axis range is calculated from the data for all ethnic groups.
• The data symbols are filled circles.
• The ethnic group labels are right-aligned.
• The help page for themtext() function shown in Appendix B may be helpful.
The following code defines a function called middle() that will be used in some of the remaining questions in this exam. Line numbers are provided in grey so that you can refer to specificlines in your answers if necessary.
The purpose of this function is to calculate a “middle” value for each age range in the group column of the smokingByAge data frame. (from page 2). For each row of the group column, if the value contains a dash, the function splits the value into pieces either side of the dash, converts those pieces into numbers, and averages the numbers. If the value does not contain a dash, the function searches for and removes any plus signs, converts what remains into a number and then calulates the average of that number and 100.
1 middle <- function() {
2 n <- nrow(smokingByAge)
3 column <- smokingByAge[[“group”]]
4 middle <- rep(NA, n)
5 for (i in 1:n) {
6 range <- column[i]
7 if (grepl(“-“, range)) {
8 bounds <- strsplit(range, “-“)[[1]]
9 boundsNum <- as. numeric(bounds)
10 middle[i] <- mean(boundsNum)
11 } else {
12 lower <- gsub(“+”, “”, range, fixed=TRUE)
13 lowerNum <- as. numeric(lower)
14 middle[i] <- mean(c(lowerNum, 100))
15 }
16 }
17 middle
18 }
The following code and output shows that the middle() function returns a numeric vector of “middle” values.
> middle()
[1] 16 . 0 21 . 0 19 . 5 29 . 5 39 . 5 49 . 5 59 . 5 69 . 5 87 . 5
5. [10 marks]
Identify all constant values in the middle() function code (from page 10) and, for each constant, describe what the constant represents.
Describe any overall assumptions that the function is making.
6. [10 marks]
Identify all local and global symbols (excluding functions) in the middle() function code (from page 10) and, for each symbol, describe the mode of the R object that is assigned to the symbol.
The data frame. shown below, vapingByAge, will be used in some of the remaining questions in this exam.
vapingByAge
group X17 X18 X19 X20 X21 X22
1 15-17 0 . 6 e 1 .7 e 2 .3 e 5 .8 e 8.3 e 15 .4
2 18-24 4 . 1 4.4 5 . 0 15 .3 23 . 0 25 . 2
3 15-24 3 . 0 3 . 6 4.3 12 .4 18.8 22 . 1
4 25-34 4 . 1 5 . 2 5 . 9 9 .8 10 . 9 14.8
5 35-44 3 . 2 5 . 0 4 . 6 5 .8 10 .3 10 .7
6 45-54 2 .4 3 . 2 3.3 6 . 5 5 .7 6 .3
7 55-64 2 . 5 2 . 2 2 . 1 2 . 6 5 . 2 4 . 2
8 65-74 1 . 1 1 . 0 1 . 5 1 . 0 1 .7 2 . 5
9 75+ S 0 . 2 e S 0 .4 e 0 .8 e 0 .7 e
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