Clust Clus e t ring 2 Nov
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1 Clustering 2 Nov
2 HAC Algorithm Start t with all objects in their own cluster. Until there is only one cluster: Among the current clusters, determine the two clusters, c i and c j, that are most similar. Replace c i and c j with a single cluster c i c j To compute the distance between two clusters Single Link: distance of two closest members of clusters Complete Link: distance of two furthest members of clusters Average Link: average distance
3 Comments on HAC HAC is a fast tool that often provides interesting views of a dataset Primarily HAC can be viewed as an intuitively appealing clustering procedure for data analysis/eploration We can create clusterings of different granularity by stopping at different levels of the dendrogram HAC often used together with visualization of the dendrogram to decide how many clusters eist in the data
4 HAC creates a Dendrogram D(C1, C2) Dendrogram draws the tree such that the height of a tree branch = the distance between the two merged clusters at that particular step The distances are always monotonically increasing This can provide some understanding about how many natural groups there are in the data A drastic height change indicates that we are merging two very different clusters together maybe a good stopping point
5 Flat vs hierarchical clustering Hierarchical clustering generates nested clusterings Flat clustering generates a single partition of the data without resorting to the hierachical procedure Representative eample: K means clustering
6 Mean Squared Error Objective Assume instances are real valued vectors Given a clustering of the data into k clusters, we can compute the centroid of each cluster If we use the mean of each cluster to represent all data points in the cluster, our mean squared error (MSE) is: One possible objective of clustering is to find a clustering that minimizes this error Note that this is a combinatorial optimization problem Difficult to find the eact solution Commonly solved using an iterative approach
7 Basic idea We assume that the number of desired clusters, k, is given Randomly choose k eamples as seeds, one per cluster. Form initial clusters based on these seeds. Iterate by repeatedly reallocating instances to different clusters to improve the overall clustering. Stop when clustering converges or after a fied number of iterations.
8 K means algorithm (MacQueen 1967) Input: data to be clustered and desired number of clusters k Output: k mutually eclusive clusters that cover all eamples
9 K Means Eample (K=2)
10 K Means Eample (K=2) Pick seeds
11 K Means Eample (K=2) Pick seeds Reassign clusters
12 K Means Eample (K=2) Pick seeds Reassign clusters Compute centroids
13 K Means Eample (K=2) Pick seeds Reassign clusters Compute centroids Reasssign clusters
14 K Means Eample (K=2) Pick seeds Reassign clusters Compute centroids Reasssign clusters Compute centroids
15 K Means Eample (K=2) Pick seeds Reassign clusters Compute centroids Reasssign clusters Compute centroids Reassign clusters
16 K Means Eample (K=2) Pick seeds Reassign clusters Compute centroids Reasssign clusters Compute centroids Reassign clusters Converged!
17 Monotonicity of K means Monotonicity Property: Each iteration of K means strictly decreases the MSE until convergence The following lemma is key to the proof: Lemma: Given a finite set C of data points the value of μ that minimizes the MSE: is:
18 Proof of monoticity Given a current set of clusters with their means, the MSE is given by : Consider the reassignment step: Since each point is only reassigned if it is closer to some other cluster than its current cluster, so we know the reassignment step will only decrease MSE Consider the re center step: From our lemma we know that μ i minimizes the distortion of c i which implies that the resulting li MSE again is decreased. d Combine the above two, we know Kmeans always decreases the MSE
19 Kmeans properties Kmeans always converges in a finite number of steps Typically converges very fast (in fewer iterations than the number of points) Time compleity: Assume computing distance between two instances is O(d) where d is the dimensionality of the vectors. Reassigning clusters: O(kn) distance computations, or O(knd). Computing centroids: Each instance vector gets added once to some centroid: O(nd). Assume these two steps are each done once for I iterations: O(Iknd). Linear in all relevant factors, assuming a fied number of iterations, more efficient than O(n 2 ) HAC.
20 More Comments Highly sensitive to the initial seeds This is because MSE has many local minimal solutions, i.e., solutions that can not be improved by local reassignments of any particular points
21 Solutions Run multiple trials and choose the one with the best MSE This is typically done in practice Heuristics: try to choose initial centers to be far apart Using furthest first traversal Start with a random initial center, set the second center to be furthest from the first center, the third center to be furthest from the first two centers, and son on One can also initialize iti with results of other clustering method, then apply kmeans s
22 Even more comments K Means is ehaustive: Cluster every data point, no notion of outlier Outliers maycause problems, why? Outliers will strongly impact the cluster centers Alternative: K medoids instead of computing the mean of each cluster, we find the medoid for each cluster, i.e., the data point that is on average closest to other objects in the cluster
23 Deciding k a model selection problem What if we don t know how many clusters there are in the data? Can we use MSE to decide k by choosing k that gives the smallest MSE? We will always favor larger k values Any quick solutions? Find the knee We will see some other model selection techniques later
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