Market segmentation
Source: Analytics Edge Unit 6 Clustering Homework
Techniques: Normalization, colMeans
Market segmentation is a strategy that divides a broad target market of customers into smaller, more similar groups, and then designs a marketing strategy specifically for each group. Clustering is a common technique for market segmentation since it automatically finds similar groups given a data set.
In this problem, we’ll see how clustering can be used to find similar groups of customers who belong to an airline’s frequent flyer program. The airline is trying to learn more about its customers so that it can target different customer segments with different types of mileage offers.
The file AirlinesCluster.csv contains information on 3,999 members of the frequent flyer program. This data comes from the textbook “Data Mining for Business Intelligence,” by Galit Shmueli, Nitin R. Patel, and Peter C. Bruce. For more information, see the website for the book.
There are seven different variables in the dataset, described below:
Balance = number of miles eligible for award travel
QualMiles = number of miles qualifying for TopFlight status
BonusMiles = number of miles earned from non-flight bonus transactions in the past 12 months
BonusTrans = number of non-flight bonus transactions in the past 12 months
FlightMiles = number of flight miles in the past 12 months
FlightTrans = number of flight transactions in the past 12 months
DaysSinceEnroll = number of days since enrolled in the frequent flyer program
Load the data
setwd("C:/Users/jzchen/Documents/Courses/Analytics Edge/Unit_6_Clustering")
airlines <- read.csv("AirlinesCluster.csv")
str(airlines)## 'data.frame': 3999 obs. of 7 variables:
## $ Balance : int 28143 19244 41354 14776 97752 16420 84914 20856 443003 104860 ...
## $ QualMiles : int 0 0 0 0 0 0 0 0 0 0 ...
## $ BonusMiles : int 174 215 4123 500 43300 0 27482 5250 1753 28426 ...
## $ BonusTrans : int 1 2 4 1 26 0 25 4 43 28 ...
## $ FlightMiles : int 0 0 0 0 2077 0 0 250 3850 1150 ...
## $ FlightTrans : int 0 0 0 0 4 0 0 1 12 3 ...
## $ DaysSinceEnroll: int 7000 6968 7034 6952 6935 6942 6994 6938 6948 6931 ...NORMALIZING THE DATA
We can normalize the variables in a data frame by using the preProcess function in the “caret” package.
library(caret)## Loading required package: lattice
## Loading required package: ggplot2
##
## Attaching package: 'ggplot2'
##
## The following object is masked _by_ '.GlobalEnv':
##
## moviespreproc <- preProcess(airlines)
airlinesNorm <- predict(preproc, airlines)
summary(airlinesNorm)## Balance QualMiles BonusMiles BonusTrans
## Min. :-0.7303 Min. :-0.1863 Min. :-0.7099 Min. :-1.20805
## 1st Qu.:-0.5465 1st Qu.:-0.1863 1st Qu.:-0.6581 1st Qu.:-0.89568
## Median :-0.3027 Median :-0.1863 Median :-0.4130 Median : 0.04145
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.00000
## 3rd Qu.: 0.1866 3rd Qu.:-0.1863 3rd Qu.: 0.2756 3rd Qu.: 0.56208
## Max. :16.1868 Max. :14.2231 Max. :10.2083 Max. : 7.74673
## FlightMiles FlightTrans DaysSinceEnroll
## Min. :-0.3286 Min. :-0.36212 Min. :-1.99336
## 1st Qu.:-0.3286 1st Qu.:-0.36212 1st Qu.:-0.86607
## Median :-0.3286 Median :-0.36212 Median :-0.01092
## Mean : 0.0000 Mean : 0.00000 Mean : 0.00000
## 3rd Qu.:-0.1065 3rd Qu.:-0.09849 3rd Qu.: 0.80960
## Max. :21.6803 Max. :13.61035 Max. : 2.02284lapply(airlinesNorm, sd)## $Balance
## [1] 1
##
## $QualMiles
## [1] 1
##
## $BonusMiles
## [1] 1
##
## $BonusTrans
## [1] 1
##
## $FlightMiles
## [1] 1
##
## $FlightTrans
## [1] 1
##
## $DaysSinceEnroll
## [1] 1HIERARCHICAL CLUSTERING
distance <- dist(airlinesNorm, method = "euclidean")
airlineHClust <- hclust(distance, method = "ward.D")
plot(airlineHClust)
Suppose that after looking at the dendrogram and discussing with the marketing department, the airline decides to proceed with 5 clusters.
airlineClusters <- cutree(airlineHClust, k = 5)
HClust1 <- subset(airlinesNorm, airlineClusters == 1 )
HClust2 <- subset(airlinesNorm, airlineClusters == 2 )
HClust3 <- subset(airlinesNorm, airlineClusters == 3 )
HClust4 <- subset(airlinesNorm, airlineClusters == 4 )
HClust5 <- subset(airlinesNorm, airlineClusters == 5 )Compare the average values in each of the variables for the 5 clusters. We may want to compute the average values of the unnormalized data so that it is easier to interpret.
