The objective of these computer exercises is to become familiar with the basic structure of R and to learn some more sophisticated tips and tricks of R. The R code to perform the analyses, together with the output, is given in a separate file. However, you are strongly advised to write the code yourself. Questions are posed to reflect on what you are doing and to study output and results in more detail.

1 Basics

If you have not done so already, start R via Starten - Alle programma’s - R - R x64 3.6.1 (more recent versions of R should also be fine). This opens the R console. R is a command line driven environment. This means that you have to type in commands (line-by-line) for it to compute or calculate something. In its simplest form R can therefore be used as a pocket calculator:

2+2
# [1] 4

Question 1. (a) Look up today’s exchange rate of the euro versus the US$ on the Internet. Use R to calculate how many US$ is 15 euro.

Answer 

# This is the current exchange rate
15 * 1.1767
# [1] 17.6505

(b) Round the number you found to the first decimal using function round (use help or ? to inspect the arguments of the function) and assign the result to an object curr.

Answer 

curr <- round(15 * 1.1767, 1)
curr
# [1] 17.7

(c) Use mode, str, and summary to inspect the object you created. These functions give a compact display of the type of object and its contents.

Answer 

mode(curr)
# [1] "numeric"
str(curr)
#  num 17.7
summary(curr)
#    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
#    17.7    17.7    17.7    17.7    17.7    17.7

Luckily R is more than just a calculator, it is a programming language with most of the elements that every programming language has: statements, objects, types, classes etc.

2 R syntax: data and functions

Question 2. (a) One of the simplest data structures in R is a vector. Use the function seq to generate a vector vec that consists of all numbers from 11 to 30 (see ?seq for documentation).

Answer 

vec <- seq(11,30)
vec
#  [1] 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(b) Select the 7th element of the vector.

Answer 

vec[7]
# [1] 17

(c) Select all elements except the 15th.

Answer 

# Use '-' to exclude elements
vec[-15]
#  [1] 11 12 13 14 15 16 17 18 19 20 21 22 23 24 26 27 28 29 30

(d) Select the 2nd and 5th element of the vector.

Answer 

# Use the concatenate function 'c'
vec[c(2,5)]
# [1] 12 15

(e) Select only the odd valued elements of the vector vec. Do this is in two steps: first create the appropriate vector of indices index using the function seq, then use index to make the selection.

Answer 

index <- seq(1,length(vec),by=2)
vec[index]
#  [1] 11 13 15 17 19 21 23 25 27 29

Until now you have only used the R console. However, RStudio (Starten - Alle programma’s - R - RStudio) provides a much nicer and richer interface when working with R. Although all exercises can be made within the basic R environment, we highly recommend to use RStudio.

Question 3. (a) What are the windows that RStudio consists of?

Answer

Script, Console, Environment, History, Files, Plots, Packages, Help, Viewer.

From now on we advise you to use RStudio. You can use either the Console or the Script window (the upper left window). We recommend you to use the Script window, since this allows you to easily save your code to a file for later usage.

(b) Use the elementary functions /, -, ^ and the functions sum and length to calculate the mean \(\bar{x}=\sum_i x_i/n\) and the standard deviation \(\sqrt{\sum_i(x_i-\bar{x})^2/(n-1)}\) of the vector vec of all numbers from 11 to 30. You can verify your answer by using the built-in R functions mean and sd.

Answer 

# Don't call the variable 'mean' since the function 'mean' already exists
xmean <- sum(vec)/length(vec)
xmean
# [1] 20.5
mean(vec)
# [1] 20.5
# Calculate the standard deviation in three (baby) steps 
numerator <- sum((vec-xmean)^2)
denominator <- (length(vec)-1)
xstd  <- sqrt(numerator/denominator)
xstd
# [1] 5.91608
sd(vec)
# [1] 5.91608

(c) Once you completed the analysis, have a look at the Environment and History windows. Do you understand their contents?

