AS6-2 航空公司的市場區隔

主要議題：依顧客屬性做市場區隔

學習重點：

• 利用集群分析做市場區隔
• 資料常態化
• 資料視覺化
• 族群特性與行銷策略
• 行銷工具vs行銷對象

rm(list=ls(all=T))
Sys.setlocale("LC_ALL","C")
options(digits=4, scipen=12)
library(dplyr)
library(caret)

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1. 資料常態化

1.1 資料摘要

Read the dataset AirlinesCluster.csv into R and call it “airlines”. Looking at the summary of airlines,

A = read.csv('data/AirlinesCluster.csv')
summary(A)

    Balance          QualMiles       BonusMiles       BonusTrans    FlightMiles   
 Min.   :      0   Min.   :    0   Min.   :     0   Min.   : 0.0   Min.   :    0  
 1st Qu.:  18528   1st Qu.:    0   1st Qu.:  1250   1st Qu.: 3.0   1st Qu.:    0  
 Median :  43097   Median :    0   Median :  7171   Median :12.0   Median :    0  
 Mean   :  73601   Mean   :  144   Mean   : 17145   Mean   :11.6   Mean   :  460  
 3rd Qu.:  92404   3rd Qu.:    0   3rd Qu.: 23800   3rd Qu.:17.0   3rd Qu.:  311  
 Max.   :1704838   Max.   :11148   Max.   :263685   Max.   :86.0   Max.   :30817  
  FlightTrans    DaysSinceEnroll
 Min.   : 0.00   Min.   :   2   
 1st Qu.: 0.00   1st Qu.:2330   
 Median : 0.00   Median :4096   
 Mean   : 1.37   Mean   :4119   
 3rd Qu.: 1.00   3rd Qu.:5790   
 Max.   :53.00   Max.   :8296   

colMeans(A) %>% sort

    FlightTrans      BonusTrans       QualMiles     FlightMiles DaysSinceEnroll 
          1.374          11.602         144.115         460.056        4118.559 
     BonusMiles         Balance 
      17144.846       73601.328 

which TWO variables have (on average) the smallest values?

Which TWO variables have (on average) the largest values?

1.2 為甚麼要做資料常態化

In this problem, we will normalize our data before we run the clustering algorithms.

Why is it important to normalize the data before clustering?

1.3 使用caret套件做資料常態化

Let’s go ahead and normalize our data. You can normalize the variables in a data frame by using the preProcess function in the “caret” package. You should already have this package installed from Week 4, but if not, go ahead and install it with install.packages(“caret”). Then load the package with library(caret).

Now, create a normalized data frame called “airlinesNorm” by running the following commands:

preproc = preProcess(airlines)

airlinesNorm = predict(preproc, airlines)

The first command pre-processes the data, and the second command performs the normalization. If you look at the summary of airlinesNorm, you should see that all of the variables now have mean zero. You can also see that each of the variables has standard deviation 1 by using the sd() function.

library(caret)
preproc = preProcess(A)
AN = predict(preproc, A)  
summary(AN)

    Balance         QualMiles        BonusMiles       BonusTrans    
 Min.   :-0.730   Min.   :-0.186   Min.   :-0.710   Min.   :-1.208  
 1st Qu.:-0.546   1st Qu.:-0.186   1st Qu.:-0.658   1st Qu.:-0.896  
 Median :-0.303   Median :-0.186   Median :-0.413   Median : 0.041  
 Mean   : 0.000   Mean   : 0.000   Mean   : 0.000   Mean   : 0.000  
 3rd Qu.: 0.187   3rd Qu.:-0.186   3rd Qu.: 0.276   3rd Qu.: 0.562  
 Max.   :16.187   Max.   :14.223   Max.   :10.208   Max.   : 7.747  
  FlightMiles      FlightTrans     DaysSinceEnroll  
 Min.   :-0.329   Min.   :-0.362   Min.   :-1.9934  
 1st Qu.:-0.329   1st Qu.:-0.362   1st Qu.:-0.8661  
 Median :-0.329   Median :-0.362   Median :-0.0109  
 Mean   : 0.000   Mean   : 0.000   Mean   : 0.0000  
 3rd Qu.:-0.106   3rd Qu.:-0.098   3rd Qu.: 0.8096  
 Max.   :21.680   Max.   :13.610   Max.   : 2.0228  

apply(AN, 2, mean) %>% round(3)

        Balance       QualMiles      BonusMiles      BonusTrans     FlightMiles 
              0               0               0               0               0 
    FlightTrans DaysSinceEnroll 
              0               0 

apply(AN, 2, sd) %>% round(3)

        Balance       QualMiles      BonusMiles      BonusTrans     FlightMiles 
              1               1               1               1               1 
    FlightTrans DaysSinceEnroll 
              1               1 

apply(AN, 2, max) %>% sort

DaysSinceEnroll      BonusTrans      BonusMiles     FlightTrans       QualMiles 
          2.023           7.747          10.208          13.610          14.223 
        Balance     FlightMiles 
         16.187          21.680 

In the normalized data, which variable has the largest maximum value?

apply(AN, 2, min) %>% sort

DaysSinceEnroll      BonusTrans         Balance      BonusMiles     FlightTrans 
        -1.9934         -1.2081         -0.7303         -0.7099         -0.3621 
    FlightMiles       QualMiles 
        -0.3286         -0.1863 

In the normalized data, which variable has the smallest minimum value?

