Strojno učenje. Metoda potpornih vektora (SVM Support Vector Machines) Tomislav Šmuc
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1 Strojno učenje Metoda potpornih vektora (SVM Support Vector Machines) Tomislav Šmuc
2 Generativni i diskriminativni modeli Diskriminativni Generativni (Učenje linije koja razdvaja klase) Učenje modela za svaku pojedinu klasu 2
3 Diskriminativni algoritmi x 2 LDA C( x) arg max ( x) k C k x 2 { x : ( x) ( x)} k l QDA 6-Mar-13 TS: Strojno učenje Linearne metode & SVM 3
4 Koja hiper-ravnina je najbolja? Separabilni problem: Između točaka možemo provudi velik broj (beskonačan) ploha koje dobro razdvajaju primjere dvije klase 4
5 Metoda potpornih vektora (metoda jezgrenih funkcija) SVM - Support Vector Machines SVM nalazi hiper-ravninu koja ima najvedu marginu razdvajanja klasa margina - udaljenost između kritičnih točaka najbliže blizu plohi razdvajanja (en. decision boundary) Intuicija: Ako ne postoje točke blizu plohe razdvajanja, to znači da nemamo nesigurnih klasifikacijskih odluka!? 5
6 Druga intuicija Ako postavimo što je mogude vedi razmak (marginu) između dviju klasa, postoji manje mogudnosti izbora za funkciju razdvajanja kapacitet modela se smanjuje (manje šanse za overfitting)!! 6
7 SVM dakle maksimiziraju marginu m oko hiperravnine (plohe) razdvajanja. (en. large margin classifiers) Potporni vektori Kod SVM - funkcija odluke definirana je preko podskupa primjera iz skupa za učenje tzv. Potpornih vektora (en. support vectors) Problem određivanja SV: kvadratni optimizacijski problem (en. quadratic programming) Max. margina 7
8 Separabilni i neseparabilni problemi Ako se pokaže da problem nije linearno separabilan: - Dozvoljene su greške uz penaliziranje Osnovni princip ostaje: - Ploha razdvajanja u principu mora što bolje razdvajati klase TS: Strojno učenje Linearne metode & 6-Mar-13 8 SVM
9 SVM linearno separabilni podaci (2 klase) Geometrijska intuicija - konveksna ljuska (Convex hull) K.LJ. najmanji konveksni skup koji sadrži sve točke skupa točaka {x i } Za dani skup točaka {x i } K.LJ. je skup točaka definiran sa: Separabilni problem Neseparabilni problem za dva skupa {x i } i {y i } vrijedi ako se njihove K.LJ. ne preklapaju onda se radi o linearno separabilnom problemu! 9
10 SVM linearno separabilni podaci (2 klase) Geometrijska intuicija - Konveksne ljuske K.LJ. najmanji konveksni skup koji sadrži sve točke Zadatak nadi konveksnu ljusku za svaku od klasa Nadi dvije najbliže točke u dvije konveksne ljuske Odrediti ravninu koja prolazi između te dvije točke 10
11 Formalizacija- određivanje maksimalne margine w: normala na hiperravninu razdvajanja x i : primjer (točka) i y i : klasa primjera i ( +1/-1) Klasifikator je određen funkcijom sign(w T x i + b) margina x i je određena sa y i (w T x i + b) 11
12 Pojam - geometrijske margine Primjeri najbliže plohi su potporni vektori. Margina ρ plohe razdvajanja je širina razdvajanja između potpornih vektora suprotnih klasa x w r ρ Udaljenost između plohe i točke x r w T x b w 12
13 Linearni SVM - izvod Pretpostavimo da vrijedi da se sve točke nalaze na distanci 1 od plohe razdvajanja Tada vrijede dva ograničenja na skupu primjera za učenje {(x i,y i )} w T x i + b 1 ako je y i = 1 w T x i + b -1 ako je y i = -1 Za potporne vektore (točke koje se nalaze na margini), ove nejednakosti su jednakosti; S obzirom da je udaljenost svakog primjera od plohe razdvajanja: T w x b r w 2 Margina je: w TS: Strojno učenje Linearne metode & 6-Mar SVM
14 Linearni SVM ρ w T x a + b = 1 Ploha razdvajanja w T x + b = 0 w T x b + b = -1 Uz ograničenja: min i=1,,n w T x i + b = 1 Dolazimo do vrijednosti za marginu razdvajanja: w T (x a x b ) = 2 ρ = x a x b = 2/ w w T x + b = 0 14
15 Linearni SVM - formulacija optimizacijskog problema Pronađi w i b tako da je 2 => maksimalna w a za sve zadane {(x i, y i )} mora vrijediti: w T x i + b 1 ako je y i =1; w T x i + b -1 ako je y i = -1 15
