Clustering

1. Clustering

2. Inter-cluster distances are maximized Intra-cluster distances are minimized The Problem of Clustering • Given a set of points, with a notion of distance between points, group the points into some number of clusters, so that members of a cluster are in some sense as close to each other as possible.

3. A bit of history

4. Clustering precipitation in Australia Applications of Cluster Analysis • Clustering for Understanding • Group related documents for browsing • Group genes and proteins that have similar functionality • Group stocks with similar price fluctuations • Segment customers into a small number of groups for additional analysis and marketing activities. • Clustering for Summarization • Reduce the size of large data sets

5. Examples Documents • Represent a document by a vector (x1, x2,…, xk), where xi = 1 iff the i th word of a vocabulary appears in the document. • Documents with similar sets of words may be about the same topic. DNAs • Objects are sequences of {C,A,T,G}. • Distance between sequences is edit distance, the minimum number of inserts and deletes needed to turn one into the other. • Note there is a “distance,” but no convenient space in which points “live.”

6. Distance Measures • Each clustering problem is based on some kind of “distance” between points. • Two major classes of distance measure: • Euclidean • Non-Euclidean • A Euclidean space has some number of real-valued dimensions. • There is a notion of “average” of two points. • A Euclidean distance is based on the locations of points in such a space. • A Non-Euclidean distance is based on properties of points, but not their “location” in a space.

7. Axioms of a Distance Measure • d is a distance measure if it is a function from pairs of points to real numbers such that: • d(x,y) > 0. • d(x,y) = 0 iff x = y. • d(x,y) = d(y,x). • d(x,y) < d(x,z) + d(z,y) (triangle inequality).

8. y = (9,8) L2-norm: dist(x,y) = (42+32) = 5 5 3 L1-norm: dist(x,y) = 4+3 = 7 4 x = (5,5) Some Euclidean Distances • L2 norm : d(x,y) = square root of the sum of the squares of the differences between x and y in each dimension. • The most common notion of “distance.” • L1 norm : sum of the differences in each dimension. • Manhattan distance = distance if you had to travel along coordinates only.

9. Non-Euclidean Distances • Jaccard distance for sets = 1 minus ratio of sizes of intersection and union. • Cosine distance = angle between vectors from the origin to the points in question. • Edit distance = number of inserts and deletes to change one string into another.

10. Jaccard Distance for Sets (Bit-Vectors) • Example: p1 = 10111; p2 = 10011. • Size of intersection = 3; size of union = 4, Jaccard similarity (not distance) = 3/4. • d(x,y) = 1 – (Jaccard similarity) = 1/4.

11. Why J.D. Is a Distance Measure • d(x,x) = 0 because xx = xx. • d(x,y) = d(y,x) because union and intersection are symmetric. • d(x,y) > 0 because |xy| < |xy|. • d(x,y) < d(x,z) + d(z,y) trickier... (1 - |x z|/|x z|) + (1 - |y z|/|y z|)  1 - |x y|/|x y| • Remember: • |a b|/|a b| = probability that minhash(a) = minhash(b) • Thus, 1 - |a b|/|a b| = probability that minhash(a)  minhash(b). • Claim: prob[mh(x)mh(y)]  prob[mh(x)mh(z)] + prob[mh(z)mh(y)] • Proof: whenever mh(x)mh(y), at least one of mh(x)mh(z) and mh(z)mh(y) must be true.

12. {a,b,d,e} {d,e,f} {a,b,c} {b,c,e,f} Similarity threshold = 1/3; distance < 2/3 Clustering with JD

13. p1  p2 p1.p2 |p2| d (p1, p2) =  = arccos(p1.p2/|p2||p1|) Cosine Distance • Think of a point as a vector from the origin (0,0,…,0) to its location. • Two points’ vectors make an angle, whose cosine is the normalized dot-product of the vectors: p1.p2/|p2||p1|. • Example: p1 = 00111; p2 = 10011. • p1.p2 = 2; |p1| = |p2| = 3. • cos() = 2/3;  is about 48 degrees.

