PROCEDURE

Library definition

# Need to employ kable
library(knitr)
# Need to employ stringr for Regular Expressions
library(stringr)
# Need to employ to use tidy data functions
library(tidyr)
library(dplyr)

(1) Create .CSV file

I have created a .CSV file named: “Villalobos-airlines.csv”

Villalobos-airlines.csv

(2) Read information from .CSV file into R.

For simplicity and reproducibility reasons, I have posted this file on my GitHub repository as follows:

url <- "https://raw.githubusercontent.com/dvillalobos/MSDA/master/607/Homework/Homework5/Villalobos-airlines.csv"

my.data <- read.csv(url, header=FALSE, sep=",", stringsAsFactors=FALSE)
my.data <- data.frame(my.data)

##        V1      V2          V3      V4        V5            V6     V7
## 1                 Los Angeles Phoenix San Diego San Francisco Seatle
## 2  ALASKA on time         497     221       212           503   1841
## 3         delayed          62      12        20           102    305
## 4                                                                   
## 5 AM WEST on time         694    4840       383           320    201
## 6         delayed         117     415        65           129     61

# Adding "Missing" titles from original file onto the Row #1
my.data$V1[1] <- "Airline"
my.data$V2[1] <- "Status"

# Assigning all the values from the row #1 as the Column Headers
names(my.data) <- my.data[1,]

# Need to eliminate Row #1 in order to keep data consistency.
my.data <- my.data[-c(1), ]

## 'data.frame':    5 obs. of  7 variables:
##  $ Airline      : chr  "ALASKA" "" "" "AM WEST" ...
##  $ Status       : chr  "on time" "delayed" "" "on time" ...
##  $ Los Angeles  : chr  "497" "62" "" "694" ...
##  $ Phoenix      : chr  "221" "12" "" "4840" ...
##  $ San Diego    : chr  "212" "20" "" "383" ...
##  $ San Francisco: chr  "503" "102" "" "320" ...
##  $ Seatle       : chr  "1841" "305" "" "201" ...

for (i in 3:dim(my.data)[2]){
  my.data[,i] <- as.integer(my.data[,i])
}

## 'data.frame':    5 obs. of  7 variables:
##  $ Airline      : chr  "ALASKA" "" "" "AM WEST" ...
##  $ Status       : chr  "on time" "delayed" "" "on time" ...
##  $ Los Angeles  : int  497 62 NA 694 117
##  $ Phoenix      : int  221 12 NA 4840 415
##  $ San Diego    : int  212 20 NA 383 65
##  $ San Francisco: int  503 102 NA 320 129
##  $ Seatle       : int  1841 305 NA 201 61

my.data <- my.data %>% drop_na()

for (i in 1:dim(my.data)[1]){
  if (i %% 2 == 0){
      my.data$Airline[i] <-  my.data$Airline[i-1]
  }
}

(3) Analysis

First: we need to transform our table by employing gather() from tidyr library.

# Tidy table by having 4 variables (Airline, Status, City, number of flights)
my.tidy.data <- my.data %>% gather("City","n flights", 3:7)

Second: Now, I will separate the values “on time” and “delayed” from the Status column into two different columns by employing the spread() function from tidyr library.

Please note that these values can be considered as two different variables.

my.tidy.data <-  my.tidy.data %>%
                 spread(Status, `n flights`)

Final Tidy Table

Some Horizontal Probabilities

From the previous “Final Tidy Table” we can find some horizontal probabilities, those can be found as follows:

Joining tables with horizontal probabilities

For this, I will join the Final Tidy Table with the respective probabilities tables in order to create a Final Horizontal Probability Table by employing inner_join() from the dplyr library.

Resulting Horizontal Probability table:

By comparing with the below table, we noticed that the values are different. That is that so far I have calculated only horizontal probabilities and further analysis can be performed.

Other probabilities

Other probabilities that can be found will be Vertical Probabilities or/and Total Probabilities by taking into consideration the total flights (“delayed” + “on time”).

Once those probabilities are found we can then start answering questions like:

1. What’s the probability that a randonmly selected flight will be delayed?

2. What’s the probability that a randonly selected flight will be from Alaska Airlines and it’s destination will be Seatle?

3. What’s the probability that out of 5 randonmly selected flights will be 2 delayed and 3 on time?