The k-means clustering algorithms aim at partitioning n observations into a fixed number of k clusters. The algorithm will find homogeneous clusters. It works by dividing a large set of points (vectors) into groups having approximately the same number of points closest to them. Each group is represented by its centroid point. This type of clustering is an unsupervised learning technique. Clustering is mainly used for exploratory data mining.

Convergence of k-means (https://en.wikipedia.org/wiki/K-means_clustering)

We can perform k-means clustering on a data matrix in R using the function **“kmeans()”**.

### Usage

kmeans(x, centers, iter.max = 10, nstart = 1, algorithm = c(“Hartigan-Wong”, “Lloyd”, “Forgy”, “MacQueen”), trace=FALSE)

# S3 method for kmeans

fitted(object, method = c(“centers”, “classes”), …)

### Dataset from clustring

#We can genrate randome data: x <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2), matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2)) colnames(x) <- c("x", "y") summary(x)

## x y ## Min. :-0.51232 Min. :-0.4576 ## 1st Qu.: 0.01864 1st Qu.: 0.0838 ## Median : 0.53680 Median : 0.4405 ## Mean : 0.51812 Mean : 0.5265 ## 3rd Qu.: 1.03567 3rd Qu.: 1.0227 ## Max. : 1.54161 Max. : 1.8433

#We also can use famous <i>Iris</i> flower data set. summary(iris)

## Sepal.Length Sepal.Width Petal.Length Petal.Width ## Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100 ## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300 ## Median :5.800 Median :3.000 Median :4.350 Median :1.300 ## Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199 ## 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800 ## Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500 ## Species ## setosa :50 ## versicolor:50 ## virginica :50 ## ## ##

### Let’s starts with single variable (“single column at once”) clustering.

**Sepal Length**

i <- grep("Sepal.Length", names(iris)) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Sepal.Length")

**Sepal Width**

i <- grep("Sepal.Width", names(iris)) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Sepal.Width")

**Petal Length**

i <- grep("Petal.Length", names(iris)) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Petal.Length")

**Petal Width**

i <- grep("Petal.Width", names(iris)) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Petal.Width")

Up to this point, it is clear that Petal feature is quite distinct comparing other plots. Now, Let’s try to cluster Sepal and Petal features combined.

**Sepal**

i <- grep("Sepal", names(iris)) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Sepal")

**Petal**

i <- grep("Petal", names(iris)) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Petal")

Now, let’s do cluster using all four feature together.

**Sepal and Petal**

i <- c(1,2,3,4) x <- iris[, i] cl <- kmeans(x, 3, nstart = 100) plot(x, col = cl$cluster, main="Sepal and Petal")

Let’s check out the centers and size of each cluster.

cl$centers

## Sepal.Length Sepal.Width Petal.Length Petal.Width ## 1 5.006000 3.428000 1.462000 0.246000 ## 2 5.901613 2.748387 4.393548 1.433871 ## 3 6.850000 3.073684 5.742105 2.071053

cl$size

## [1] 50 62 38

### Evaluation of the model.

Finally, summarize our model.

print(cl)

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

A (*between_SS / total_SS* = **88.4 %**) indicates good fit .

**Note: **This is a guest post, and opinion in this article is of the guest writer. If you have any issues with any of the articles posted at www.marktechpost.com please contact at asif@marktechpost.com

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Very nice thanks for post

Hi Nilesh, Nice work with the computer generated Bible. I had a similar thought with another piece…

Very nice

Thank you

Good work