16 Linearni SVM - formulacija optimizacijskog problema Bolja formulacija kao minimizacijski problem (kvadratni optimizacijski problem uz ograničenja): min w = max (2/ w ) Nadi w i b - takve da je: Φ(w) =½ w T w je minimalno; A za sve {(x i,y i )} vrijedi : y i (w T x i + b) 1 16
17 Rješavanje optimizacijskog problema => optimizacija uz ograničenja Nađi w i b - takve da je: Φ(w) =½ w T w - minimalno; A za sve {(x i,y i )} vrijedi : y i (w T x i + b) 1 Ova formulacija minimizacija kvadratne funkcije uz linearna ograničenja (nejednakosti) Dobro poznati i rješivi problem puno (više ili manje složenih) algoritama za njihovo rješavanje (MATLAB, MATHEMATICA) Standardno za SVM - prevođenje u dualni problem u kojem se tzv. Lagrangeovim multiplikatorima (težinski faktori koji penaliziraju svako ograničenje iz primalnog problema) 17
18 Lagrange-ovi multiplikatori => optimizacija uz ograničenja x * 18
19 Lagrange-ovi multiplikatori => optimizacija uz ograničenja x A 19
20 Lagrange-ovi multiplikatori => optimizacija uz ograničenja Slučaj sa ograničenjem u obliku nejednakosti f(x) Slučaj sa ograničenjem u obliku jednakosti f(x) f(x) b) Ograničenje nije aktivno Min f(x) (neograničeni minimum) Min L(x) (ograničeni minimum f(x) 20
21 Lagrange-ovi multiplikatori => optimizacija uz ograničenja 21
22 Lagrange-ova funkcija i min-max dualnost Dualna formulacija: TS: Strojno učenje Linearne metode & 6-Mar SVM
23 SVM - optimizacijski problem Dualna formulacija: Nadi λ 1 λ N tako da je: Q(λ) =Σ λ i - ½ΣΣ λ i λ j y i y j x it x j - maksimalno Te da vrijedi (1) Σ λ i y i = 0 - i,j =1,N (2) λ i 0 α i TS: Strojno učenje Linearne metode & 6-Mar SVM
24 Rješenje optimizacijskog problema w =Σ λ i y i x i b= y k - w T x k za bilo koji x k za koji je α k 0 Svaki α i koji je različit od 0 zapravo znači da je taj x i potporni vektor. Konačno klasifikacijska funkcija ima ovaj oblik: f(x) = Σ λ i y i x it x + b Rješenje je proporcionalno sumi umnožaka - unutarnjih produkta x it x između nove točke (primjera) x i svih potpornih vektora x i! Treba zapamtiti i da je u rješavanju optimizacijskog problema također korišten produkt x it x j između svih parova točaka skupa za učenje. 24
25 Što ako problem nije linearno separabilan? Dodajemo nove varijable ξ i (en. slack variables) - dozvoljavanje krivih klasifikacija za teške ili šumovite primjere ξ i ξ i 25
26 Izvod SVM s mekanom marginom (en. Soft margin) Stara formulacija: Nađi w i b - takve da je: Φ(w) =½ w T w is minimalno; A za sve {(x i,y i )} vrijedi : y i (w T x i + b) 1 Nova formulacija sa ξ i : Nađi w i b - takve da je: Φ(w) =½ w T w + CΣξ i - minimalno; A za sve {(x i,y i )} vrijedi: y i (w T x i + b) 1 ξ i te da vrijedi ξ i 0 za sve i 26
27 Rješenje - SVM s mekanom marginom (en. Soft margin) Dualni problem za SVM s mekanom marginom : Nađi λ 1 λ N tako da je max Q(λ) =Σ λ i - ½ΣΣ λ i λ j y i y j x it x j Te da vrijedi (1) Σ λ i y i = 0 (2) 0 λ i C λ i I dalje vrijedi: x i sa λ i >0 su potporni vektori. U dualnoj formulaciji nema slack varijabli, samo ograničenje na λ (λ < C) Rješenje dualnog problema (soft margin): w =Σ λ i y i x i b= y k (1- ξ k ) - w T x k gdje je k = argmax λ k k w nije potreban kod procesa klasifikacije kao i prije! f(x) = λ i y i x it x + b 27
28 Klasifikacija Uz novu točku (x 1,x 2 ), Odrediti projekciju na normalu plohe razdvajanja: 2 dimenzije: p = w 1 x 1 +w 2 x 2 +b. To ustvari znači (wx + b = Σ λ i y i x it x + b) Dodatno => ako stavimo neku graničnu mjeru t >0 (pouzdanost) f > t: => 1 f < -t: => -1 Inače: suzdržan 28
29 Linearni SVM - sažetak Klasifikator je posebno određena ploha razdvajanja (en. separating hyperplane) Najvažnije točke - točke iz skupa za učenje koje definiraju plohu razdvajanja - potporni vektori Potporni vektori se pronalaze optimizacijskim algoritmima Rješavani problem je kvadratni optimizacijski problem s linearnim ograničenjima. U dualnoj formulaciji problema i rješenja potporni vektori odnosno točke iz skupa za učenje se pojavljuju u skalarnim produktima: Nađi λ 1 λ N tako da je Q(λ) =Σ λ i - ½ΣΣ λ i λ j y i y j x it x j (1) Σ λ i y i = 0 - maksimalno i vrijedi: (2) 0 λ i C λ i f(x) = Σ λ i y i x it x + b 29
