Conditional Random Fields for Sequential Supervised Learning

Transcription

1 Conditional Random Fields for Sequential Supervised Learning Thomas G. Dietterich Adam Ashenfelter Department of Computer Science Oregon State University Corvallis, Oregon Supported by NSF IIS

2 Outline Goal: Off-the-Shelf Sequential Supervised Learning Candidate Methods Training CRFs by Gradient Boosting Concluding Remarks

3 Many Application Problems Require Sequential Learning Part-of-speech Tagging Information Extraction from the Web Text-to-Speech Mapping

4 Part-of-Speech Tagging Given an English sentence, can we assign a part of speech to each word? Do you want fries with that? <verb pron verb noun prep pron>

5 Information Extraction from the Web <dl><dt><b>srinivasan Seshan</b> (Carnegie Mellon University) <dt><a href= ><i>making Virtual Worlds Real</i></a><dt>Tuesday, June 4, 2002<dd>2:00 PM, 322 Sieg<dd>Research Seminar * * * name name * * affiliation affiliation affiliation * * * * title title title title * * * date date date date * time time * location location * event-type event-type

6 Text-to-Speech Mapping photograph =>

7 Sequential Supervised Learning (SSL) Given: A set of training examples of the form (X i,y i ), where X i = x i,1,, x i,t i and Y i = y i,1,, y i,t i are sequences of length T i Find: A function f for predicting new sequences: Y = f(x).

8 Examples as Sequential Supervised Learning Domain Input X i Output Y i Part-of-speech sequence of sequence of Tagging words parts of speech Information Extraction Test-to-speech Mapping sequence of tokens sequence of letters sequence of field labels {name, } sequence phonemes

9 Goal: Off-the-Shelf Learning Methods for SSL No existing machine learning, data mining, and statistical packages supports SSL No existing method meets all of the requirements needed for an off-theshelf method

10 Requirements for Off-the-Shelf Methods Accuracy Model both x y and y t y t+1 relationships Support rich X y t features Avoid label bias problem Computational efficiency Easy to use No parameter tuning required

11 Two Kinds of Relationships y 1 y 2 y 3 x 1 x 2 x 3 Vertical relationship between the x t s and y t s Example: Friday is usually a date Horizontal relationships among the y t s Example: name is usually followed by affiliation SSL can (and should) exploit both kinds of information

12 Example of y y relationships Consider the text-to-speech problem: photograph => /f-ot@graf-/ photography =>/f-@tagr@f-i/ The letter y changes the pronunciation of all vowels in the word! x y relationships are not sufficient: o is pronounced as /O/, /@/, and /A/ need context to tell which is correct

13 Rich X y Relationships Generative models such as HMMs model each x t as being generated by a single y t Can t incorporate the context around x t Example: Decide how to pronounce h based on surrounding letters th, ph, sh, ch. Can t include global features Example: Sentence begins with question word

14 Existing Methods Sliding windows Recurrent sliding windows Hidden Markov models Maximum entropy Markov models Input/Output Markov models Conditional Random Fields Maximum Margin Markov Networks

15 Sliding Windows Do you want fries with that Do you verb Do you want pron you want fries verb want fries with noun fries with that prep with that pron

16 Properties of Sliding Windows Converts SSL to ordinary supervised learning Only captures the relationship between (part of) X and y t. Does not explicitly model relations among the y t s Assumes each window is independent

17 Recurrent Sliding Windows Do you want fries with that Do you verb Do you want verb pron you want fries pron verb want fries with verb noun fries with that noun prep with that prep pron

18 Recurrent Sliding Windows Key Idea: Include y t as input feature when computing y t+1. During training: Use the correct value of y t Or train iteratively (especially recurrent neural networks) During evaluation: Use the predicted value of y t

19 Properties of Recurrent Sliding Windows Captures relationship among the y s, but only in one direction! Results on text-to-speech: Method Direction Words Letters sliding window none 12.5% 69.6% recurrent s. w. left-right 17.0% 67.9% recurrent s. w. right-left 24.4% 74.2%

