Cluster Analysis with R
Gabriel Martos
Clustering wines
K-Means
This first example is to learn to make cluster analysis with R. The library rattle is loaded in order to use the data set wines.
# install.packages('rattle')
data(wine, package='rattle')
head(wine)## Type Alcohol Malic Ash Alcalinity Magnesium Phenols Flavanoids
## 1 1 14.23 1.71 2.43 15.6 127 2.80 3.06
## 2 1 13.20 1.78 2.14 11.2 100 2.65 2.76
## 3 1 13.16 2.36 2.67 18.6 101 2.80 3.24
## 4 1 14.37 1.95 2.50 16.8 113 3.85 3.49
## 5 1 13.24 2.59 2.87 21.0 118 2.80 2.69
## 6 1 14.20 1.76 2.45 15.2 112 3.27 3.39
## Nonflavanoids Proanthocyanins Color Hue Dilution Proline
## 1 0.28 2.29 5.64 1.04 3.92 1065
## 2 0.26 1.28 4.38 1.05 3.40 1050
## 3 0.30 2.81 5.68 1.03 3.17 1185
## 4 0.24 2.18 7.80 0.86 3.45 1480
## 5 0.39 1.82 4.32 1.04 2.93 735
## 6 0.34 1.97 6.75 1.05 2.85 1450In this data set we observe the composition of different wines. Given a set of observations \((x_1, x_2, ., x_n)\), where each observation is a \(d\)-dimensional real vector, \(k\)-means clustering aims to partition the n observations into (\(k \leq n\)) \(S = \{S_1, S_2, ., S_k\}\) so as to minimize the within-cluster sum of squares (WCSS). In other words, its objective is to find::
\[ \underset{\mathbf{S}} {\operatorname{arg\,min}} \sum_{i=1}^{k} \sum_{\mathbf x_j \in S_i} \left\| \mathbf x_j - \boldsymbol\mu_i \right\|^2 \]
where \(\mu_i\) is the mean of points in \(S_i\). The clustering optimization problem is solved with the function kmeans in R.
wine.stand <- scale(wine[-1]) # To standarize the variables
# K-Means
k.means.fit <- kmeans(wine.stand, 3) # k = 3In k.means.fit are contained all the elements of the cluster output:
attributes(k.means.fit)## $names
## [1] "cluster" "centers" "totss" "withinss"
## [5] "tot.withinss" "betweenss" "size" "iter"
## [9] "ifault"
##
## $class
## [1] "kmeans"# Centroids:
k.means.fit$centers## Alcohol Malic Ash Alcalinity Magnesium Phenols Flavanoids
## 1 0.1644 0.8691 0.1864 0.5229 -0.07526 -0.97658 -1.21183
## 2 0.8329 -0.3030 0.3637 -0.6085 0.57596 0.88275 0.97507
## 3 -0.9235 -0.3929 -0.4931 0.1701 -0.49033 -0.07577 0.02075
## Nonflavanoids Proanthocyanins Color Hue Dilution Proline
## 1 0.72402 -0.7775 0.9389 -1.1615 -1.2888 -0.4059
## 2 -0.56051 0.5787 0.1706 0.4727 0.7771 1.1220
## 3 -0.03344 0.0581 -0.8994 0.4605 0.2700 -0.7517# Clusters:
k.means.fit$cluster## [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## [36] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 1 3 3 3 3 3 3 3 3
## [71] 3 3 3 2 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3
## [106] 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 2 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1
## [141] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [176] 1 1 1# Cluster size:
k.means.fit$size## [1] 51 62 65A fundamental question is how to determine the value of the parameter \(k\). If we looks at the percentage of variance explained as a function of the number of clusters: One should choose a number of clusters so that adding another cluster doesn’t give much better modeling of the data. More precisely, if one plots the percentage of variance explained by the clusters against the number of clusters, the first clusters will add much information (explain a lot of variance), but at some point the marginal gain will drop, giving an angle in the graph. The number of clusters is chosen at this point, hence the “elbow criterion”.
wssplot <- function(data, nc=15, seed=1234){
wss <- (nrow(data)-1)*sum(apply(data,2,var))
for (i in 2:nc){
set.seed(seed)
wss[i] <- sum(kmeans(data, centers=i)$withinss)}
plot(1:nc, wss, type="b", xlab="Number of Clusters",
ylab="Within groups sum of squares")}
wssplot(wine.stand, nc=6) 
Which is the optimal value for k in this case? Why?