Question 4. R comes with many predefined datasets. You can type data() to get the whole list. The islands dataset gives the areas of the world’s major landmasses exceeding 10,000 square miles. It can be loaded by typing:

# This could in this case actually be skipped since the package datasets is 
# already loaded
data(islands)

(a) Inspect the object using the functions head, str etc. Also have a look at the help file using help(islands).

(b) How many landmasses in the world exceeding 10,000 square miles are there?

Answer 

# From ?islands: "Description: The areas in thousands of square miles of 
# the landmasses which exceed 10,000 square miles." So all of them:
length(islands)
# [1] 48

(c) Make a Boolean vector that has the value TRUE for all landmasses exceeding 20,000 square miles.

Answer 

# Remember that area was given in thousands of square miles
islands.more20 <- (islands > 20)

(d) Select the islands with landmasses exceeding 20,000 square miles.

Answer 

# Use the Boolean vector islands.more20 to select the ones with value TRUE
islands[islands.more20] 
#          Africa      Antarctica            Asia       Australia          Baffin 
#           11506            5500           16988            2968             184 
#           Banks          Borneo         Britain         Celebes           Celon 
#              23             280              84              73              25 
#            Cuba           Devon       Ellesmere          Europe       Greenland 
#              43              21              82            3745             840 
#      Hispaniola        Hokkaido          Honshu         Iceland         Ireland 
#              30              30              89              40              33 
#            Java           Luzon      Madagascar        Mindanao        Moluccas 
#              49              42             227              36              29 
#      New Guinea New Zealand (N) New Zealand (S)    Newfoundland   North America 
#             306              44              58              43            9390 
#   Novaya Zemlya        Sakhalin   South America         Sumatra        Tasmania 
#              32              29            6795             183              26 
#        Victoria 
#              82

(e) Make a character vector that only contains the names of the islands.

Answer 

islands.names <- names(islands)

(f) The Moluccas have mistakenly been counted as a single island. The largest island of the Moluccas, Halmahera, only has about 7,000 square miles. Remove Moluccas from the data. Hint: an elegant solution uses the Boolean operator !=.

Answer 

notMoluccas <- islands.names != "Moluccas"
islands.withoutMoluccas <- islands[notMoluccas]
# Another solution is:
islands.withoutMoluccas <- subset(islands,islands.names!="Moluccas")

We will work with a data set that gives the survival status of passengers on the Titanic. See titanic3info.txt for a description of the variables. Some background information on the data can be found on titanic.html.

Question 5. (a) Download the titanic data set titanic3.dta in STATA format from the course website and import it into R. Give the data set an appropiate name as an R object. E.g., we can call it titanic3 (but feel free to give it another name). Before importing the data, you first have to load the foreign package in which the function read.dta is defined. Moreover, you probably will need to point R to the right folder where you saved the file titanic3.dta. You can do this by changing the so-called working directory, which will be explained later. Using the basic R environment you can do this via the menu (File - Change dir), and similarly for RStudio (Session - Set Working Directory - Choose Directory).

Answer 

library(foreign)
titanic3 <- read.dta("titanic3.dta", convert.underscore=TRUE)

We take a look at the values in the titanic data set.

(b) First, make the whole data set visible in a spreadsheet like format (how this is done depends on whether base R or RStudio is used).

Answer

This can be done using View on the command line. You can also use the menu: in R use Edit - Data editor, in RStudio click on the name of the data set in the Environment window.

(c) Often, it is sufficient to show a few records of the total data set. This can be done via selections on the rows, but another option is to use the functions head and tail. Use these functions to inspect the first 4 and last 4 records (instead of 6 which is the default).

Answer

The function dim can be used to find the number of rows (passengers in this case) and columns (variables) in the data.

dim(titanic3)

An alternative is the function str. This function shows information per column (type of variable, first records, levels in case of factor variables) as well as the total number of rows and columns.

str(titanic3)

(d) Issue both commands and study the output.

Answer 

dim(titanic3)
# [1] 1309   17

(e) Summarize the data set by using the summary function.

