Support Vector and Kernel Methods

Transcription

1 SIGIR 2003 Tutorial Support Vector and Kernel Methods Thorsten Joachims Cornell University Computer Science Department 0

2 Linear Classifiers Rules of the Form: weight vector w, threshold hx ( ) = sign w i x i + b = i = 1 Geometric Interpretation (Hyperplane): N N 1 if w i x i i = 1 b 1 else + b > 0 w b 14

3 Optimal Hyperplane (SVM Type 1) Assumption: The training examples are linearly separable. 19

4 Maximizing the Margin δ The hyperplane with maximum margin <~ (roughly, see later) ~> The hypothesis space with minimal VC-dimension according to SRM Support Vectors: Examples with minimal distance. 21

5 Example: Optimal Hyperplane vs. Perceptron Percent Training/Testing Errors Perceptron with eta=0.1 "perceptron_iter_trainerror.dat" "perceptron_iter_testerror.dat" hard_margin_svm_testerror.dat Iterations Train on 1000 pos / 1000 neg examples for acq (Reuters-21578). 24

6 Non-Separable Training Samples For some training samples there is no separating hyperplane! Complete separation is suboptimal for many training samples! => minimize trade-off between margin and training error. 25

7 Soft-Margin Separation Idea: Maximize margin and minimize training error simultanously. Hard Margin: Soft Margin: minimize s. t. Pwb (, ) = 1 --w w 2 y i [ w x i + b] 1 minimize Hard Margin (separable) Pwbξ (,, ) w w + C ξ i s. t. + b and = n i = 1 y i [ w x i ] 1 ξ i ξ i 0 δ Soft Margin (training error) ξ j ξ i 26

8 Controlling Soft-Margin Separation Soft Margin: minimize ξ i Pwbξ (,, ) w w + C ξ i s. t. + b and i = 1 is an upper bound on the number of training errors. C is a parameter that controls trade-off between margin and error. = y i [ w x i ] 1 ξ i ξ i 0 n Large C ξ i δ Small C ξ j 27

9 Example Reuters acq : Varying C Percent Training/Testing Errors "svm_trainerror.dat" "svm_testerror.dat" hard-margin SVM C Observation: Typically no local optima, but not necessarily... 28

10 Properties of the Soft-Margin Dual OP Dual OP: maximize s. t. D( α) = α i i = 1 typically single solution (i. e. wb, is unique) one factor α i for each training example influence of single training example limited by C 0 < α i < C <=> SV with ξ i = 0 α i = C <=> SV with ξ i > 0 α i = 0 else based exclusively on inner product between training examples n n n n 1 -- α 2 i α j y i y j ( x i x j ) i = 1 j = 1 α i y i = 0 und 0 α i C i = 1 ξ j ξ i 37

11 Primal <=> Dual Theorem: The primal OP and the dual OP have the same solution. Given the solution α i of the dual OP, w = α i y i x i b i = 1 is the solution of the primal OP. n = 1 -- ( w 2 0 x pos + w 0 x neg ) Theorem: For any set of feasible points Pwb (, ) D( α). => two alternative ways to represent the learning result weight vector and threshold wb, vector of influences α 1, α, n 36

12 Non-Linear Problems ==> Problem: some tasks have non-linear structure no hyperplane is sufficiently accurate How can SVMs learn non-linear classification rules? 38

13 Example Input Space: = (, ) (2 Attributes) x x 1 x 2 Feature Space: Φ x = (,, 2x1, 2x 2, 2x 1 x 2, 1) (6 Attributes) ( ) x 1 2 x2 2 40

14 Extending the Hypothesis Space Idea: Input Space ==> Find hyperplane in feature space! Φ Feature Space Example: a b c ==> The separating hyperplane in features space is a degree two polynomial in input space. Φ a b c aa ab ac bb bc cc 39

15 Kernels Problem: Very many Parameters! Polynomials of degree p over N attributes in input space lead to ON ( p ) attributes in feature space! Solution: [Boser et al., 1992] The dual OP need only inner products => Kernel Functions Kx ( i, x j ) = Φ( x i ) Φ( x j ) Example: For Φ x = (,, 2x1, 2x 2, 2x 1 x 2, 1) calculating ( ) x 1 2 x2 2 Kx ( i, x j ) = [ x i x j + 1] 2 = Φ( x i ) Φ( x j ) gives inner product in feature space. We do not need to represent the feature space explicitly! 41