Library clusters allow us to represent (with the aid of PCA) the cluster solution into 2 dimensions:
library(cluster)
clusplot(wine.stand, k.means.fit$cluster, main='2D representation of the Cluster solution',
color=TRUE, shade=TRUE,
labels=2, lines=0)
In order to evaluate the clustering performance we build a confusion matrix:
table(wine[,1],k.means.fit$cluster)##
## 1 2 3
## 1 0 59 0
## 2 3 3 65
## 3 48 0 0Hierarchical clustering:
Hierarchical methods use a distance matrix as an input for the clustering algorithm. The choice of an appropriate metric will influence the shape of the clusters, as some elements may be close to one another according to one distance and farther away according to another.
d <- dist(wine.stand, method = "euclidean") # Euclidean distance matrix.We use the Euclidean distance as an input for the clustering algorithm (Ward’s minimum variance criterion minimizes the total within-cluster variance):
H.fit <- hclust(d, method="ward")## The "ward" method has been renamed to "ward.D"; note new "ward.D2"The clustering output can be displayed in a dendrogram
plot(H.fit) # display dendogram
groups <- cutree(H.fit, k=3) # cut tree into 5 clusters
# draw dendogram with red borders around the 5 clusters
rect.hclust(H.fit, k=3, border="red") 
The clustering performance can be evaluated with the aid of a confusion matrix as follows:
table(wine[,1],groups)## groups
## 1 2 3
## 1 58 1 0
## 2 7 58 6
## 3 0 0 48Study case I: EUROPEAN PROTEIN CONSUMPTION
We consider 25 European countries (n = 25 units) and their protein intakes (in percent) from nine major food sources (p = 9). The data are listed below.
url = 'http://www.biz.uiowa.edu/faculty/jledolter/DataMining/protein.csv'
food <- read.csv(url)
head(food)## Country RedMeat WhiteMeat Eggs Milk Fish Cereals Starch Nuts
## 1 Albania 10.1 1.4 0.5 8.9 0.2 42.3 0.6 5.5
## 2 Austria 8.9 14.0 4.3 19.9 2.1 28.0 3.6 1.3
## 3 Belgium 13.5 9.3 4.1 17.5 4.5 26.6 5.7 2.1
## 4 Bulgaria 7.8 6.0 1.6 8.3 1.2 56.7 1.1 3.7
## 5 Czechoslovakia 9.7 11.4 2.8 12.5 2.0 34.3 5.0 1.1
## 6 Denmark 10.6 10.8 3.7 25.0 9.9 21.9 4.8 0.7
## Fr.Veg
## 1 1.7
## 2 4.3
## 3 4.0
## 4 4.2
## 5 4.0
## 6 2.4We start first, clustering on just Red and White meat (p=2) and k=3 clusters.
set.seed(123456789) ## to fix the random starting clusters
grpMeat <- kmeans(food[,c("WhiteMeat","RedMeat")], centers=3, nstart=10)
grpMeat## K-means clustering with 3 clusters of sizes 8, 12, 5
##
## Cluster means:
## WhiteMeat RedMeat
## 1 12.062 8.838
## 2 4.658 8.258
## 3 9.000 15.180
##
## Clustering vector:
## [1] 2 1 3 2 1 1 1 2 3 2 1 3 2 1 2 1 2 2 2 2 3 3 2 1 2
##
## Within cluster sum of squares by cluster:
## [1] 39.46 69.86 35.67
## (between_SS / total_SS = 75.7 %)
##
## Available components:
##
## [1] "cluster" "centers" "totss" "withinss"
## [5] "tot.withinss" "betweenss" "size" "iter"
## [9] "ifault"## list of cluster assignments
o=order(grpMeat$cluster)
data.frame(food$Country[o],grpMeat$cluster[o])## food.Country.o. grpMeat.cluster.o.
