CODEWITHSUNDEEP
rcluster-analysisk-means
Chris
Chris

Reputation: 1237

Determining optimum number of clusters for k-means with a large dataset

I have a matrix of 62 columns and 181408 rows that I am going to be clustering using k-means. What I would ideally like is a method of identifying what the optimum number of clusters should be. I have tried implementing the gap statistic technique using clusGap from the cluster package (reproducible code below), but this produces several error messages relating to the size of the vector (122 GB) and memory.limitproblems in Windows and a "Error in dist(xs) : negative length vectors are not allowed" in OS X. Does anyone has any suggestions on techniques that will work in determining optimum number of clusters with a large dataset? Or, alternatively, how to make my code function (and does not take several days to complete)? Thanks. 

library(cluster)
inputdata<-matrix(rexp(11247296, rate=.1), ncol=62)
clustergap <- clusGap(inputdata, FUN=kmeans, K.max=12, B=10)

Upvotes: 2

Views: 4558

Answers (3)

user9562553
user9562553

Reputation:

If you don't know the numbers of the clusters k to provide as parameter to k-means so there are three ways to find it automaticaly:

  • G-means algortithm: it discovers the number of clusters automatically using a statistical test to decide whether to split a k-means center into two. This algorithm takes a hierarchical approach to detect the number of clusters, based on a statistical test for the hypothesis that a subset of data follows a Gaussian distribution (continuous function which approximates the exact binomial distribution of events), and if not it splits the cluster. It starts with a small number of centers, say one cluster only (k=1), then the algorithm splits it into two centers (k=2) and splits each of these two centers again (k=4), having four centers in total. If G-means does not accept these four centers then the answer is the previous step: two centers in this case (k=2). This is the number of clusters your dataset will be divided into. G-means is very useful when you do not have an estimation of the number of clusters you will get after grouping your instances. Notice that an inconvenient choice for the "k" parameter might give you wrong results. The parallel version of g-means is called p-means. G-means sources: source 1 source 2 source 3

  • x-means: a new algorithm that efficiently, searches the space of cluster locations and number of clusters to optimize the Bayesian Information Criterion (BIC) or the Akaike Information Criterion (AIC) measure. This version of k-means finds the number k and also accelerates k-means.

  • Online k-means or Streaming k-means: it permits to execute k-means by scanning the whole data once and it finds automaticaly the optimal number of k. Spark implements it.

Upvotes: 2

kRazzy R
kRazzy R

Reputation: 1589

This is from RBloggers. https://www.r-bloggers.com/k-means-clustering-from-r-in-action/

You could do the following: 

data(wine, package="rattle")
head(wine)
df <- scale(wine[-1])
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(df)

this will create a plot like this.
From this you can choose the value of k to be either 3 or 4. i.e
enter image description here

there is a clear fall in 'within groups sum of squares' when moving from 1 to 3 clusters. After three clusters, this decrease drops off, suggesting that a 3-cluster solution may be a good fit to the data.

But like Anony-Mouse pointed out, the curse of dimensionality affects due to the fact that euclidean distance being used in k means.
I hope this answer helps you to a certain extent.

Upvotes: 0

Has QUIT--Anony-Mousse
Has QUIT--Anony-Mousse

Reputation: 77495

At 62 dimensions, the result will likely be meaningless due to the curse of dimensionality.

k-means does a minimum SSQ assignment, which technically equals minimizing the squared Euclidean distances. However, Euclidean distance is known to not work well for high dimensional data.

Upvotes: 2

Related Questions