Introduction
East-West Airlines is trying to learn more about its customers. Key issues are their flying patterns, earning and use of frequent flyer rewards, and use of the airline credit card. The task is to identify customer segments via clustering. The file EastWestAirlines.xlsx contains information on 4000 passengers who belong to an airline’s frequent flier program. For each passenger the data include information on their mileage history and on different ways they accrued or spent miles in the last year. The goal is to try to identify clusters of passengers that have similar characteristics for the purpose of targeting different segments for different types of mileage offers.
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| # Library imports
library(readxl)
library(dplyr)
library(ggplot2)
library(cluster)
# Setting the seed
seed <- 425
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| # Importing, modifying, and checking the data set
data <- read_excel("EastWestAirlines.xlsx", sheet = "data")
data <- data[,-1] # Dropping the unnecessary index column
data.scaled <- apply(data, # Scaling the data between 0 and 1
MARGIN = 2,
FUN = function(X) (X - min(X))/diff(range(X)))
summary(data.scaled) # Checking the modified version
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| ## Balance Qual_miles cc1_miles cc2_miles
## Min. :0.00000 Min. :0.00000 Min. :0.0000 Min. :0.000000
## 1st Qu.:0.01087 1st Qu.:0.00000 1st Qu.:0.0000 1st Qu.:0.000000
## Median :0.02528 Median :0.00000 Median :0.0000 Median :0.000000
## Mean :0.04317 Mean :0.01293 Mean :0.2649 Mean :0.007252
## 3rd Qu.:0.05420 3rd Qu.:0.00000 3rd Qu.:0.5000 3rd Qu.:0.000000
## Max. :1.00000 Max. :1.00000 Max. :1.0000 Max. :1.000000
## cc3_miles Bonus_miles Bonus_trans Flight_miles_12mo
## Min. :0.000000 Min. :0.00000 Min. :0.00000 Min. :0.00000
## 1st Qu.:0.000000 1st Qu.:0.00474 1st Qu.:0.03488 1st Qu.:0.00000
## Median :0.000000 Median :0.02720 Median :0.13953 Median :0.00000
## Mean :0.003063 Mean :0.06502 Mean :0.13491 Mean :0.01493
## 3rd Qu.:0.000000 3rd Qu.:0.09026 3rd Qu.:0.19767 3rd Qu.:0.01009
## Max. :1.000000 Max. :1.00000 Max. :1.00000 Max. :1.00000
## Flight_trans_12 Days_since_enroll Award
## Min. :0.00000 Min. :0.0000 Min. :0.0000
## 1st Qu.:0.00000 1st Qu.:0.2807 1st Qu.:0.0000
## Median :0.00000 Median :0.4936 Median :0.0000
## Mean :0.02592 Mean :0.4963 Mean :0.3703
## 3rd Qu.:0.01887 3rd Qu.:0.6979 3rd Qu.:1.0000
## Max. :1.00000 Max. :1.0000 Max. :1.0000
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Hierarchical Clustering
Applying hierarchical clustering with Euclidean distance and complete linkage:
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| # Applying hierarchical clustering
dm_a <- dist(data.scaled,
method = "euclidean") # Dissimilarity matrix (Euclidean distance)
hc.complete.a <- hclust(dm_a,
method = "complete") # Hierarchical clustering using complete linkage
sil.widths.a <- c() # Empty vector to store silhouette widths
# for different values of K
for (i in 2:10) { # The for loop that generates cluster sets and
# calculates their respective silhouette widths
clust <- cutree(hc.complete.a, k = i)
sil <- silhouette(clust, dm_a)
sil.widths.a[i-1] <- mean(sil[, c("sil_width")])
}
results_a <- data.frame(k = 2:10, sil.widths.a) # Storing the results in a data frame
results_a # Viewing the results
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| ## k sil.widths.a
## 1 2 0.4704091
## 2 3 0.5318787
## 3 4 0.4645971
## 4 5 0.4167248
## 5 6 0.4172319
## 6 7 0.4161869
## 7 8 0.3912069
## 8 9 0.3519030
## 9 10 0.3522325
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As can be seen in the above table, K = 3 yields the highest silhouette width, which is 0.53, implying that 3 is the appropriate number of clusters to be used.