tapply(airlines$Balance, airlineClusters, mean)## 1 2 3 4 5
## 57866.90 110669.27 198191.57 52335.91 36255.91tapply(airlines$QualMiles, airlineClusters, mean)## 1 2 3 4 5
## 0.6443299 1065.9826590 30.3461538 4.8479263 2.5111773tapply(airlines$BonusMiles, airlineClusters, mean)## 1 2 3 4 5
## 10360.124 22881.763 55795.860 20788.766 2264.788tapply(airlines$BonusTrans, airlineClusters, mean)## 1 2 3 4 5
## 10.823454 18.229287 19.663968 17.087558 2.973174tapply(airlines$FlightMiles, airlineClusters, mean)## 1 2 3 4 5
## 83.18428 2613.41811 327.67611 111.57373 119.32191tapply(airlines$FlightTrans, airlineClusters, mean)## 1 2 3 4 5
## 0.3028351 7.4026975 1.0688259 0.3444700 0.4388972tapply(airlines$DaysSinceEnroll, airlineClusters, mean)## 1 2 3 4 5
## 6235.365 4402.414 5615.709 2840.823 3060.081Instead of using tapply, We could have alternatively used colMeans and subset, as follows:
colMeans(subset(airlines, clusterGroups == 1))## Balance QualMiles BonusMiles BonusTrans
## 72404.48606 126.39417 17189.49176 11.58048
## FlightMiles FlightTrans DaysSinceEnroll
## 443.78200 1.33270 4142.10456colMeans(subset(airlines, clusterGroups == 2))## Balance QualMiles BonusMiles BonusTrans
## 78082.02966 210.45552 16977.70344 11.68209
## FlightMiles FlightTrans DaysSinceEnroll
## 520.98102 1.52669 4030.41163colMeans(subset(airlines, clusterGroups == 3))## Balance QualMiles BonusMiles BonusTrans
## NaN NaN NaN NaN
## FlightMiles FlightTrans DaysSinceEnroll
## NaN NaN NaNcolMeans(subset(airlines, clusterGroups == 4))## Balance QualMiles BonusMiles BonusTrans
## NaN NaN NaN NaN
## FlightMiles FlightTrans DaysSinceEnroll
## NaN NaN NaNcolMeans(subset(airlines, clusterGroups == 5))## Balance QualMiles BonusMiles BonusTrans
## NaN NaN NaN NaN
## FlightMiles FlightTrans DaysSinceEnroll
## NaN NaN NaNThis only requires 5 lines of code instead of the 7 above. But an even more compact way of finding the centroids would be to use the function “split” to first split the data into clusters, and then to use the function “lapply” to apply the function “colMeans” to each of the clusters:
lapply(split(airlines, clusterGroups), colMeans)
How would you describe the customers in Cluster 1? Infrequent but loyal customers.
How would you describe the customers in Cluster 3? Customers who have accumulated a large amount of miles, mostly through non-flight transactions.
K-means clustering
set.seed(88)
airlinesKMC <- kmeans(airlinesNorm, centers = 5)
str(airlinesKMC)## List of 9
## $ cluster : int [1:3999] 4 4 4 4 1 4 3 4 2 3 ...
## $ centers : num [1:5, 1:7] 1.4444 1.0005 -0.0558 -0.1333 -0.4058 ...
## ..- attr(*, "dimnames")=List of 2
## .. ..$ : chr [1:5] "1" "2" "3" "4" ...
## .. ..$ : chr [1:7] "Balance" "QualMiles" "BonusMiles" "BonusTrans" ...
## $ totss : num 27986
## $ withinss : num [1:5] 4948 3624 2054 2040 2321
## $ tot.withinss: num 14987
## $ betweenss : num 12999
## $ size : int [1:5] 408 141 993 1182 1275
## $ iter : int 4
## $ ifault : int 0
## - attr(*, "class")= chr "kmeans"Subset the data to 5 clusters
airlinesKcluster1 <- subset(airlinesNorm, airlinesKMC$cluster == 1)
airlinesKcluster2 <- subset(airlinesNorm, airlinesKMC$cluster == 2)
airlinesKcluster3 <- subset(airlinesNorm, airlinesKMC$cluster == 3)
airlinesKcluster4 <- subset(airlinesNorm, airlinesKMC$cluster == 4)
airlinesKcluster5 <- subset(airlinesNorm, airlinesKMC$cluster == 5)Now, compare the cluster centroids to each other either by dividing the data points into groups and then using tapply, or by looking at the output of kmeansClust\(centers, where "kmeansClust" is the name of the output of the kmeans function. (Note that the output of kmeansClust\)centers will be for the normalized data. If you want to look at the average values for the unnormalized data, you need to use tapply like we did for hierarchical clustering.)
Do you expect Cluster 1 of the K-Means clustering output to necessarily be similar to Cluster 1 of the Hierarchical clustering output?
tapply(airlines$Balance, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 219161.40 174431.51 67977.44 60166.18 32706.67tapply(airlines$QualMiles, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 539.57843 673.16312 34.99396 55.20812 126.46667tapply(airlines$BonusMiles, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 62474.483 31985.085 24490.019 8709.712 3097.478tapply(airlines$BonusTrans, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 21.524510 28.134752 18.429003 8.362098 4.284706tapply(airlines$FlightMiles, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 623.8725 5859.2340 289.4713 203.2589 181.4698tapply(airlines$FlightTrans, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 1.9215686 17.0000000 0.8851964 0.6294416 0.5403922tapply(airlines$DaysSinceEnroll, airlinesKMC$cluster, mean)## 1 2 3 4 5
## 5605.051 4684.901 3416.783 6109.540 2281.055The clusters are not displayed in a meaningful order, so while there may be a cluster produced by the k-means algorithm that is similar to Cluster 1 produced by the Hierarchical method, it will not necessarily be shown first.