Answer 

summary(titanic3[,c("survived","pclass","home.dest","dob")])
#     survived     pclass     home.dest              dob        
#  Min.   :0.000   1st:323   Length:1309        Min.   :-46647  
#  1st Qu.:0.000   2nd:277   Class :character   1st Qu.:-31672  
#  Median :0.000   3rd:709   Mode  :character   Median :-27654  
#  Mean   :0.382                                Mean   :-28341  
#  3rd Qu.:1.000                                3rd Qu.:-25097  
#  Max.   :1.000                                Max.   :-17488  
#                                               NA's   :263

This gives a summary of all the variables in the data set. We can see that for categorical variables like pclass a frequency distribution is given, whereas for continuous variables like age numerical summaries are given. Still, for some variables we do not automatically get what we would like or expect:

  • The variable survived is dichotomous with values zero and one, which are interpreted as numbers.
  • The categorical variable pclass is represented differently from the categorical variable home.dest.
  • The variable dob, which gives the date of birth, is represented by large negative numbers.

In subsequent exercises, we will shed further light on these anomalies and we will try to repair some of these.

Question 6. We can also summarize specific columns (variables) of a data.frame. There are many ways to summarize a variable, depending on its structure and variability. For continuous variables, the same summary function can be used, but other options are the functions mean, quantile, IQR, sd and var. For categorical summaries, one may use summary and table. Note that missing values are treated differently depending on the function used.

(a) Summarize the age variable in the titanic data set (the age column is selected via titanic3$age). Give the 5%, 25%, 50%, 75% and 95% quantiles and the inter-quartile range. You may have to take a look at the help files for the functions.

Answer 

# Note that you have to convert percentages into probabilities
quantile(titanic3$age, probs=c(0.05,0.25,0.5,0.75,0.95), na.rm=TRUE)
#  5% 25% 50% 75% 95% 
#   5  21  28  39  57
IQR(titanic3$age, na.rm=TRUE)
# [1] 18

(b) Summarize the survived variable using the table function. Also give a two-by-two table for sex and survival status using the same table function.

Answer 

table(titanic3$survived)
# 
#   0   1 
# 809 500
table(titanic3$sex, titanic3$survived)
#         
#            0   1
#   female 127 339
#   male   682 161

Question 7 (OPTIONAL).

Instead of the file in STATA format, we also made available part of the titanic data set in tab-delimited format (titanic3select.txt). Download this file from the course website and then import it using the command read.table. Note that this file has been manipulated in Excel and that importing the data is not as straightforward as you would have hoped. You will need to tweak the arguments of read.table and/or make some changes to the file manually.

Answer

Running the following line of code:

titanic3.select <- read.table("titanic3select.txt",sep="\t",header=TRUE)

throws an error message:

Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec, : line 16 did not have 18 elements Error during wrapup: cannot open the connection

This error message can in principle be corrected by adding the argument fill=TRUE, which adds blank fields if rows have unequal length. However, running

titanic3.select <- read.table("titanic3select.txt",sep="\t",header=TRUE,fill=TRUE)

throws a warning message:

Warning message: In scan(file = file, what = what, sep = sep, quote = quote, dec = dec, : EOF within quoted string

It turns out that the data was not correctly imported, which is often easily diagnosed by inspecting the dimensions:

dim(titanic3.select)

Indeed the number of rows is 25 instead of the expected 30. This is caused by a trailing ’ at row 25 for variable home_dest. This can be corrected by changing the default set of quoting characters:

titanic3.select <- read.table("titanic3select.txt", sep="\t", header=TRUE, fill=TRUE, quote="\"")

A closer inspection of the data shows that the last column only contains NAs caused by trailing spaces. On the course website there is a corrected version of the file where the last column and the trailing ’ at row 25 were deleted.

titanic3.select <- read.table("titanic3select_corrected.txt",sep="\t",header=TRUE)