16 Training: maximize s. t. SVM with Kernels D( α) = α i n n i = 1 n n 1 -- α 2 i α j y i y j Kx ( i, x j ) i = 1 j = 1 α i y i = 0 und 0 α i C i = 1 Classification: For new example x hx ( ) = sign α i y i Kx ( i, x) x i SV + b New hypotheses spaces through new Kernels: Linear: Polynomial: Radial Basis Functions: Sigmoid: Kx ( i, x j ) = x i x j Kx ( i, x j ) = [ x i x j + 1] d Kx ( i, x j ) = tanh( γ( x i x j ) + c) 2 σ 2 Kx ( i, x j ) = exp( x i x j ) 42

17 Example: SVM with Polynomial of Degree 2 Kernel: Kx ( i, x j ) = [ x i x j + 1] 2 plot by Bell SVM applet 43

18 Example: SVM with RBF-Kernel Kernel: 2 σ 2 Kx ( i, x j ) = exp( x i x j ) plot by Bell SVM applet 44

19 Two Reasons for Using a Kernel (1) Turn a linear learner into a non-linear learner (e.g. RBF, polynomial, sigmoid) (2) Make non-vectorial data accessible to learner (e.g. string kernels for sequences) 51

20 Given: Summary What is an SVM? Training examples ( x 1, y 1 ),, ( x n, y n ) x i R N yi {, 1 1} Hypothesis space according to kernel Kx ( i, x j ) Parameter C for trading-off training error and margin size Training: Finds hyperplane in feature space generated by kernel. The hyperplane has maximum margin in feature space with minimal training error (upper bound ξ i ) given C. The result of training are,,. They determine wb,. α 1 α n Classification: For new example hx ( ) = sign α i y i Kx ( i, x) x i SV + b 52

21 Part 2: How to use an SVM effectively and efficiently? normalization of the input vectors selecting C handling unbalanced datasets selecting a kernel multi-class classification selecting a training algorithm 53

22 How to Assign Feature Values? Things to take into consideration: importance of feature is monotonic in its absolute value the larger the absolute value, the more influence the feature gets typical problem: number of doors [0-5], price [ ] want relevant features large / irrelevant features low (e.g. IDF) normalization to make features equally important by mean and variance: x norm = x mean ( X ) var( X) by other distribution normalization to bring feature vectors onto the same scale directional data: text classification x by normalizing the length of the vector x norm = according to some norm x changes whether a problem is (linearly) separable or not scale all vectors to a length that allows numerically stable training 57

23 Selecting a Kernel Things to take into consideration: kernel can be thought of as a similarity measure examples in the same class should have high kernel value examples in different classes should have low kernel value ideal kernel: equivalence relation Kx ( i, x j ) = sign( y i y j ) normalization also applies to kernel relative weight for implicit features normalize per example for directional data Kx ( i, x j ) = Kx ( i, x j ) Kx ( i, x i ) Kx ( j, x j ) potential problems with large numbers, for example polynomial kernel (, ) = [ x i x j + 1] d for large d Kx i x j 58

24 Common Method Selecting Regularization Parameter C a reasonable starting point and/or default value is search for C on a log-scale, for example C [ 10 4 C def,, 10 4 Cdef] selection via cross-validation or via approximation of leave-one-out [Jaakkola&Haussler,1999][Vapnik&Chapelle,2000][Joachims,2000] Note optimal value of C scales with the feature values 1 C def = Kx ( i, x i ) 59

25 Problem Selecting Kernel Parameters results often very sensitive to kernel parameters (e.g. variance γ in RBF kernel) need to simultaneously optimize C, since optimal C typically depends on kernel parameters Common Method search for combination of parameters via exhaustive search selection of kernel parameters typically via cross-validation Advanced Approach avoiding exhaustive search for improved search efficiency [Chapelle et al, 2002] 60

26 Handling Multi-Class / Multi-Label Problems Standard classification SVM addresses binary problems y { 1, 1} Multi-class classification: y { 1,, k} one-against-rest decomposition into k binary problems learn one binary SVM h () i per class with y i = 1 assign new example to y = argmax[ h () i ( x) ] pairwise decomposition into kk ( 1) binary problems learn one binary SVM h () i per class pair y i, j = 1 assign new example by majority vote reducing number of classifications [Platt et al., 2000] multi-class SVM [Weston & Watkins, 1998] multi-class SVM via ranking [Crammer & Singer, 2001] () 1 if( y = i) else ( ) 1 if( y = i) if( y = j) 55