In this document, we are going to explore the K-means Clustering algorithm in R. We will be working with the iris dataset.

K-means Clustering is used with unlabeled data, but in this case, we have a labeled dataset so we have to use the iris data without the Species column. In this way, algorithm will cluster the data and we will be able to compare the predicted results with the original results, getting the accuracy of the model.

library(ggplot2)

## Registered S3 methods overwritten by 'ggplot2':
##   method         from 
##   [.quosures     rlang
##   c.quosures     rlang
##   print.quosures rlang

df <- iris

head(iris)

##   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1          5.1         3.5          1.4         0.2  setosa
## 2          4.9         3.0          1.4         0.2  setosa
## 3          4.7         3.2          1.3         0.2  setosa
## 4          4.6         3.1          1.5         0.2  setosa
## 5          5.0         3.6          1.4         0.2  setosa
## 6          5.4         3.9          1.7         0.4  setosa

Let’s make a scatterplot.

ggplot(df, aes(Petal.Length, Petal.Width)) + geom_point(aes(col=Species), size=4)

As we can see, setosa is going to be clustered easier. Meanwhile, there is noise between versicolor and virginica even when they look like perfectly clustered.

Let’s run the model. kmeans is installed in the base package from R, so we don’t have to install any package.

In the kmeans function, it is necessary to set center, which is the number of groups we want to cluster to. In this case, we know this value will be 3. Let’s set that.

, but let’s see how we would build the model if we didn’t know it.

set.seed(101)
irisCluster <- kmeans(df[,1:4], center=3, nstart=20)
irisCluster

## K-means clustering with 3 clusters of sizes 38, 62, 50
## 
## Cluster means:
##   Sepal.Length Sepal.Width Petal.Length Petal.Width
## 1     6.850000    3.073684     5.742105    2.071053
## 2     5.901613    2.748387     4.393548    1.433871
## 3     5.006000    3.428000     1.462000    0.246000
## 
## Clustering vector:
##   [1] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
##  [36] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
##  [71] 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 1
## [106] 1 2 1 1 1 1 1 1 2 2 1 1 1 1 2 1 2 1 2 1 1 2 2 1 1 1 1 1 2 1 1 1 1 2 1
## [141] 1 1 2 1 1 1 2 1 1 2
## 
## Within cluster sum of squares by cluster:
## [1] 23.87947 39.82097 15.15100
##  (between_SS / total_SS =  88.4 %)
## 
## Available components:
## 
## [1] "cluster"      "centers"      "totss"        "withinss"    
## [5] "tot.withinss" "betweenss"    "size"         "iter"        
## [9] "ifault"

We can compare the predicted clusters with the original data.

table(irisCluster$cluster, df$Species)

##    
##     setosa versicolor virginica
##   1      0          2        36
##   2      0         48        14
##   3     50          0         0

We can plot out these clusters.

library(cluster)
clusplot(iris, irisCluster$cluster, color=T, shade=T, labels=0, lines=0)

We can see the setosa cluster perfectly explained, meanwhile virginica and versicolor have a little noise between their clusters.

As I said before, we will not always have the labeled data. If we would want to know the exactly number of centers, we should have built the elbow method.

tot.withinss <- vector(mode="character", length=10)
for (i in 1:10){
  irisCluster <- kmeans(df[,1:4], center=i, nstart=20)
  tot.withinss[i] <- irisCluster$tot.withinss
}

Let’s visualize it.

plot(1:10, tot.withinss, type="b", pch=19)

As we saw, the optimal number of clusters is 3.