Question 8. We have a further look at the output from the summary function, which was not always what we would like to see. First, give the commands (sapply will be explained later):

sapply(titanic3, mode)
#      pclass    survived        name         sex         age       sibsp 
#   "numeric"   "numeric" "character"   "numeric"   "numeric"   "numeric" 
#       parch      ticket        fare       cabin    embarked        boat 
#   "numeric" "character"   "numeric"   "numeric"   "numeric"   "numeric" 
#        body   home.dest         dob      family      agecat 
#   "numeric" "character"   "numeric"   "numeric"   "numeric"
sapply(titanic3, is.factor)
#    pclass  survived      name       sex       age     sibsp     parch    ticket 
#      TRUE     FALSE     FALSE      TRUE     FALSE     FALSE     FALSE     FALSE 
#      fare     cabin  embarked      boat      body home.dest       dob    family 
#     FALSE      TRUE      TRUE      TRUE     FALSE     FALSE     FALSE      TRUE 
#    agecat 
#      TRUE

We see that survived has mode “numeric” and is not a factor (is.factor(titanic3$survived) gives FALSE). It is interpreted in the same way as the truly continuous variable age.

(a) Add a variable status to the titanic data set, which gives the survival status of the passengers as a factor variable with labels “no” and “yes” describing whether individuals survived. Make the value “no” the first level (see titanic3info.txt for the reason).

Answer 

# titanic3info.txt: Survival (0 = No; 1 = Yes)
titanic3$status <- factor(titanic3$survived, labels=c("no","yes"))

(b) Variable pclass has mode “numeric” and is a factor, whereas home.dest has mode “character” and is not a factor. Give the commands

as.numeric(head(titanic3$pclass))
as.numeric(head(titanic3$home.dest))

and explain the difference in output.

Answer 

as.numeric(head(titanic3$pclass))
# [1] 1 1 1 1 1 1
as.numeric(head(titanic3$home.dest))
# [1] NA NA NA NA NA NA

We do not change the variables in this dataset that have mode “character”. They are variables that have many different values, and there is no reason to convert them into factors.

Question 9. (a) Take a look at the name (name), and the home town/destination (home.dest) of all passengers who were older than 70 years. Use the appropriate selection functions.

Answer 

# Note that the function 'subset' also leaves out the passsengers for which the
# variable 'age' is missing 
subset(titanic3,age>70)[,c("name","home.dest")]
#                                 name               home.dest
# 10           Artagaveytia, Mr. Ramon     Montevideo, Uruguay
# 15   Barkworth, Mr. Algernon Henry W           Hessle, Yorks
# 62    Cavendish, Mrs. Tyrell William Little Onn Hall, Staffs
# 136        Goldschmidt, Mr. George B            New York, NY
# 728             Connors, Mr. Patrick                        
# 1236             Svensson, Mr. Johan
# Another solution is
subset(titanic3,age>70,select=c(name,home.dest))
#                                 name               home.dest
# 10           Artagaveytia, Mr. Ramon     Montevideo, Uruguay
# 15   Barkworth, Mr. Algernon Henry W           Hessle, Yorks
# 62    Cavendish, Mrs. Tyrell William Little Onn Hall, Staffs
# 136        Goldschmidt, Mr. George B            New York, NY
# 728             Connors, Mr. Patrick                        
# 1236             Svensson, Mr. Johan
# Yet another solution is
titanic3[titanic3$age>70 & !is.na(titanic3$age),c("name","home.dest")]
#                                 name               home.dest
# 10           Artagaveytia, Mr. Ramon     Montevideo, Uruguay
# 15   Barkworth, Mr. Algernon Henry W           Hessle, Yorks
# 62    Cavendish, Mrs. Tyrell William Little Onn Hall, Staffs
# 136        Goldschmidt, Mr. George B            New York, NY
# 728             Connors, Mr. Patrick                        
# 1236             Svensson, Mr. Johan

(b) There is one person from Uruguay in this group. Select the single record from that person. Did this person travel with relatives?

Answer 

subset(titanic3, name=="Artagaveytia, Mr. Ramon")
#    pclass survived                    name  sex age sibsp parch   ticket
# 10    1st        0 Artagaveytia, Mr. Ramon male  71     0     0 PC 17609
#       fare cabin  embarked boat body           home.dest       dob family
# 10 49.5042       Cherbourg        22 Montevideo, Uruguay -43359.75     no
#     agecat status
# 10 (40,80]     no
# No he didn't travel with relatives, since the variable 'family' is 'no'

(c) Make a table of survivor status by sex, but only for the first class passengers. Use the xtabs function.

