Read in training and test data.

train <- read.csv("train.csv",header=TRUE,row.names=1,stringsAsFactors=FALSE)
test <- read.csv("test.csv",header=TRUE,row.names=1,stringsAsFactors=FALSE)

Set aside the dependent variable (“Survived”).

survived <- train$Survived
train <- train[,setdiff(colnames(train),"Survived")]

Show that name and ticket have too many unique values.

length(unique(train$Name))

## [1] 891

length(unique(train$Ticket))

## [1] 681

Show that cabin has too many empty values.

length(which(train$Cabin == ""))

## [1] 687

table(ifelse(train$Cabin == "","Cabin empty","Cabin given"),train$Pclass)

##              
##                 1   2   3
##   Cabin empty  40 168 479
##   Cabin given 176  16  12

Show that both train and test have a few zero values for fare, but no or almost no NA values originally. Then, convert to NA if zero.

length(which(train$Fare == 0))

## [1] 15

length(which(is.na(train$Fare) == TRUE))

## [1] 0

length(which(test$Fare == 0))

## [1] 2

length(which(is.na(test$Fare) == TRUE))

## [1] 1

train$Fare[train$Fare == 0] <- NA
test$Fare[test$Fare == 0] <- NA

Show that both train and test have the same few extreme values and next-to-highest values for fare. Also show similar range and distribution.

head(train$Fare[order(train$Fare,decreasing=TRUE)])

## [1] 512.3292 512.3292 512.3292 263.0000 263.0000 263.0000

head(test$Fare[order(test$Fare,decreasing=TRUE)])

## [1] 512.3292 263.0000 263.0000 262.3750 262.3750 262.3750

range(train$Fare,na.rm=TRUE)

## [1]   4.0125 512.3292

range(test$Fare,na.rm=TRUE)

## [1]   3.1708 512.3292

par(mfrow=c(1,2))
hist(train$Fare,labels=TRUE,xlab="Fare",ylab="Passengers",main="Train")
hist(test$Fare,labels=TRUE,xlab="Fare",ylab="Passengers",main="Test")

Show the distribution of fares within each class in the training data, after truncate outliers.

train$Fare[train$Fare > 500 & is.na(train$Fare) == FALSE] <- 263
test$Fare[test$Fare > 500 & is.na(test$Fare) == FALSE] <- 263

par(mfrow=c(2,2))

for(class in 1:3)
{
    hist(train$Fare[train$Pclass == class],xlab="Fare",ylab="Passengers",main=paste0("Class=",class),labels=TRUE)
}

Show that very few first and second class passengers came aboard at Queensland.

table(train$Embarked,train$Pclass)

##    
##       1   2   3
##       2   0   0
##   C  85  17  66
##   Q   2   3  72
##   S 127 164 353

table(test$Embarked,test$Pclass)

##    
##       1   2   3
##   C  56  11  35
##   Q   1   4  41
##   S  50  78 142

Show that first-class passengers who embarked at Cherbourg tended to have higher fares than those at Southampton.

train$Embarked[train$Embarked == ""] <- NA
boxplot(Fare ~ Embarked,xlab="",ylab="Fare",data=train[train$Embarked != "Q" & train$Pclass == 1,],main="Train, 1st class passengers")

boxplot(Fare ~ Embarked,xlab="",ylab="Fare",data=test[test$Embarked != "Q" & test$Pclass == 1,],main="Test, 1st class passengers")

Show that the passengers who embarked at a unknown port are more likely from Cherbourg based on fare.

boxplot(Fare ~ Embarked,xlab="",ylab="Fare",data=train[train$Embarked != "Q" & train$Pclass == 1,],main="Train, 1st class passengers")
abline(h=80,lty=2)
legend("topright",legend="Fare for unknown port",lty=2)

For each sex and class, compare survival rates between ports.

for(class in 1:3)
{
    for(sex in c("male","female"))
    {
        print(paste0("Class=",class,";Sex=",sex))
        survival_rates = table(train$Embarked[train$Pclass == class & train$Sex == sex],survived[train$Pclass == class & train$Sex == sex])
        print(survival_rates)
        print(round(survival_rates[,2]*100/rowSums(survival_rates)))
    }
}

## [1] "Class=1;Sex=male"
##    
##      0  1
##   C 25 17
##   Q  1  0
##   S 51 28
##  C  Q  S 
## 40  0 35 
## [1] "Class=1;Sex=female"
##    
##      0  1
##   C  1 42
##   Q  0  1
##   S  2 46
##   C   Q   S 
##  98 100  96 
## [1] "Class=2;Sex=male"
##    
##      0  1
##   C  8  2
##   Q  1  0
##   S 82 15
##  C  Q  S 
## 20  0 15 
## [1] "Class=2;Sex=female"
##    
##      0  1
##   C  0  7
##   Q  0  2
##   S  6 61
##   C   Q   S 
## 100 100  91 
## [1] "Class=3;Sex=male"
##    
##       0   1
##   C  33  10
##   Q  36   3
##   S 231  34
##  C  Q  S 
## 23  8 13 
## [1] "Class=3;Sex=female"
##    
##      0  1
##   C  8 15
##   Q  9 24
##   S 55 33
##  C  Q  S 
## 65 73 38

For each sex and class, compare survival rates by number of relatives (SibSp + Parch).

num_relatives <- train$SibSp + train$Parch
table(num_relatives)

## num_relatives
##   0   1   2   3   4   5   6   7  10 
## 537 161 102  29  15  22  12   6   7

num_relatives[num_relatives > 3] <- 4

for(class in 1:3)
{
    for(sex in c("male","female"))
    {
        print(paste0("Class=",class,";Sex=",sex))
        survival_rates = table(num_relatives[train$Pclass == class & train$Sex == sex],survived[train$Pclass == class & train$Sex == sex])
        print(survival_rates)
        print(round(survival_rates[,2]*100/rowSums(survival_rates)))
    }
}

## [1] "Class=1;Sex=male"
##    
##      0  1
##   0 50 25
##   1 19 12
##   2  6  5
##   3  0  3
##   4  2  0
##   0   1   2   3   4 
##  33  39  45 100   0 
## [1] "Class=1;Sex=female"
##    
##      0  1
##   0  1 33
##   1  0 39
##   2  0 13
##   3  2  2
##   4  0  4
##   0   1   2   3   4 
##  97 100 100  50 100 
## [1] "Class=2;Sex=male"
##    
##      0  1
##   0 65  7
##   1 14  1
##   2  9  8
##   3  3  1
##  0  1  2  3 
## 10  7 47 25 
## [1] "Class=2;Sex=female"
##    
##      0  1
##   0  3 29
##   1  2 17
##   2  1 13
##   3  0  9
##   4  0  2
##   0   1   2   3   4 
##  91  89  93 100 100 
## [1] "Class=3;Sex=male"
##    
##       0   1
##   0 232  32
##   1  23   5
##   2  17   8
##   3   2   1
##   4  26   1
##  0  1  2  3  4 
## 12 18 32 33  4 
## [1] "Class=3;Sex=female"
##    
##      0  1
##   0 23 37
##   1 14 15
##   2 10 12
##   3  1  5
##   4 24  3
##  0  1  2  3  4 
## 62 52 55 83 11