Natural Language Processing using Machine Learning
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1 Natural Language Processing using Machine Learning Miguel Almeida, André Martins, Afonso Mendes Priberam Labs December 18, 2012
2
3 One of the easiest NLP problems One of the simplest classifiers: Naïve Baes Also used for spam detection Relies on two simple concepts: Baes Rule Conditional independence
4 (Baes rule) For an random variables A and B: P A B = P B A P(A) P(B)
5 (conditional independence) Independence between variables A and B: Knowing A does not give information about B and vice-versa P A, B = P A P(B) Conditional independence of A and B, given C: If we know C, knowing A does not give information about B and vice-versa P A, B C = P A C P(B C)
6 Training data: Wikipedia 3.3 GB of Portuguese text (PT) 5.6 GB of Spanish text (ES) 8.4 GB of French text (FR) Some preprocessing involved Remove XML markup to keep onl the text Remove uninformative sections (e.g. references) Transform everthing to lowercase
7 x = input (string) Example: x = eu fui = output (language) belongs to {PT, ES, FR} Eas to add more languages (use more Wikipedias...) Our goal: given the string x, find the language which is most likel maximize P( x) Known as Maximum A Posteriori (MAP) estimator
8 x = string = language Goal: maximize P( x) Baes Rule = argmax P( x) = argmax P x P() P(x) P(x) does not depend on = argmax P(x )P()
9 x = string = language Goal: find = argmax P(x )P() How do we compute P()? How do we compute P(x )?
10 How do we find P()? (called prior) In this case, essentiall two choices: All languages have the same prior (uniform prior) P( = PT) = P( = ES) = P( = FR) = 1/3 Estimate prior from the data P() a (size of data of language ) In our case, we use the uniform prior Since we want the argmax, we can forget about the prior argmax P x P = argmax P x 1 3 = argmax P(x )
11 (MAP with uniform prior = ML) Maximum A Posteriori Estimator = argmax P( x) Baes Rule = argmax = argmax P x P() P(x) P x P Drop P(x) Uniform Prior Maximum Likelihood Estimator = argmax P x
12 How do we find P(x )? (called class conditional) For example, what s P( eu fui PT)? Mabe count how often eu fui appears in the PT data... What about this one? P( eu fui à praia com os meus amigos, mas começou a chover por isso fomos ao cinema ver o Shrek, que é um filme de animação PT) Probabl never appears in the training set for an language! Most non-small sentences would get P(x ) = 0 for ever What would be the best???
13 Slight change of notation: P "eu fui" PT) = P("eu_", "u_f", "_fu", "fui" PT) i.e. we represent the sentence with all its triplets this is completel equivalent to the original formulation Naïve Baes: assume conditional independence P "eu fui" PT) = P("eu_", "u_f", "_fu", "fui" PT) = P("eu_" PT) P("u_f" PT) P("_fu" PT) P("fui" PT)
14 We just need to estimate probabilities of the form P( abc ), where abc are an three characters Can be estimated from train data just b counting: Example: P(" abc" PT) fui appears 10 2 times in PT train data there are 10 6 triplets in PT train data then, P( fui PT) = 10-4 #"abc"in PT train data # triplets in PT train data
15 No problem with long sentences! P( eu fui à praia com os meus amigos, mas começou a chover por isso fomos ao cinema ver o Shrek, que é um filme de animação PT) = = P( eu_ PT) P( u_f PT) P( _fu PT)... P( açã PT) P( ção PT) eu_ probabl appears in all languages same for u_f, _fu, fui, and so on if a few triplets do not appear in a language, that can be solved with smoothing
16 log (log trick) Each P( abc ) probabilit of the order of 10-4 to 10-7 Sentence with N characters has (N-2) triplets Sentence with 60 characters (10-12 words) has probabilit of order (10-4 to 10-7 ) 58 = to Ver eas to get underflow errors! Solution: use log-probabilities, log( ) = -406*log(10) = = , no risk of underflow, and same argmax: arg max P( x ) arg max log Products of probabilities become sums of log-probabilities P log P( x ) ("eu_" PT) P("u_f" PT) P("_fu" PT) P("fui" PT) P("eu_" PT) log P("u_f" PT) log P("_fu" PT) log P("fui" PT)
17 Demo time! Feel free to suggest a few sentences to test...
18 Wh is não sei Portuguese? log[p(x )] PT ES FR não -7,561-14,777-15,513 ão_ -5,655-10,812-11,252 o_s -6,779-7,234-9,674 _se -6,000-5,997-6,571 sei -9,464-10,188-8,589 não sei -35,459-49,008-51,599 Best: PT, second best: ES large log-ratio high confidence in result log ratio log P x = PT P x = ES = log (P x = PT log x = ES =
19 Wh is eu fui French? log[p(x )] PT ES FR eu_ -7,417-11,610-8,198 u_f -10,024-10,196-9,014 _fu -7,960-7,067-8,366 fui -12,456-13,531-11,640 eu fui -37,857-42,404-37,218 Best: FR, second best: PT small log-ratio low confidence in result log ratio = log (P x = FR log x = PT = 0,639
20 Naïve Baes (summar) Goal: maximize P( x) Baes Rule, drop P(x) from argmax, uniform prior maximize P(x ) Assume features conditionall independent: P x 1, x 2,, x N = P x 1 P x 2 P(x N ) Advantage: number of parameters to estimate P( fui ): eas to estimate from train data (just count) P( eu fui à praia com... ): hard (usuall impossible) to estimate directl Usuall NOT a good model of the data! Is ( _fu PT) reall independent of ( fui PT)? Sometimes, the best model which can be used in reasonable time... In this case, it works well even though it is not a perfect model
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