Answer 

xtabs(~sex+status, data=titanic3, subset=(pclass=="1st"))
#         status
# sex       no yes
#   female   5 139
#   male   118  61

3 Graphics

3.1 Basic graphics

The graphics subsystem of R offers a very flexible toolbox for high-quality graphing. There are typically three steps to producing useful graphics:

  • Creating the basic plot
  • Enhancing the plot with labels, legends, colors etc.
  • Exporting the plot from R for use elsewhere

In the graphics model that R uses, a figure region consists of a central plotting region surrounded by margins. The basic graphics command is plot. The following piece of code illustrates the most common options:

plot(3,3,main="Main title",sub="subtitle",xlab="x-label",ylab="y-label")
text(3.5,3.5,"text at (3.5,3.5)")
abline(h=3.5,v=3.5)
for (side in 1:4) mtext(-1:4,side=side,at=2.3,line=-1:4)
mtext(paste("side",1:4),side=1:4,line=-1,font=2)

In this figure one can see that

  • Sides are labelled clockwise starting with x-axis
  • Coordinates in the margins are specified in lines of text
  • Default margins are not all wide enough to hold all numbers -1,…,4

You might want to use the help function to investigate some of the other functions and options.

Plots can be further finetuned with the par function. For instance, the default margin sizes can be changed using par. The default settings are

par("mar")
# [1] 5.1 4.1 4.1 2.1

This explains why only side 1 in the figure had a wide enough margin. This can be remedied by setting

par(mar=c(5,5,5,5))

before plotting the figure.

Question 10. (a) Try making this figure yourself by executing the code shown above.

(b) Save the figure you just made to a file. For this you have to know that R sends graphics to a device. The default device on most operating systems is the screen, for example the “Plots” window in RStudio. Saving a figure, therefore, requires changing R’s current device. See help(device) for the options. Save the figure you just made to a png and a pdf file. Don’t forget to close the device afterwards.

Answer 

png(file="figure.png")
# Add code here to plot the figure (including par(mar=c(5,5,5,5)))
dev.off()
pdf(file="figure.pdf")
# Add code here to plot the figure (including par(mar=c(5,5,5,5)))
dev.off()

(c) When saving a figure to file, default values for the figure size are used. Save the figure to a pdf file with width and height of 21 inches.

Answer 

pdf(file="figureLarge.pdf",width=21,height=21)
# Add code here to plot the figure (including par(mar=c(5,5,5,5)))
dev.off()

Question 11. The quakes data set gives location and severity of earthquakes off Fuji. First load the data:

data(quakes)

(a) Make a scatterplot of latitude versus longitude. You should obtain a graph similar to:

Answer 

plot(quakes$long, quakes$lat)

(b) Use cut to divide the magnitude of quakes into three categories (cutoffs at 4.5 and 5) and use ifelse to divide depth into two categories (cutoff at 400). Hint: have a careful look at the (admittedly complicated) help page of cut, in particular the arguments breaks and include.lowest.

Answer 

mag.cat <- with(quakes,cut(mag, breaks = c(4, 4.5, 5, 7),
                           include.lowest = TRUE ))
# More generic using the functions 'min' and 'max' that return the 
# minimum and maximum, respectively, of a vector
mag.cat <- with(quakes,cut(mag, breaks = c(min(mag), 4.5, 5, max(mag)),
                           include.lowest = TRUE ))
levels(mag.cat) <- c("low", "medium", "high")
depth.cat <- factor(ifelse(quakes$depth>400, "deep", "shallow"))

(c) Redraw the plot, with symbols by magnitude and colors by depth. You should obtain a graph similar to:

Answer 

# Note that for 'col' the factor 'depth.cat' is interpreted as numeric (deep=1 and 
# shallow=2) and that the first two colours of palette() are used. Strangely enough 
# for 'pch' this is not the case and the factor ''mag.cat' has to be converted to a 
# numeric vector using the function 'as.numeric'.
plot(quakes$long, quakes$lat, pch=as.numeric(mag.cat), col=depth.cat)

(d) The magnitude of the earthquakes is given in Richter scale. Calculate the energy released in each earthquake as \(10^{(3/2)}\) to the power of the Richter measurement.