14. Why C.D. Is a Distance Measure • d(x,x) = 0 because arccos(1) = 0. • d(x,y) = d(y,x) by symmetry. • d(x,y) > 0 because angles are chosen to be in the range 0 to 180 degrees. • Triangle inequality: physical reasoning. If I rotate an angle from x to z and then from z to y, I can’t rotate less than from x to y.

15. Edit Distance • The edit distance of two strings is the number of inserts and deletes of characters needed to turn one into the other. Equivalently: • d(x,y) = |x| + |y| - 2|LCS(x,y)| • LCS = longest common subsequence = any longest string obtained both by deleting from x and deleting from y. Example • x = abcde; y = bcduve. • Turn x into y by deleting a, then inserting u and v after d. • Edit distance = 3. • Or, LCS(x,y) = bcde. • Note: |x| + |y| - 2|LCS(x,y)| = 5 + 6 –2*4 = 3 = edit dist.

16. Why Edit Distance Is a Distance Measure • d(x,x) = 0 because 0 edits suffice. • d(x,y) = d(y,x) because insert/delete are inverses of each other. • d(x,y) > 0: no notion of negative edits. • Triangle inequality: changing x to z and then to y is one way to change x to y.

17. Hierarchical Clustering • Produces a set of nested clusters organized as a hierarchical tree • Can be visualized as a dendrogram • A tree like diagram that records the sequences of merges or splits

18. Agglomerative Clustering Algorithm Compute the proximity matrix Let each data point be a cluster Repeat Merge the two closest clusters Update the proximity matrix Until only a single cluster remains • Key operation is the computation of the proximity of two clusters • Different approaches to defining the distance between clusters distinguish the different algorithms

19. p1 p2 p3 p4 p5 . . . p1 p2 p3 p4 p5 . . . Starting Situation • Start with clusters of individual points and a proximity matrix Proximity Matrix

20. C1 C2 C3 C4 C5 C1 C2 C3 C4 C5 Intermediate Situation • After some merging steps, we have some clusters C3 C4 Proximity Matrix C1 C5 C2

21. C1 C2 C3 C4 C5 C1 C2 C3 C4 C5 Intermediate Situation • We want to merge the two closest clusters (C2 and C5) and update the proximity matrix. C3 C4 Proximity Matrix C1 C5 C2

22. After Merging • The question is “How do we update the proximity matrix?” C2 U C5 C1 C3 C4 C1 ? ? ? ? ? C2 U C5 C3 C3 ? C4 ? C4 Proximity Matrix C1 C2 U C5

23. p1 p2 p3 p4 p5 . . . p1 p2 p3 p4 p5 . . . How to Define Inter-Cluster Similarity Similarity? • MIN • MAX • Group Average Proximity Matrix

24. p1 p2 p3 p4 p5 . . . p1 p2 p3 p4 p5 . . . How to Define Inter-Cluster Similarity • MIN • MAX • Group Average Proximity Matrix

25. p1 p2 p3 p4 p5 . . . p1 p2 p3 p4 p5 . . . How to Define Inter-Cluster Similarity • MIN • MAX • Group Average Proximity Matrix

26. p1 p2 p3 p4 p5 . . . p1 p2 p3 p4 p5 . . . How to Define Inter-Cluster Similarity • MIN • MAX • Centroid distance Proximity Matrix

27. Cluster Similarity: MIN • Similarity of two clusters is based on the two most similar (closest) points in the different clusters

28. 5 1 3 5 2 1 2 3 6 4 4 Hierarchical Clustering: MIN Nested Clusters Dendrogram

29. Two Clusters Strength of MIN Original Points Can handle non-globular shapes

30. Original Points Four clusters Three clusters: The yellow points got wrongly merged with the red ones, as opposed to the green one. Limitations of MIN Sensitive to noise and outliers

31. Cluster Similarity: MAX • Similarity of two clusters is based on the two least similar (most distant) points in the different clusters

32. 4 1 2 5 5 2 3 6 3 1 4 Hierarchical Clustering: MAX Nested Clusters Dendrogram