## 1 Austria 1
## 2 Czechoslovakia 1
## 3 Denmark 1
## 4 E Germany 1
## 5 Hungary 1
## 6 Netherlands 1
## 7 Poland 1
## 8 W Germany 1
## 9 Albania 2
## 10 Bulgaria 2
## 11 Finland 2
## 12 Greece 2
## 13 Italy 2
## 14 Norway 2
## 15 Portugal 2
## 16 Romania 2
## 17 Spain 2
## 18 Sweden 2
## 19 USSR 2
## 20 Yugoslavia 2
## 21 Belgium 3
## 22 France 3
## 23 Ireland 3
## 24 Switzerland 3
## 25 UK 3To see a graphical representation of the clustering solution we plot cluster assignments on Red and White meat on a scatter plot:
plot(food$Red, food$White, type="n", xlim=c(3,19), xlab="Red Meat", ylab="White Meat")
text(x=food$Red, y=food$White, labels=food$Country,col=grpMeat$cluster+1)
Next, we cluster on all nine protein groups and prepare the program to create seven clusters. The resulting clusters, shown in color on a scatter plot of white meat against red meat (any other pair of features could be selected), actually makes lot of sense. Countries in close geographic proximity tend to be clustered into the same group.
## same analysis, but now with clustering on all
## protein groups change the number of clusters to 7
set.seed(123456789)
grpProtein <- kmeans(food[,-1], centers=7, nstart=10)
o=order(grpProtein$cluster)
data.frame(food$Country[o],grpProtein$cluster[o])## food.Country.o. grpProtein.cluster.o.
## 1 Austria 1
## 2 E Germany 1
## 3 Netherlands 1
## 4 W Germany 1
## 5 Portugal 2
## 6 Spain 2
## 7 Albania 3
## 8 Greece 3
## 9 Italy 3
## 10 USSR 3
## 11 Czechoslovakia 4
## 12 Hungary 4
## 13 Poland 4
## 14 Bulgaria 5
## 15 Romania 5
## 16 Yugoslavia 5
## 17 Denmark 6
## 18 Finland 6
## 19 Norway 6
## 20 Sweden 6
## 21 Belgium 7
## 22 France 7
## 23 Ireland 7
## 24 Switzerland 7
## 25 UK 7library(cluster)
clusplot(food[,-1], grpProtein$cluster, main='2D representation of the Cluster solution', color=TRUE, shade=TRUE, labels=2, lines=0)
Alternatively we can implement a Hierarchical approach. We use the agnes function in the package cluster. Argument diss=FALSE indicates that we use the dissimilarity matrix that is being calculated from raw data. Argument metric=“euclidian” indicates that we use Euclidean distance. No standardization is used and the link function is the “average” linkage.
foodagg=agnes(food,diss=FALSE,metric="euclidian")
plot(foodagg, main='Dendrogram') ## dendrogram
groups <- cutree(foodagg, k=4) # cut tree into 3 clusters
rect.hclust(foodagg, k=4, border="red") 
Study case II: Customer Segmentation
Customer segmentation is as simple as it sounds: grouping customers by their characteristics - and why would you want to do that? To better serve their needs!
Our example is to do with e-mail marketing. We use the dataset from this link
offers<-read.table('offers.csv', sep = ';', header=T)
head(offers)## OfferID Campaign Varietal MinimumQt Discount Origin
## 1 1 January Malbec 72 56 France
## 2 2 January Pinot Noir 72 17 France
## 3 3 February Espumante 144 32 Oregon
## 4 4 February Champagne 72 48 France
## 5 5 February Cabernet Sauvignon 144 44 New Zealand
## 6 6 March Prosecco 144 86 Chile
## PastPeak
## 1 FALSE
## 2 FALSE
## 3 TRUE
## 4 TRUE
## 5 TRUE
## 6 FALSEtransactions<-read.table('transactions.csv', sep = ';', header=T)
head(transactions)## CustomerLastName OfferID
## 1 Smith 2
## 2 Smith 24
## 3 Johnson 17
## 4 Johnson 24
## 5 Johnson 26
## 6 Williams 18Step 1: Organizing the information
We have two data sets: one for the offers and the other for the transactions. First what we need to do is create a transaction matrix. That means, we need to put the offers we mailed out next to the transaction history of each customer. This is easily achieved with a pivot table.
# Create transaction matrix (a pivot table like in Excel way!)
library(reshape)
pivot<-melt(transactions[1:2])## Using CustomerLastName as id variablespivot<-(cast(pivot,value~CustomerLastName,fill=0,fun.aggregate=function(x) length(x)))
pivot<-cbind(offers,pivot[-1])
# write.csv(file="pivot.csv",pivot) # to save your data
cluster.data<-pivot[,8:length(pivot)]
cluster.data<-t(cluster.data)
head(cluster.data)## 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
## Adams 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
## Allen 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## Anderson 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
## Bailey 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## Baker 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
## Barnes 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
## 26 27 28 29 30 31 32
## Adams 0 0 0 1 1 0 0
## Allen 0 1 0 0 0 0 0
## Anderson 1 0 0 0 0 0 0
## Bailey 0 0 0 0 1 0 0
## Baker 0 0 0 0 0 1 0
## Barnes 0 0 0 0 0 1 0In the clustering data set, rows represents costumers and columns are different wine brands/types.