Answer 

energy <- (10^(3/2))^quakes$mag

(e) The cex argument of plot can be used to scale the plot symbols. We will scale the plot symbols so that the surface of the plot symbol is proportional to the released energy. Calculate plot symbol size as the square root of the energy divided by the median square root of the energy (to get median symbol size 1).

Answer 

symbolsize <- sqrt(energy)/median(sqrt(energy))

(f) Plot the magnitude of earthquakes again, but with plot symbols sized by energy. Keep the coloring by depth. You should obtain a graph similar to:

Answer 

plot(quakes$long, quakes$lat, cex=symbolsize, col=depth.cat)

Question 12. Different plots can be combined in a single window (device) via , e.g., par(mfrow=c(..)) or layout(..). Combine the histogram and the two boxplots for the titanic data from the lecture into a single plot. Find a nice layout of your final plot, see help files for setting proper margins, etc. You should obtain a graph similar to:

Answer 

par(mfrow=c(3,1))
# Make the margins smaller than default
par(mar=c(2,4,2,2))
hist(titanic3$age,breaks=15,freq=FALSE)
boxplot(titanic3$fare,ylim=c(0,300),ylab="fare")
boxplot(fare~pclass,data=titanic3,ylim=c(0,300),ylab="fare")

4 Structure of R

Question 13. Investigate which environments are in the search path. Take a look at the objects that exist in the workspace. Which is the current working directory?

Answer

Question 14. The function mean is defined in the base package, which is included in the search path at startup. From the search command, we can see where in the search path the base package is located. Issue the command

ls("package:base", pattern="mean")

Instead of the argument package:base, we could have given the position in the search path of the base package (that is ls(10, pa="mean") if base is in position 10). Have a look at the different names of the mean function.

Question 15. Save the titanic data set in R binary format with extension “.RData”.

Answer 

save(titanic3, file="Titanic.RData")

5 Data manipulation

Question 16. Sort the titanic data set according to the age of the passengers and store the result in a separate object, e.g. named data.sorted. Have a look at the 10 youngest and the 10 oldest individuals. For reasons of space, restrict to the first 5 columns when showing the results. What do you notice with respect to passenger class and age?

Answer 

# If not installed yet, first install the package dplyr that contains the function 
# arrange
#install.packages("dplyr")
library(dplyr)
data.sorted <- arrange(titanic3, age)
head(data.sorted[,1:5],10)
#      pclass survived                              name    sex    age
# 764     3rd        1 Dean, Miss. Elizabeth Gladys \\"M female 0.1667
# 748     3rd        0   Danbom, Master. Gilbert Sigvard   male 0.3333
# 1241    3rd        1   Thomas, Master. Assad Alexander   male 0.4167
# 428     2nd        1         Hamalainen, Master. Viljo   male 0.6667
# 658     3rd        1            Baclini, Miss. Eugenie female 0.7500
# 659     3rd        1     Baclini, Miss. Helene Barbara female 0.7500
# 1112    3rd        0    Peacock, Master. Alfred Edward   male 0.7500
# 360     2nd        1     Caldwell, Master. Alden Gates   male 0.8333
# 549     2nd        1   Richards, Master. George Sibley   male 0.8333
# 612     3rd        1         Aks, Master. Philip Frank   male 0.8333
# Use is.na to exclude the passengers for whom 'age' is missing
tail(subset(data.sorted,!is.na(age))[,1:5],10)
#      pclass survived                            name    sex  age
# 595     2nd        0           Wheadon, Mr. Edward H   male 66.0
# 286     1st        0              Straus, Mr. Isidor   male 67.0
# 82      1st        0    Crosby, Capt. Edward Gifford   male 70.0
# 507     2nd        0     Mitchell, Mr. Henry Michael   male 70.0
# 728     3rd        0            Connors, Mr. Patrick   male 70.5
# 10      1st        0         Artagaveytia, Mr. Ramon   male 71.0
# 136     1st        0       Goldschmidt, Mr. George B   male 71.0
# 1236    3rd        0             Svensson, Mr. Johan   male 74.0
# 62      1st        1  Cavendish, Mrs. Tyrell William female 76.0
# 15      1st        1 Barkworth, Mr. Algernon Henry W   male 80.0