20 Evaluation of Methods Issue SW RSW HMM MEMM IOHMM CRF x t y t y t y t+1 NO Partly YES YES YES YES X y t rich? YES YES NO YES YES YES label bias ok? YES YES YES NO NO YES efficient? YES YES YES YES? NO NO

21 Conditional Random Fields y 1 y 2 y 3 x 1 x 2 x 3 The y t s form a Markov Random Field conditioned on X: P(Y X) Lafferty, McCallum, & Pereira (2001)

22 Markov Random Fields Graph G = (V,E) Each vertex v V represents a random variable y v. Each edge represents a direct probabilistic dependency. P(Y) = 1/Z exp [ c c (c(y))] c indexes the cliques in the graph c is a potential function c(y) selects the random variables participating in clique c.

23 A Simple MRF y 1 y 2 y 3 Cliques: singletons: {y 1 }, {y 2 }, {y 3 } pairs (edges); {y 1,y 2 }, {y 2,y 3 } P( y 1,y 2,y 3 ) = 1/Z exp[ 1 (y 1 ) + 2 (y 2 ) + 3 (y 3 ) + 12 (y 1,y 2 ) + 23 (y 2,y 3 )]

24 CRF Potential Functions are Conditioned on X y 1 y 2 y 3 x 1 x 2 x 3 t (y t,x) t,t+1 (y t,y t+1,x)

25 CRF Potentials are Log Linear Models t (y t,x) = b b g b (y t,x) t,t+1 (y t,y t+1,x) = a a f a (y t,y t+1,x) where g b and f a are user-defined boolean functions ( features ) Example: g 23 = [x t = o and y t = /@/]

26 Training CRFs Let = { 1, 2,, 1, 2, } be all of our parameters Let F be our CRF, so F (Y,X) = P(Y X) Define the loss function L(Y,F (Y,X)) to be the Negative Log Likelihood L(Y,F (Y,X)) = log F (Y,X) Goal: Find to minimize loss (maximize likelihood)

27 Algorithms Iterative Scaling Gradient Descent Functional Gradient Descent Gradient tree boosting

28 Gradient Descent Search From calculus we know that the minimum loss will be where d L(Y,F (Y,X)) d = L(Y,F (Y,X)) = 0 Method: := L(Y,F (Y,X))

29 Gradient Descent with Set of Training Examples We have N training examples (X i,y i ) Negative log likelihood of all N examples is the sum of the neg log likelihoods of each example The gradient of the negative log likelihood is the sum of the gradients of the neg log likelihoods of each example.

30 Gradients from Each Example example gradient (X 1,Y 1 ) L(Y 1,F (Y 1,X 1 )) (X 2,Y 2 ) L(Y 2,F (Y 2,X 2 )) (X 3,Y 3 ) L(Y 3,F (Y 3,X 3 )) (X 4,Y 4 ) L(Y 4,F (Y 4,X 4 )) := i L(Y i, F (Y i,x i ))

31 Problem: Gradient Descent is Very Slow Lafferty et al. employed modified iterative scaling but reported that it was very slow. We (and others) implemented conjugate gradient search, which is faster, but not fast enough For text-to-speech: 16 parallel processors, 40 hours per line search.

32 Functional Gradient Descent (Breiman, et al.) Standard gradient descent: final = M where m = m-1 i L(Y i, F m-1 (Y i,x i )) Functional Gradient Descent: F final = F M where m = h m, and h m is a function that approximates F i L(Y i,f m-1 (Y i,x i ))

33 Functional Gradient Descent (2) example functional gradient functional gradient example (X 1,Y 1 ) F L(Y 1,F m-1 (Y 1,X 1 )) = g 1 (X 1,g 1 ) (X 2,Y 2 ) F L(Y 2,F m-1 (Y 2,X 2 )) = g 2 (X 2,g 2 ) (X 3,Y 3 ) F L(Y 3,F m-1 (Y 3,X 3 )) = g 3 (X 3,g 3 ) (X 4,Y 4 ) F L(Y 4,F m-1 (Y 4,X 4 )) = g 4 (X 4,g 4 ) Fit h to minimize i [h(x i ) g i ] 2

