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k-Medoids – A Representative Object-Based Technique in Data Mining

K-MEDOIDS  is a partitional clustering algorithm that is slightly modified from the k-means algorithm. They both attempt to minimize the squared error but the k-medoids algorithm is more robust to noise than k-means algorithm. In k-means algorithm, one chooses means as the centroids but in k-medoids, data points are chosen to be the medoids(median). A medoid can be defined as the object of a cluster, whose average is dissimilar to all the objects in the cluster is minimal.

Algorithm:

Step 1:Initialize select k random points out of the n data points as the medoids.



Step 2: Associate each data point to the closest medoid by using any common distance metric methods.

Step 3: While the cost decreases, For each medoid m, for each data o point which is not a medoid:



 

In this image we have all diamonds in black except three diamonds:

 

Depending on the distance, the color of the diamonds changes. Distance calculated and black color is changed to the respective color bases on the minimal distance. In this image we are having 3 clusters.

 Now changing the position of the color diamonds.

 

The cluster groups will be changing according to the minimal distance.

 

Cost Calculation:

The dissimilarity of the medoid(ci) and object (oi) is calculated by using E=| pi-ci |. The cost in the k-medoids algorithm is given as:

C

Example:

x-axis y-axis
7 6
2 6
3 8
8 5
7 4
4 7
6 2

7

6

3

4
3 4

Let us randomly choose 2 medoids, let us take (3,4) & (7,4) as our medoids points. Now calculating the distance from each point,

So, now after the clustering. The clusters formed are:

 {(3,4); (2,6); (3,8); (4,7)}
 {(7,4); (6,2); (6,4); (7,3);
  (8,5); (7,6)}

Calculating the cost which is nothing but the sum of distance of each non-selected point from the selected point.

Total cost = cost((3,4),(2,6)) + cost((3,4),(3,8)) 
             + cost((3,4),(4,7)) + cost((7,4),(6,2))
             + cost((7,4),(6,4)) + cost((7,4),(7,3)) 
             + cost((7,4),(8,5)) + cost((7,4),(7,6))
            = 3 + 4 + 4 + 3 + 1 + 1 + 2 + 2 
            = 20

Now let us choose some other point to be a medoid instead of (7,4). Let us randomly choose (7,3). Now (3,4),(7,3) be new medoid points

x-axis y-axis
7 6
2 6
3 8
8 5
7 4

4

6

7

2
7 3
6 4
3 4

Now repeat as earlier, Calculating the distance from each point

clusters are = {(3,4); (2,6); (3,8); 
               (4,7)} & {(7,3); (7,6); 
               (8,5); (6,2); (7,4); (6,4)}
Calculating the total cost = cost((3,4),(2,6)) + 
                             ((3,4),(3,8)) + ((3,4),(4,7)) + 
                              cost((7,3),(7,6)) + cost((7,3),(8,5)) 
                              + cost((7,3),(6,2)) + cost((7,3),(7,4)) 
                              + cost((7,3),(6,4))
                           = 3 + 4 + 4 + 3 + 3 + 2 +1 + 2 
                           = 22

Total cost when (7,3) is the medoid is greater than(>) total cost when (7,4) was the medoid. Hence (7,4) is chosen instead of (7,3)  as the medoid. Hence the clusters finally obtained are:- {(3,4); (2,6); (3,8); (4,7)} and {(7,4); (6,2); (6,4); (7,3); (8,5); (7,6)}  

Advantages:

Disadvantages:

Article Tags :
Data Mining
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