33. Original Points Four clusters Three clusters: The yellow points get now merged with the green one. Strengths of MAX Less susceptible respect to noise and outliers

34. Two Clusters Limitations of MAX Original Points Tends to break large clusters

35. Cluster Similarity: Centroid distance • Consider the distance between cluster centroids.

36. And in the Non-Euclidean Case? • The only “locations” we can talk about are the points themselves. • i.e., there is no “average” of two points. • Approach 1: clustroid = point “closest” to other points. • Treat clustroid as if it were centroid, when computing intercluster distances. • “Closest” Point? • Possible meanings: • Smallest maximum distance to the other points. • Smallest average distance to other points.

37. Example clustroid 1 2 6 4 3 clustroid 5 intercluster distance

38. k – Means Algorithm(s) • Start by picking k, the number of clusters. • Initialize clusters by picking one point per cluster. • Pick one point at random, • Then k -1 other points, each as far away as possible from the previous points • each with the maximum possible minimum distance to the previously selected points. • Example • Suppose we pick A first. • E is the furthest from A, so we pick it next. • For the third point, the minimum distances to A or E are as follows B: 3.00, C: 2.83, D: 3.16, F: 2.00 The winner is D. Thus, D becomes the third seed.

39. Populating Clusters • For each point, place it in the cluster whose current centroid it is nearest. • After all points are assigned, fix the centroids of the k clusters. • Optional: reassign all points to their closest centroid. • Sometimes moves points between clusters. Repeat for a couple of rounds.

40. Reassigned points Clusters after first round Example: Assigning Clusters 2 4 x 6 3 1 8 7 5 x

41. Best value of k Average distance to centroid k Getting k Right • Try different k, looking at the change in the average distance to centroid, as k increases. • Average falls rapidly until right k, then changes little.

42. Too few; many long distances to centroid. Example: Picking k x xx x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

43. Just right; distances rather short. Example: Picking k x xx x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

44. Too many; little improvement in average distance. Example: Picking k x xx x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

45. BFR Algorithm • BFR (Bradley-Fayyad-Reina) is a variant of k -means designed to handle very large (disk-resident) data sets. • The goal is not to assign every point to a cluster, but to determine where the centroids of the clusters are. • We can assign the points to clusters by another pass through the data, assigning each point to its nearest centroid and writing out the cluster number with the point. • Points are read one main-memory-full at a time. • Most points from previous memory loads are summarized by simple statistics. • To begin, from the initial load we select the initial k centroids by some sensible approach.

46. Initialization: k -Means Possibilities include: • Take a small random sample and cluster optimally. • Take a sample; pick a random point, and then k – 1 more points, each as far from the previously selected points as possible.

47. Three Classes of Points • The discard set : points close enough to a centroid to be summarized. • The compression set : groups of points that are close together but not close to any centroid. They are summarized, but not assigned to a cluster. • The retained set : isolated points.

48. Summarizing Sets of Points For each cluster, the discard set is summarized by: • The number of points, N. • The vector SUM, whose ith component is the sum of the coordinates of the points in the ith dimension. • The vector SUMSQ: ith component = sum of squares of coordinates in ith dimension. • 2d + 1 values represent any number of points. (d = number of dimensions). • Averages in each dimension (centroid coordinates) can be calculated easily as SUMi/N. • SUMi = ith component of SUM. • Variance of a cluster’s discard set in dimension i can be computed by: • (SUMSQi /N ) – (SUMi /N )2 • And the standard deviation is the square root of that. • The same statistics can represent any compression set.

49. Points in the RS Compressed sets. Their points are in the CS. A cluster. Its points are in the DS. The centroid “Galaxies” Picture

50. Processing a “Memory-Load” of Points • Find those points that are “sufficiently close” to a cluster centroid; add those points to that cluster and the DS. • Adjust statistics of the clusters to account for the new points. • Use any main-memory clustering algorithm to cluster the remaining points and the old RS. • Clusters go to the CS; outlying points to the RS. • Consider merging compressed sets in the CS. • If this is the last round, merge all compressed sets in the CS and all RS points into their nearest cluster.