34 Friedman s Gradient Boosting Algorithm F 0 = argmin i L(Y i, ) For m = 1,, M do g i := F L(Y i,f m-1 (Y i,x i )), i = 1,, N fit regression tree h := argmin f i [f(x i ) g i ] 2 m = argmin i L(Y i, F m-1 (Y i,x i ) + h(x i )) F m = F m-1 + m h m

35 Regression Trees x 1 = o yes no x 2 = t x 1 = a yes no yes no h= 0.3 h= -2.1 h= 0.8 Very fast and effective algorithms

36 Application to CRF Training Recall CRF model: (y t-1,y t,x) = a a f a (y t-1,y t,x) (y t,x) = b b g b (y t,x)] Represent (y t-1,y t,x) + (y t,x) by a set of K functions (one per class label): (l,k,x) + (k,x) = F k (l,x), k = 1,, K where F k (l,x) = m m h k,m (l,x) Each h k,m is a regression tree that tests the features {f a, g b } of the CRF The values in the leaves of the tree become the weights a and b

37 Sum of Regression Trees is Equivalent to CRF Circled Path is equivalent to expression of the form a f a a = f a = s 1 & s 4 & s 18

38 Resulting CRF Model P(Y X) = 1/Z * exp[ t F yt (y t-1,x)]

39 Forward-Backward Algorithm: Recursive Computation of Z Let (k,1) = exp F k (,X) (k,t) = k [exp F k (k,x)] (k,t 1) (k,t) = 1 (k,t) = k [exp F k (k,x)] (k,t+1) Z = k (k,t) = (,0)

40 Functional Gradient Computation Let w t (X i ) be the window of X i used by the features at time t. We get one training example for each k, l, i, and t: g k,l,i,t = log L(Y i, P(Y i X i )) F k (l,w t (X i )) Training example for F k : ( l,w t (X i ), g k,l,i,t )

41 Functional Gradient Computation (2) g k,l,i,t = log L(Y i, P(Y i X i )) F k (l,w t (X i )) = F k (l,w t (X i )) t F yt (y t-1,w t (X i )) log Z = I[y t-1 =l,y t =k] 1 Z F k (l,w t (X i )) Z

42 Functional Gradient Computation (3) 1 Z F k (l,w t (X i )) Z = 1 Z F k (l,w t (X i )) u ( v [exp F u (v,w t (X i ))] (v,t-1)) (u,t) = F k (l,w t (X i ) (l,t-1) (k,t) Z = P(y i,t =k, y i,t-1 =l X i )

43 Functional Gradient Computation (4) g k,l,i,t = log L(Y i, P(Y i X i )) F k (l,w t (X i )) = I[y i,t-1 =l, y i,t =k] P(y i,t =k, y i,t-1 =l X i ) This is our residual on the probability scale

44 Training Procedure Initialize F k = 0; k=1,,k For m = 1,, M For i = 1,, N Compute (k,t), (k,t), Z via forward/backward for (X i,y i ) Compute gradients for F k according to g k,l,i,t = I[y i,t = k, y i,t-1 = l] (l,t 1) [exp F k (l,x i )] (k,t)/z Fit regression trees h k,m to ( l,w t (X i ), g k,l,i,t ) pairs Update: F k := F k +h k,m

45 Initial Results: Training Times Gradient Boosting 1 processor: 100 iterations requires 6 hours (compared to 16*40*100 = 64,000 hours for conjugate gradient) However: Full Gradient Boosting algorithm was not implemented

46 Results: Whole words correct 5-letter window Viterbi beam width 20.

47 Whole Words: Window Sizes of 3 and 7

48 Predicting Single Letters

49 Why Gradient Boosting is More Effective Each step is large: Each iteration adds one regression tree to the potential function for each class Parameters are introduced only as necessary Combinations of features are constructed

50 Concluding Remarks Many machine learning applications can be formalized as Sequential Supervised Learning Similar issues arise in other complex learning problems (e.g., spatial and relational data) Many methods have been developed specifically for SSL, but none is perfect Gradient boosting may provide a general, offthe-shelf way of fitting CRFs.