R语言代写

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Homework 1 Solutions

Part 1

Homework 1 Solutions Gabriel Young (gjy2107) Febuary, 8th 2018

i. Looking at Titanic.txt I see that the data is tab delimited and therefore I use read.table(). Since the first row of data in the text file holds the column names, I use header = TRUE.

ii. The function dim() provides the dimension of its input object. dim(titanic)

## [1] 891 12

str(titanic)

## ‘data.frame’: 891 obs. of 12 variables: ## $PassengerId:int 12345678910…

setwd(“~/Desktop/Data”)
titanic <- read.csv(“Titanic.txt”, header = TRUE, as.is = TRUE)

## $Survived ## $ Pclass ## $ Name
## $ Sex

:int 0111000011…
:int 3131331332…
: chr “Braund, Mr. Owen Harris” “Cumings, Mrs. John Bradley (Florence Briggs Thayer)” : chr “male” “female” “female” “female” …
:num 2238263535NA5422714…
:int 1101000301…
:int 0000000120…
: chr “A/5 21171” “PC 17599” “STON/O2. 3101282” “113803” …
: num 7.25 71.28 7.92 53.1 8.05 …
: chr “” “C85” “” “C123” … 

##  $ Age
##  $ SibSp
##  $ Parch
##  $ Ticket
##  $ Fare
##  $ Cabin
##  $ Embarked   : chr  "S" "C" "S" "S" ...

iii. There are multiple ways to do this. In the following, I add a new column called Survived.Word with each entry equal to “survived”. Then I reassign the values in the variable Survived.Word to “died” in the rows where Survived equals 0.

Alternatively, we could use ifelse() to complete this. The first argument is a logical vector that’s TRUE when Survived equals 1 and FALSE when Survived equals 0.

titanic$Survived.Word <- ifelse(titanic$Survived == 1, “surived”, “died”) 

titanic$Survived.Word <- "survived"
titanic$Survived.Word[titanic$Survived == 0] <- "died"

1

Part 2

i. To solve this problem we create a sub-matrix of the variables of entry and then pass this sub-matrix into the apply() command. 

##   Survived        Age       Fare
##  0.3838384         NA 32.2042080

The mean of “Survived” is the proportion of passengers that survived the disaster. The mean of the Age variable is NA, because some of the passenger’s ages are unknown (i.e. Age also has some missing values).

ii. As in the last question, we can calculate the proportion of survivors by taking the mean of the Survived variable, but here we filter to only include female passengers.

round(mean(titanic$Survived[titanic$Sex == “female”]), 2) ## [1] 0.74

iii. To answer this question we create a sub-matrix survivors which only includes the rows of titanic corresponding to those surviving the disater. Then we calculate the proportion as the number of female passengers in the survivors matrix divided by the total number of people in the survivors matrix.

## [1] 0.68

Alternatively, we can use the table() command. 

##
## female   male
##   0.68   0.32

iv.

sub_mat <- titanic[, c(“Survived”, “Age”, “Fare”)] apply(sub_mat, 2, mean)

survivors <- titanic[titanic$Survived == 1, ]
proportion <- sum(survivors$Sex == “female”)/length(survivors$Sex) round(proportion, 2)

survivors <- titanic[titanic$Survived == 1, ]
proportion <- table(survivors$Sex)/length(survivors$Sex)[1] round(proportion, 2)

classes <- sort(unique(titanic$Pclass)) Pclass.Survival <- vector(“numeric”, length = 3) names(Pclass.Survival) <- classes

for (i in 1:3) {
thisclass <- titanic[titanic$Pclass == i, ] Pclass.Survival[i] <- round(mean(thisclass$Survived), 2)

}

2

v. 

##    1    2    3
## TRUE TRUE TRUE

vi. 

Pclass.Survival

##    1    2    3
## 0.63 0.47 0.24

There does appear to be a relationship between survival and class. We can see from the previous question that the survival rate decreases with ticket class, meaning fewer members of the lower class survived than members of the upper class.

Pclass.Survival2 <- round(tapply(titanic$Survived, titanic$Pclass, mean), 2) Pclass.Survival == Pclass.Survival2

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