Step 2: Distances and Clusters
We will use \(k = 4\) indicating that we will use 4 clusters. This is somewhat arbitrary, but the number you pick should be representative of the number of segments you can handle as a business. So 100 segments does not make sense for an e-mail marketing campaign.
We need to calculate how far away each customer is from the cluster’s mean. To do this we could use many distances/dissimilarity index, one of which is the Gower dissimilarity.
library(cluster)
D=daisy(cluster.data, metric='gower')## Warning: binary variable(s) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
## 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32
## treated as interval scaledAfter the creation of a distance matrix, we implement a Ward’s hierarchical clustering procedure:
H.fit <- hclust(D, method="ward")## The "ward" method has been renamed to "ward.D"; note new "ward.D2"plot(H.fit) # display dendrogram
groups <- cutree(H.fit, k=4) # cut tree into 4 clusters
# draw dendogram with red borders around the 4 clusters
rect.hclust(H.fit, k=4, border="red") 
# 2D representation of the Segmentation:
clusplot(cluster.data, groups, color=TRUE, shade=TRUE,
labels=2, lines=0, main= 'Customer segments')
Top get the top deals we will have to do a little bit of data manipulation. First we need to combine our clusters and transactions. Notably the lengths of the ‘tables’ holding transactions and clusters are different. So we need a way to merge the data . so we use the merge() function and give our columns sensible names:
# Merge Data
cluster.deals<-merge(transactions[1:2],groups,by.x = "CustomerLastName", by.y = "row.names")
colnames(cluster.deals)<-c("Name","Offer","Cluster")
head(cluster.deals)## Name Offer Cluster
## 1 Adams 18 1
## 2 Adams 29 1
## 3 Adams 30 1
## 4 Allen 9 2
## 5 Allen 27 2
## 6 Anderson 24 3We then want to repeat the pivoting process to get Offers in rows and clusters in columns counting the total number of transactions for each cluster. Once we have our pivot table we will merge it with the offers data table like we did before:
# Get top deals by cluster
cluster.pivot<-melt(cluster.deals,id=c("Offer","Cluster"))
cluster.pivot<-cast(cluster.pivot,Offer~Cluster,fun.aggregate=length)
cluster.topDeals<-cbind(offers,cluster.pivot[-1])
head(cluster.topDeals)## OfferID Campaign Varietal MinimumQt Discount Origin
## 1 1 January Malbec 72 56 France
## 2 2 January Pinot Noir 72 17 France
## 3 3 February Espumante 144 32 Oregon
## 4 4 February Champagne 72 48 France
## 5 5 February Cabernet Sauvignon 144 44 New Zealand
## 6 6 March Prosecco 144 86 Chile
## PastPeak 1 2 3 4
## 1 FALSE 0 8 2 0
## 2 FALSE 0 3 7 0
## 3 TRUE 1 2 0 3
## 4 TRUE 0 8 0 4
## 5 TRUE 0 4 0 0
## 6 FALSE 1 5 0 6#### And finally we can export the data in excel format with the command:
#### write.csv(file="topdeals.csv",cluster.topDeals,row.names=F)Study case III: Social Network Clustering Analysis
For this analysis, we will be using a dataset representing a random sample of 30.000 U.S. high school students who had profiles on a well-known Social Network in from 2006 to 2009.