30 Nelinearni SVM Za problem koji su linearno separabilni (i uz razumni šum) Linearni SVM je OK : No, što ako to ne vrijedi? 0 x 0 x mapiranje podataka u (još) više-dimenzionalni prostor? : x 2 0 TS: Strojno učenje SVM 6-Mar x
31 Nelinearni SVM: prostori novih dimenzija (atributa) Osnovna ideja: originalni prostor varijabli možemo mapirati u novi-višedimenzionalni prostor - gdje de naš problem biti (linearno) separabilan (rješiv) Φ: x φ(x) 31
32 Kernel trik Linearni SVM - skalarni produkt između vektora K(x i,x j )=x it x j Transformacija kojom svaku točku mapiramo u neki novi prostor Φ: x φ(x) Sada bi skalarni produkt trebao izgledati drukčije: K(x i,x j )= φ(x i ) T φ(x j ) Kernel funkcija - funkcija koja korespondira nutarnjem produktu u nekom ekspandiranom prostoru novih varijabli 32
33 Kernel trik Primjer: originalni prostor 2-dimenzionalni vektori x=[x 1 x 2 ]; Neka je K(x i,x j )=(1 + x it x j ) 2, Pokažimo da vrijedi K(x i,x j )= φ(x i ) T φ(x j ), ako je φ(x) = [1, x 12, 2 x 1 x 2, x 2 2, 2x 1 2x 2 ] K(x i,x j )=(1 + x it x j ) 2,= 1+ x i12 x j x i1 x j1 x i2 x j2 + x i22 x j x i1 x j1 + 2x i2 x j2 = = [1 x i1 2 2 x i1 x i2 x i2 2 2x i1 2x i2 ] T [1 x j1 2 2 x j1 x j2 x j2 2 2x j1 2x j2 ] = φ(x i ) T φ(x j ) 33
34 Kerneli Zašto? Omogudavaju pretvaranje neseparabilnih problema u separabilne. Mapiranje u bolje (reprezentacijski) višedimenzionalne prostore Uobičajeni kerneli Linearni K(x i,x j ) = (x it x j ) Polinomijalni K(x i,x j ) = (1+x it x j ) d RBF K xi x 2 2 ( xi,x j ) e j Sigmoid T K( x,x ) tanh( x x r) i j i j 34
35 Linearni SVM efekt različitih vrijednosti parametra C Prisjetimo se gdje smo vidjeli C Φ(w) =½ w T w + CΣξ i 35
36 Odabir kernela efekti Polinomni kerneli (C=const.) Linearni kernel polinom 2 stupnja polinom 5 stupnja 36
37 Odabir parametara kernela efekti Gaussian kernel, C=const. 37
38 Odabir parametara parametara kernela i faktora C RBF kernel, C=1, =1 RBF kernel, C=10, =1 RBF kernel, C=100, =1 RBF kernel, C=1, =10 RBF kernel, C=10, =10 RBF kernel, C=100, =10 38
39 SVM - Sa dvije klase na više klasa Vedina SVM metoda je za binarne klasifikacijske probleme Uobičajen pristup za višeklasne probleme je preko redukcije problema na više binarnih klasifikacijskih problema, npr. za K-klasni problem: Jedan-nasuprot-ostalih => (K binarnih klasifikatora) Svaki-protiv-svakog => (K*(K-1)/2 binarnih klasifikatora) ECOC Error Correcting Output Codes ( u jednom od slijededih predavanja) 39
40 SVM Regresija Problem 40
41 Karakteristike SVM-a Jedna od najboljih metoda strojnog učenja ukupno gledajudi Popularna i vrlo dobri rezultati text mining U praksi mnoge druge metode rade otprilike isto dobro Postoji mnogo komparacija koje to pokazuju Važno je: Razumijevanje i iskustvo za efektivno korištenju Odabir kernela, odabir parametara (C, parametri kernela) Dobro je kombinirati selekciju varijabli i SVM u praktičnim primjerima RapidMiner, Matlab, R (više SVM varijanti + vizualizacija) 41
42 Sažetak Definiranje plohe razdvajanja preko potpornih vektora Potporni vektor = kritične točke (primjeri) najbliži plohi razdvajanja Pojam kernela: Modan alat za mapiranje u visoko-dimenzionalne prostore, kao i (re)definiranje metrika sličnosti Samostalni paketi i okruženja koja imaju SVM: SVM Torch; SVMLight; LibSVM RapidMiner, WEKA, MATLAB, R
43 Literatura SVM literatura i materijali The Elements of Statistical Learning Hastie, Tibshirani, Friedman (1st ed - ch. 11) A Tutorial on Support Vector Machines for Pattern Recognition (1998) Christopher J. C. Burges Christopher M. Bishop Pattern recognition and machine learning, Springer, Kristin P. Bennet and Campbell Support Vector Machines: Hype or Hallelujah? SIGKDD Explorations, vol. 2, 2000 A User's Guide to Support Vector Machines, Ben-Hur & Weston T. Joachims, Learning to Classify Text using Support Vector Machines. Kluwer, Classic Reuters data set: /resources /testcollections/reuters21578/ 43
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