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2. 層級式集群分析

2.1 依據樹狀圖和應用需求決定群數

Compute the distances between data points (using euclidean distance) and then run the Hierarchical clustering algorithm (using method=“ward.D”) on the normalized data. It may take a few minutes for the commands to finish since the dataset has a large number of observations for hierarchical clustering.

Then, plot the dendrogram of the hierarchical clustering process. Suppose the airline is looking for somewhere between 2 and 10 clusters.

d = dist(AN,method="euclidean")
hc = hclust(d, method='ward.D')
plot(hc)

According to the dendrogram, which of the following is NOT a good choice for the number of clusters?

2.2 分割群組

Suppose that after looking at the dendrogram and discussing with the marketing department, the airline decides to proceed with 5 clusters. Divide the data points into 5 clusters by using the cutree function.

kg = cutree(hc, k=5)
table(kg)

kg
   1    2    3    4    5 
 776  519  494  868 1342 

How many data points are in Cluster 1?

2.3 從區隔變數的平均值推論族群特性

Now, use tapply to compare the average values in each of the variables for the 5 clusters (the centroids of the clusters). You may want to compute the average values of the unnormalized data so that it is easier to interpret. You can do this for the variable “Balance” with the following command:

tapply(airlines$Balance, clusterGroups, mean)

sapply(split(A,kg), colMeans) %>% round(2) 

                       1         2         3        4        5
Balance         57866.90 110669.27 198191.57 52335.91 36255.91
QualMiles           0.64   1065.98     30.35     4.85     2.51
BonusMiles      10360.12  22881.76  55795.86 20788.77  2264.79
BonusTrans         10.82     18.23     19.66    17.09     2.97
FlightMiles        83.18   2613.42    327.68   111.57   119.32
FlightTrans         0.30      7.40      1.07     0.34     0.44
DaysSinceEnroll  6235.36   4402.41   5615.71  2840.82  3060.08

Compared to the other clusters, Cluster 1 has the largest average values in which variables (if any)? Select all that apply.

How would you describe the customers in Cluster 1?

2.4 Cluster 2

par(cex=0.8)
split(AN,kg) %>% sapply(colMeans) %>% round(2)

                    1    2     3     4     5
Balance         -0.16 0.37  1.24 -0.21 -0.37
QualMiles       -0.19 1.19 -0.15 -0.18 -0.18
BonusMiles      -0.28 0.24  1.60  0.15 -0.62
BonusTrans      -0.08 0.69  0.84  0.57 -0.90
FlightMiles     -0.27 1.54 -0.09 -0.25 -0.24
FlightTrans     -0.28 1.59 -0.08 -0.27 -0.25
DaysSinceEnroll  1.03 0.14  0.72 -0.62 -0.51

par(cex=0.8)
split(AN,kg) %>% sapply(colMeans) %>% barplot(beside=T,col=rainbow(7))
legend('topright',legend=colnames(A),fill=rainbow(7))

Compared to the other clusters, Cluster 2 has the largest average values in which variables (if any)? Select all that apply.

How would you describe the customers in Cluster 2?

2.5 Cluster 3

Compared to the other clusters, Cluster 3 has the largest average values in which variables (if any)? Select all that apply.

How would you describe the customers in Cluster 3?

2.6 Cluster 4

Compared to the other clusters, Cluster 4 has the largest average values in which variables (if any)? Select all that apply.

How would you describe the customers in Cluster 4?

2.7 Cluster 5

Compared to the other clusters, Cluster 5 has the largest average values in which variables (if any)? Select all that apply.

How would you describe the customers in Cluster 5?

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3. K-Means集群分析

3.1 K-Means集群分析

Now run the k-means clustering algorithm on the normalized data, again creating 5 clusters. Set the seed to 88 right before running the clustering algorithm, and set the argument iter.max to 1000.

set.seed(88)
km = kmeans(AN, 5, iter.max = 1000)
kg2 = km$cluster
table(kg2)

kg2
   1    2    3    4    5 
 408  141  993 1182 1275 

How many clusters have more than 1,000 observations?

par(cex=0.8)
km$centers %>% round(2) %>% t %>% barplot(beside=T,col=rainbow(7))
legend('topright',legend=colnames(A),fill=rainbow(7))

3.2 Hierarchical和K-Means集群的對應關係

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.)

table(Hierarchical=kg, KMeans=kg2)

            KMeans
Hierarchical    1    2    3    4    5
           1    4    0   98  673    1
           2   92  137  105   92   93
           3  300    4  132   58    0
           4   12    0  653   30  173
           5    0    0    5  329 1008

Do you expect Cluster 1 of the K-Means clustering output to necessarily be similar to Cluster 1 of the Hierarchical clustering output?

【討論問題】

請你們為這五個族群各起一個名稱

請你們為這五個族群各設計一個行銷策略

統計上最好的分群也是實務上最好的分群嗎？

除了考慮群間和群間距離之外，實務上的分群通常還需要考慮那些因數？

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