From the top 500 words appearing across all pages, 36 words were chosen to represent five categories of interests, namely extracurricular activities, fashion, religion, romance, and antisocial behavior. The 36 words include terms such as football, sexy, kissed, bible, shopping, death, and drugs. The final dataset indicates, for each person, how many times each word appeared in the person’s SNS profile.
teens <- read.csv("snsdata.csv")
head(teens,3)## gradyear gender age friends basketball football soccer softball
## 1 2006 M 18.98 7 0 0 0 0
## 2 2006 F 18.80 0 0 1 0 0
## 3 2006 M 18.34 69 0 1 0 0
## volleyball swimming cheerleading baseball tennis sports cute sex sexy
## 1 0 0 0 0 0 0 0 0 0
## 2 0 0 0 0 0 0 1 0 0
## 3 0 0 0 0 0 0 0 0 0
## hot kissed dance band marching music rock god church jesus bible hair
## 1 0 0 1 0 0 0 0 0 0 0 0 0
## 2 0 0 0 0 0 2 2 1 0 0 0 6
## 3 0 0 0 2 0 1 0 0 0 0 0 0
## dress blonde mall shopping clothes hollister abercrombie die death drunk
## 1 0 0 0 0 0 0 0 0 0 0
## 2 4 0 1 0 0 0 0 0 0 0
## 3 0 0 0 0 0 0 0 0 1 0
## drugs
## 1 0
## 2 0
## 3 0dim(teens)## [1] 30000 40Let’s also take a quick look at the specifics of the data. The first several lines of the str() output are as follows:
str(teens)## 'data.frame': 30000 obs. of 40 variables:
## $ gradyear : int 2006 2006 2006 2006 2006 2006 2006 2006 2006 2006 ...
## $ gender : Factor w/ 2 levels "F","M": 2 1 2 1 NA 1 1 2 1 1 ...
## $ age : num 19 18.8 18.3 18.9 19 ...
## $ friends : int 7 0 69 0 10 142 72 17 52 39 ...
## $ basketball : int 0 0 0 0 0 0 0 0 0 0 ...
## $ football : int 0 1 1 0 0 0 0 0 0 0 ...
## $ soccer : int 0 0 0 0 0 0 0 0 0 0 ...
## $ softball : int 0 0 0 0 0 0 0 1 0 0 ...
## $ volleyball : int 0 0 0 0 0 0 0 0 0 0 ...
## $ swimming : int 0 0 0 0 0 0 0 0 0 0 ...
## $ cheerleading: int 0 0 0 0 0 0 0 0 0 0 ...
## $ baseball : int 0 0 0 0 0 0 0 0 0 0 ...
## $ tennis : int 0 0 0 0 0 0 0 0 0 0 ...
## $ sports : int 0 0 0 0 0 0 0 0 0 0 ...
## $ cute : int 0 1 0 1 0 0 0 0 0 1 ...
## $ sex : int 0 0 0 0 1 1 0 2 0 0 ...
## $ sexy : int 0 0 0 0 0 0 0 1 0 0 ...
## $ hot : int 0 0 0 0 0 0 0 0 0 1 ...
## $ kissed : int 0 0 0 0 5 0 0 0 0 0 ...
## $ dance : int 1 0 0 0 1 0 0 0 0 0 ...
## $ band : int 0 0 2 0 1 0 1 0 0 0 ...
## $ marching : int 0 0 0 0 0 1 1 0 0 0 ...
## $ music : int 0 2 1 0 3 2 0 1 0 1 ...
## $ rock : int 0 2 0 1 0 0 0 1 0 1 ...
## $ god : int 0 1 0 0 1 0 0 0 0 6 ...
## $ church : int 0 0 0 0 0 0 0 0 0 0 ...
## $ jesus : int 0 0 0 0 0 0 0 0 0 2 ...
## $ bible : int 0 0 0 0 0 0 0 0 0 0 ...
## $ hair : int 0 6 0 0 1 0 0 0 0 1 ...
## $ dress : int 0 4 0 0 0 1 0 0 0 0 ...
## $ blonde : int 0 0 0 0 0 0 0 0 0 0 ...
## $ mall : int 0 1 0 0 0 0 2 0 0 0 ...
## $ shopping : int 0 0 0 0 2 1 0 0 0 1 ...
## $ clothes : int 0 0 0 0 0 0 0 0 0 0 ...
## $ hollister : int 0 0 0 0 0 0 2 0 0 0 ...
## $ abercrombie : int 0 0 0 0 0 0 0 0 0 0 ...
## $ die : int 0 0 0 0 0 0 0 0 0 0 ...
## $ death : int 0 0 1 0 0 0 0 0 0 0 ...
## $ drunk : int 0 0 0 0 1 1 0 0 0 0 ...
## $ drugs : int 0 0 0 0 1 0 0 0 0 0 ...As we had expected, the data include 30,000 teenagers with four variables indicating personal characteristics and 36 words indicating interests. Note that there are some NA’s in the variable gender.