The next step is to compare the cluster centroids to characterize the different clusters and try to give each cluster a label. To check the final state of the clusters, the clustering with the best k value can be performed as follows:
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| # Performing the best clustering
hc.best.a <- cutree(hc.complete.a,
k = results_a[which.max(results_a$sil.widths.a), 1])
table(hc.best.a)
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| ## hc.best.a
## 1 2 3
## 2526 1469 4
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The above results indicate that two of the clusters dominates the data set, while the other one includes only 4 observations.
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| # Creating a table that displays the centroids of two clusters
data.w.clusters <- mutate(data, Cluster = hc.best.a)
first.cluster <- data.w.clusters[data.w.clusters$Cluster == 1, ]
second.cluster <- data.w.clusters[data.w.clusters$Cluster == 2, ]
third.cluster <- data.w.clusters[data.w.clusters$Cluster == 3, ]
centroids <- data.frame("1" = colMeans(first.cluster),
"2" = colMeans(second.cluster),
"3" = colMeans(third.cluster) )
print(round(centroids, 2))
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| ## X1 X2 X3
## Balance 59791.06 97189.59 131999.50
## Qual_miles 88.19 239.73 347.00
## cc1_miles 1.70 2.67 2.50
## cc2_miles 1.02 1.01 1.00
## cc3_miles 1.01 1.01 1.00
## Bonus_miles 10324.88 28739.99 65634.25
## Bonus_trans 9.19 15.59 69.25
## Flight_miles_12mo 230.44 801.79 19960.00
## Flight_trans_12 0.67 2.45 49.25
## Days_since_enroll 3824.89 4628.76 2200.25
## Award 0.00 1.00 1.00
## Cluster 1.00 2.00 3.00
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To obtain more insights about the clusters, centroids can be compared to the quartiles:
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| ## Balance Qual_miles cc1_miles cc2_miles
## Min. : 0 Min. : 0.0 Min. :1.00 Min. :1.000
## 1st Qu.: 18528 1st Qu.: 0.0 1st Qu.:1.00 1st Qu.:1.000
## Median : 43097 Median : 0.0 Median :1.00 Median :1.000
## Mean : 73601 Mean : 144.1 Mean :2.06 Mean :1.015
## 3rd Qu.: 92404 3rd Qu.: 0.0 3rd Qu.:3.00 3rd Qu.:1.000
## Max. :1704838 Max. :11148.0 Max. :5.00 Max. :3.000
## cc3_miles Bonus_miles Bonus_trans Flight_miles_12mo
## Min. :1.000 Min. : 0 Min. : 0.0 Min. : 0.0
## 1st Qu.:1.000 1st Qu.: 1250 1st Qu.: 3.0 1st Qu.: 0.0
## Median :1.000 Median : 7171 Median :12.0 Median : 0.0
## Mean :1.012 Mean : 17145 Mean :11.6 Mean : 460.1
## 3rd Qu.:1.000 3rd Qu.: 23801 3rd Qu.:17.0 3rd Qu.: 311.0
## Max. :5.000 Max. :263685 Max. :86.0 Max. :30817.0
## Flight_trans_12 Days_since_enroll Award
## Min. : 0.000 Min. : 2 Min. :0.0000
## 1st Qu.: 0.000 1st Qu.:2330 1st Qu.:0.0000
## Median : 0.000 Median :4096 Median :0.0000
## Mean : 1.374 Mean :4119 Mean :0.3703
## 3rd Qu.: 1.000 3rd Qu.:5790 3rd Qu.:1.0000
## Max. :53.000 Max. :8296 Max. :1.0000
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The three obtained clusters seem to be representing customer groups with different activity rates.