Question 17. Give a summary of the fare paid for the three passenger classes separately, using the summary function, the subset function for selecting the appropriate rows, as well as one of the mechanisms for selecting columns.

Answer 

summary(subset(titanic3, pclass=="1st")$fare)
#    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
#    0.00   30.70   60.00   87.51  107.66  512.33
summary(subset(titanic3, pclass=="2nd")$fare)
#    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
#    0.00   13.00   15.05   21.18   26.00   73.50
summary(subset(titanic3, pclass=="3rd")$fare)
#    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
#    0.00    7.75    8.05   13.30   15.25   69.55       1

Remark

Instead of writing three lines of code, applying the summary function to each of the three subsets, we could write a for loop. When using a for loop, we need to explicitly write the print command:

for(i in 1:3){print(summary(subset(titanic3, pclass==levels(pclass)[i])$fare))}

The function aggregate is yet another option:

aggregate(fare~pclass,data=titanic3,FUN=summary)

Later, we will see even more efficient ways.

Question 18. Create an extra variable that categorizes the variable sibsp (the number of siblings/spouses aboard) into three groups: 0, 1 and more than 1. Also, create an extra factor variable named paid, which shows whether the individual paid for the trip (i.e. whether fare >0). Preferably, use the within or the transform function. Check whether the results are correct.

Answer 

titanic3 <- within(titanic3, {
   sibspcat <- cut(sibsp, breaks=c(0,1,2,9),include.lowest=TRUE,
                   right=FALSE, labels=c("0","1","2-8"))
   paid <- factor(fare>0, labels=c("no","yes"))
} )

titanic3 <- transform(titanic3,
   sibspcat = cut(sibsp, breaks=c(0,1,2,9),include.lowest=TRUE,
     right=FALSE, labels=c("0","1","2-8")),
   paid = factor(fare>0, labels=c("no","yes"))
 )

6 Documentation and help

Question 19. R is also very flexible with respect to manipulation of character data, such as words. In this exercise, you will see an example.

(a) Create a character vector of length three that consists of the words “Academic”, “Medical” and “Center”. Give the object the name AMC. Check the mode of the object.

Answer 

AMC <- c("Academic","Medical","Center") # a character vector named AMC
mode(AMC)
# [1] "character"
is.character(AMC)
# [1] TRUE

(b) Next, we want to abbreviate the word and obtain AMC as output. Try to find the appropriate functions using the commands

help.search("abbreviate")
help.search("combine")
help.search("quote")

Answer 

#help(abbreviate)
abbreviate(AMC,1) # selects first character
# Academic  Medical   Center 
#      "A"      "M"      "C"
#help(paste)
paste(abbreviate(AMC,1),collapse="") # gives AMC
# [1] "AMC"

Finally, if you want to remove the quotes give the following command

noquote(paste(abbreviate(AMC,1),collapse="")) # removes quotes

Question 20. Try to find more information on a topic of interest and how R can be of help. If you do not have any idea, you can search for the keyword “crosstab”.

Answer 

help(crosstab)
help.search("crosstab")
RSiteSearch("crosstab")
install.packages("sos")
library(sos)
findFn("crosstab")

7 Statistical analysis

Question 21. Fit a linear model that predicts fare as a function of age. Since fare has a very skewed distribution, we use the transformed variable log10(fare+1). Consider the following issues.

(a) Fit the model and store the results in an R object. Summarize the model using the summary function.