summary(teens$age)## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 3 16 17 18 18 107 5086We will skip all the data with missing values:
teens = na.omit(teens)
dim(teens)## [1] 24005 40We’ll start our cluster analysis by considering only the 36 features that represent the number of times various interests appeared on the SNS profiles of teens. For convenience, let’s make a data frame containing only these features:
interests <- teens[5:40]To apply z-score standardization to the interests data frame, we can use the scale() function with lapply(), as follows:
interests_z <- as.data.frame(lapply(interests, scale))To divide teens into five clusters, we can use the following command:
teen_clusters <- kmeans(interests_z, 5)number of examples falling in each of the groups. If the groups are too large or too small, then they are not likely to be very useful. To obtain the size of the kmeans() clusters, use the teen_clusters$size component as follows:
teen_clusters$size## [1] 403 17255 4783 717 847For a more in-depth look at the clusters, we can examine the coordinates of the cluster centroids using the teen_clusters$centers component, which is as follows for the first eight features:
teen_clusters$centers## basketball football soccer softball volleyball swimming cheerleading
## 1 0.1510 -0.004452 0.01377 -0.03832 0.004122 0.03706 0.001678
## 2 -0.1636 -0.171249 -0.09280 -0.11707 -0.116837 -0.09624 -0.116317
## 3 0.4979 0.521693 0.29185 0.38495 0.378986 0.27075 0.336997
## 4 0.1605 0.249815 0.12107 0.04462 0.200136 0.21498 0.380099
## 5 0.3143 0.333318 0.13348 0.19162 0.068687 0.23196 0.144012
## baseball tennis sports cute sex sexy hot
## 1 0.029672 0.04294 0.01238 0.02468 0.026179 -0.04248 0.07039
## 2 -0.109056 -0.05172 -0.12781 -0.18574 -0.096249 -0.08776 -0.13528
## 3 0.344627 0.14798 0.31350 0.52770 -0.008816 0.21064 0.36860
## 4 0.008058 0.09947 0.08630 0.40237 0.015712 0.13078 0.41142
## 5 0.254633 0.11345 0.75449 0.45156 1.984811 0.50795 0.29264
## kissed dance band marching music rock god church
## 1 -0.02317 -0.01394 0.14685 0.08007 0.2349 0.12825 2.2406 1.24023
## 2 -0.13438 -0.16607 -0.09301 -0.05726 -0.1539 -0.12691 -0.1063 -0.14412
## 3 -0.04044 0.49912 0.25190 0.19532 0.3165 0.23156 0.1348 0.39146
## 4 0.03950 0.20828 -0.10137 -0.09403 0.1072 0.05519 -0.0184 -0.02148
## 5 2.94352 0.39499 0.48827 0.10495 1.1448 1.17016 0.3546 0.15357
## jesus bible hair dress blonde mall shopping
## 1 2.332629 6.048477 0.05815 0.02987 -0.003915 -0.06875 -0.01053
## 2 -0.074454 -0.109423 -0.20514 -0.15210 -0.027665 -0.18923 -0.23212
## 3 0.059948 -0.105090 0.22836 0.43163 0.028245 0.49390 0.68176
## 4 0.006561 -0.073294 0.41466 0.12938 0.058491 0.63797 0.76795
## 5 0.062839 0.006795 2.51095 0.53741 0.356443 0.55865 0.23374
## clothes hollister abercrombie die death drunk drugs
## 1 0.04932 -0.09458 -0.08999 0.22400 0.28806 0.065740 0.08217
## 2 -0.19026 -0.15715 -0.15109 -0.09936 -0.08237 -0.088794 -0.11443
## 3 0.38665 -0.05650 -0.07429 0.02297 0.09233 -0.008421 -0.07876
## 4 0.54243 4.06769 3.90321 0.04736 0.08796 0.037217 0.02999
## 5 1.20992 0.12212 0.23620 1.74789 0.94516 1.793662 2.71150The cluster characterization can be obtained with pie charts:
par(mfrow=c(2,2))
pie(colSums(interests[teen_clusters$cluster==1,]),cex=0.5)
pie(colSums(interests[teen_clusters$cluster==2,]),cex=0.5)
pie(colSums(interests[teen_clusters$cluster==3,]),cex=0.5)
pie(colSums(interests[teen_clusters$cluster==4,]),cex=0.5)
Can you give a name to every cluster? Why this clustering information it could be useful?