The third cluster have extremely high activity, when the averages of activity in the last year, bonus usage, and traveled miles of these customers are considered. Most of the features have higher average than the third quartile for this group. It can also be said that they are relatively new customers.
The first cluster consists of regular customers who travel occasionally. The second cluster have higher bonus rates and traveled distances in the last year, along with longer relationships with the company.
Based on these insights, the customers in the first, second, third clusters can be labelled as usual customers, loyal customers, and super customers, respectively.
To check the stability of the clusters, a random 5% of the data (200 observations) can be removed, and the analysis can be repeated with the modified data set.
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| # Dropping 200 random rows from the data set
set.seed(seed)
drop <- as.numeric(sample.int(3999, 200, replace = FALSE))
sliced.data <- data.scaled[-drop, ]
# Repeating the analysis
dm_b <- dist(sliced.data,
method = "euclidean")
hc.complete.b <- hclust(dm_b,
method = "complete")
sil.widths.b <- c()
for (i in 2:10) {
clust <- cutree(hc.complete.b, k = i)
sil <- silhouette(clust, dm_b)
sil.widths.b[i-1] <- mean(sil[, c("sil_width")])
}
results_b <- data.frame(k = 2:10, sil.widths.b)
results_b
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| ## k sil.widths.b
## 1 2 0.4702675
## 2 3 0.5338807
## 3 4 0.4796374
## 4 5 0.3835390
## 5 6 0.3782369
## 6 7 0.3918081
## 7 8 0.3687406
## 8 9 0.3698451
## 9 10 0.3717370
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| # Checking the new clusters
hc.best.b <- cutree(hc.complete.b,
k = results_b[which.max(results_b$sil.widths.b), 1])
table(hc.best.b)
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| ## hc.best.b
## 1 2 3
## 2387 1408 4
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Excluding 200 random observations did not cause dramatic changes in the clustering results. Considering the general non-robustness problem with clustering methods, it can be concluded that the current model is not fragile against the changes in the data set and gives useful insights.
K-Means Clustering
Using K-Means algorithm with different number of clusters and determining the best number of clusters using the silhouette index:
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| # Applying k-means clustering
dm_c <- dist(data.scaled,
method = "euclidean")
sil.widths.c <- c()
for (i in 2:10) {
clust <- kmeans(data.scaled,
centers = i,
nstart = 50)
sil <- silhouette(clust$cl, dm_c)
sil.widths.c[i-1] <- mean(sil[, c("sil_width")])
}
results_c <- data.frame(k = 2:10, sil.widths.c)
results_c
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| ## k sil.widths.c
## 1 2 0.5359305
## 2 3 0.4648453
## 3 4 0.4748377
## 4 5 0.4430672
## 5 6 0.4037918
## 6 7 0.3909657
## 7 8 0.3803502
## 8 9 0.3416435
## 9 10 0.3452961
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| # Checking the clusters obtained by using the best k value
km.best <- kmeans(data.scaled,
centers = results_c[which.max(results_c$sil.widths.c), 1],
nstart = 50)
table(km.best$cl)
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| ##
## 1 2
## 1481 2518
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Setting K = 2 yields the highest silhouette width, which is 0.54, implying that 2 is the appropriate number of clusters to be used. It seems like K-Means clustering algorithm suggests that super customer is not an important insight.
Since major part of the customers fall into the first cluster in hierarchically clustered data, setting these customers as target audience is reasonable. Increasing the number of reward point campaigns might potentially convert these usual customers to loyal customers. Special campaigns and discounts for the loyal customers would also be useful to keep these customers loyal, or event convert them to super customers.
For the clusters obtained by K-Means algorithm, similar strategy can be used by picking the second cluster as the main target audience due to the higher number of observations fall into this set. The absence of super customers is not a decision criteria since their amount is negligible.
IE 425 - Data Mining
Boğaziçi University - Industrial Engineering Department
GitHub Repository