Answer 

fit.fare <- lm(log10(fare+1) ~ age, data=titanic3)
summary(fit.fare)
# 
# Call:
# lm(formula = log10(fare + 1) ~ age, data = titanic3)
# 
# Residuals:
#      Min       1Q   Median       3Q      Max 
# -1.44366 -0.33354 -0.07742  0.27298  1.34630 
# 
# Coefficients:
#              Estimate Std. Error t value Pr(>|t|)    
# (Intercept) 1.1651838  0.0293881  39.648  < 2e-16 ***
# age         0.0056833  0.0008869   6.408 2.23e-10 ***
# ---
# Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# 
# Residual standard error: 0.4123 on 1043 degrees of freedom
#   (264 observations deleted due to missingness)
# Multiple R-squared:  0.03788, Adjusted R-squared:  0.03696 
# F-statistic: 41.06 on 1 and 1043 DF,  p-value: 2.228e-10

(b) One of the assumptions of a standard linear model is that the residuals have approximately normal distribution. Make a histogram of the residuals, using the functions resid and hist. You should obtain a graph similar to:

Answer 

fit.fare <- lm(log10(fare+1) ~ age, data=titanic3)
hist(resid(fit.fare))

(c) Make a plot of the residuals against the fitted values, using (with fit.fare the name of the object that contains the linear model fit):

plot(resid(fit.fare)~fitted(fit.fare))

(d) Make a scatterplot of fare against age and add the linear regression line. A fitted regression line is added to a plot via abline(fit.fare). You should obtain a graph similar to:

Answer 

plot(log10(fare+1)~age,data=titanic3)
abline(fit.fare)

(e) Does the object have a class? If so, which are the generic functions that have a method for this class?

Answer 

class(fit.fare)
# [1] "lm"
methods(class="lm")
#  [1] add1           alias          anova          case.names     coerce        
#  [6] confint        cooks.distance deviance       dfbeta         dfbetas       
# [11] drop1          dummy.coef     effects        extractAIC     family        
# [16] formula        hatvalues      influence      initialize     kappa         
# [21] labels         logLik         model.frame    model.matrix   nobs          
# [26] plot           predict        print          proj           qr            
# [31] residuals      rstandard      rstudent       show           simulate      
# [36] slotsFromS3    summary        variable.names vcov          
# see '?methods' for accessing help and source code

8 Programming and ply functions

Functions from the apply family are convenient shorthands for repetitions.

Question 22. Use apply to calculate the mean of the variables age, fare, and body of titanic3.

Answer 

apply(subset(titanic3,select=c("age","fare","body")),2,mean,na.rm=TRUE)
#       age      fare      body 
#  29.88113  33.29548 160.80992

Question 23. The chickwts data describes chicken weights by feed type. First load the data:

data(chickwts)

(a) Calculate the mean weight for each feed type.

Answer 

tapply(chickwts$weight, chickwts$feed, mean)
#    casein horsebean   linseed  meatmeal   soybean sunflower 
#  323.5833  160.2000  218.7500  276.9091  246.4286  328.9167

(b) Count the number of chicks with weight over 300 grams,

Answer 

sum(chickwts$weight > 300)
# [1] 26

Further abstraction of the R code is possible through functions. Functions lead to more compact code that is easier to understand and also avoid duplication of the same code over and over again.

name <- function(arg_1,arg_2, ...) expr
  • expr is, in general, a grouped expression containing the arguments arg_1, arg_2 …
  • A call to the function is of the form name(expr_1,expr_2, ...)
  • The value of the expression expr is the value returned by the function

Question 24. (a) For the chickwts data, write a function that takes a vector x as input and returns the number of observations in x greater than 300.

Answer 

greater300 <- function(x) {
  sum(x > 300)
}

(b) Calculate the number of chicks with weight over 300 grams for each feed type.

Answer 

tapply(chickwts$weight, chickwts$feed, greater300)
#    casein horsebean   linseed  meatmeal   soybean sunflower 
#         8         0         1         5         3         9