I understand the difference between k medoid and k means. But can you give me an example with a small data set where the k medoid output is different from k means output.


3 Answers 3


k-medoid is based on medoids (which is a point that belongs to the dataset) calculating by minimizing the absolute distance between the points and the selected centroid, rather than minimizing the square distance. As a result, it's more robust to noise and outliers than k-means.

Here is a simple, contrived example with 2 clusters (ignore the reversed colors) Kmeans vs. Kmedoids

As you can see, the medoids and centroids (of k-means) are slightly different in each group. Also you should note that every time you run these algorithms, because of the random starting points and the nature of the minimization algorithm, you will get slightly different results. Here is another run:

enter image description here

And here is the code:

x <- rbind(matrix(rnorm(100, mean = 0.5, sd = 4.5), ncol = 2),
           matrix(rnorm(100, mean = 0.5, sd = 0.1), ncol = 2))
colnames(x) <- c("x", "y")

# using 2 clusters because we know the data comes from two groups cl <- kmeans(x, 2) kclus <- pam(x,2)
par(mfrow=c(1,2)) plot(x, col = kclus$clustering, main="Kmedoids Cluster") points(kclus$medoids, col = 1:3, pch = 10, cex = 4) plot(x, col = cl$cluster, main="Kmeans Cluster") points(cl$centers, col = 1:3, pch = 10, cex = 4)

  • 1
    $\begingroup$ @frc, if you think someone's answer is incorrect, don't edit it to correct it. You can leave a comment (once your rep is >50), &/or downvote. Your best option is to post your own answer w/ what you believe to be the correct information (cf, here). $\endgroup$ Nov 22, 2016 at 16:18
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    $\begingroup$ K-medoids minimizes an arbitrarily chosen distance (not necessarily an absolute distance) between clustered elements and the medoid. Actually the pam method (an example implementation of K-medoids in R) used above, by default uses the Euclidean distance as a metric. K-means always uses the squared Euclidean. The medoids in K-medoids are chosen out of the cluster elements, not out of a whole points space as centroids in K-means. $\endgroup$
    – hanna
    Nov 27, 2016 at 16:40
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    $\begingroup$ I have not enough reputation to comment, but wanted to mention that there is a mistake in the plots of Ilanman's answer: he ran the whole code, such that the data was modified. If you run only the clustering part of the code, the clusters are quite stables, more stable for PAM than for k-means by the way. $\endgroup$ Jun 14, 2017 at 10:40

A medoid has to be a member of the set, a centroid does not.

Centroids are typically discussed in the context of solid, continuous objects, but there's no reason to believe that the extension to discrete samples would require the centroid to be a member of the original set.


Both k-means and k-medoids algorithms are breaking the dataset up into k groups. Also, they are both trying to minimize the distance between points of the same cluster and a particular point which is the center of that cluster. In contrast to the k-means algorithm, k-medoids algorithm chooses points as centers that belong to the dastaset. The most common implementation of k-medoids clustering algorithm is the Partitioning Around Medoids (PAM) algorithm. PAM algorithm uses a greedy search which may not find the global optimum solution. Medoids are more robust to outliers than centroids, but they need more computation for high dimensional data.


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