clustering1

clustering1 - Clustering Preliminaries Applications...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Clustering Preliminaries Applications Euclidean/NonEuclidean Spaces Distance Measures 1 The Problem of Clustering x 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. 2 Example x x x x x x x x xx x x x x x x x x xx x x x x x x x xx x x x x x x x x x x x x 3 Problems With Clustering x Clustering in two dimensions looks easy. x Clustering small amounts of data looks easy. x And in most cases, looks are not deceiving. 4 The Curse of Dimensionality x Many applications involve not 2, but 10 or 10,000 dimensions. x Highdimensional spaces look different: almost all pairs of points are at about the same distance. Example: assume random points within a bounding box, e.g., values between 0 and 1 in each dimension. 5 Example: SkyCat x A catalog of 2 billion "sky objects" represents objects by their radiation in 9 dimensions (frequency bands). x Problem: cluster into similar objects, e.g., galaxies, nearby stars, quasars, etc. x Sloan Sky Survey is a newer, better version. 6 Example: Clustering CD's (Collaborative Filtering) x Intuitively: music divides into categories, and customers prefer a few categories. x Represent a CD by the customers who bought it. x Similar CD's have similar sets of customers, and viceversa. 7 But what are categories really? The Space of CD's x Think of a space with one dimension for each customer. x A CD's point in this space is (x1, x2,..., xk), where xi = 1 iff the i th customer bought the CD. Compare with the "shingle/signature" matrix: rows = customers; cols. = CD's. 8 Values in a dimension may be 0 or 1 only. Space of CD's (2) x For Amazon, the dimension count is tens of millions. x An option: use minhashing/LSH to get Jaccard similarity between "close" CD's. x 1 minus Jaccard similarity can serve as a (nonEuclidean) distance. 9 Example: Clustering Documents x Represent a document by a vector (x1, x2,..., xk), where xi = 1 iff the i th word (in some order) appears in the document. x Documents with similar sets of words may be about the same topic. 10 It actually doesn't matter if k is infinite; i.e., we don't limit the set of words. Example: Gene Sequences x Objects are sequences of {C,A,T,G}. x Distance between sequences is edit distance, the minimum number of inserts and deletes needed to turn one into the other. x Note there is a "distance," but no convenient space in which points "live." 11 Distance Measures x Each clustering problem is based on some kind of "distance" between points. x Two major classes of distance measure: 1. Euclidean 2. NonEuclidean 12 Euclidean Vs. NonEuclidean x A Euclidean space has some number of realvalued dimensions and "dense" points. There is a notion of "average" of two points. A Euclidean distance is based on the locations of points in such a space. x A NonEuclidean distance is based on properties of points, but not their "location" in a space. 13 Axioms of a Distance Measure x d is a distance measure if it is a function from pairs of points to real numbers such that: 1. 2. 3. 4. 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 ). 14 Some Euclidean Distances x L2 norm : d(x,y) = square root of the sum of the squares of the differences between x and y in each dimension. x L1 norm : sum of the differences in each dimension. Manhattan distance = distance if you had to travel along coordinates only. 15 The most common notion of "distance." Examples of Euclidean Distances L2norm: dist(x,y) = (42+32) = 5 y = (9,8) 5 4 3 x = (5,5) L1norm: dist(x,y) = 4+3 = 7 16 Another Euclidean Distance x L norm : d(x,y) = the maximum of the differences between x and y in any dimension. x Note: the maximum is the limit as n goes to of what you get by taking the n th power of the differences, summing and taking the n th root. 17 NonEuclidean Distances x Jaccard distance for sets = 1 minus ratio of sizes of intersection and union. x Cosine distance = angle between vectors from the origin to the points in question. x Edit distance = number of inserts and deletes to change one string into another. 18 Jaccard Distance for BitVectors x Example: p1 = 10111; p2 = 10011. Size of intersection = 3; size of union = 4, Jaccard similarity (not distance) = 3/4. x Need to make a distance function satisfying triangle inequality and other laws. x d(x,y) = 1 (Jaccard similarity) works. 19 Why J.D. Is a Distance Measure x d(x,x) = 0 because xx = xx. x d(x,y) = d(y,x) because union and intersection are symmetric. x d(x,y) > 0 because |xy| < |xy|. x d(x,y) < d(x,z) + d(z,y) trickier next slide. 20 Triangle Inequality for J.D. 1 |x z| + 1 |y z| > 1 |x y| |x z| |y z| |x y| x Remember: |a b|/|a b| = probability that minhash(a) = minhash(b). x Thus, 1 |a b|/|a b| = probability that minhash(a) minhash(b). 21 Triangle Inequality (2) x Observe that prob[minhash(x) minhash(y)] < prob[minhash(x) minhash(z)] + prob[minhash(z) minhash(y)] x Clincher: whenever minhash(x) minhash(y), at least one of minhash(x) minhash(z) and minhash(z) minhash(y) must be true. 22 Cosine Distance x Think of a point as a vector from the origin (0,0,...,0) to its location. x 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. 23 CosineMeasure Diagram p1 Why? Next slide p1.p2 |p2| p2 dist(p1, p2) = = arccos(p1.p2/|p2||p1|) 24 Why? Dot product is invariant under rotation, so pick convenient coordinate system. p1.p2 = x1x2. |p2| = x2. x1 =x1x2/x2 = p1.p2/|p2| 25 p1 = (x1,y1) x1 p2 = (x2,0) Why C.D. Is a Distance Measure x d(x,x) = 0 because arccos(1) = 0. x d(x,y) = d(y,x) by symmetry. x d(x,y) > 0 because angles are chosen to be in the range 0 to 180 degrees. x 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. 26 Edit Distance x The edit distance of two strings is the number of inserts and deletes of characters needed to turn one into the other. x Equivalently: d(x,y) = |x| + | y| 2|LCS(x,y)|. LCS = longest common subsequence = longest string obtained both by deleting from x and deleting from y. 27 Example x x = abcde ; y = bcduve. x Turn x into y by deleting a, then inserting u and v after d. x Or, LCS(x,y) = bcde. x |x| + |y| 2|LCS(x,y)| = 5 + 6 2*4 = 3. 28 Editdistance = 3. Why E.D. Is a Distance Measure x d(x,x) = 0 because 0 edits suffice. x d(x,y) = d(y,x) because insert/delete are inverses of each other. x d(x,y) > 0: no notion of negative edits. x Triangle inequality: changing x to z and then to y is one way to change x to y. 29 Variant Edit Distance x Allow insert, delete, and mutate. Change one character into another. x Minimum number of inserts, deletes, and mutates also forms a distance measure. 30 ...
View Full Document

Ask a homework question - tutors are online