Decision Trees. Ruy Luiz Milidiú
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1 Decision Trees Ruy Luiz Milidiú
2 Resumo Objetivo Examinar o conceito de Árvores de Decisão e suas aplicações Sumário Surpresa e Informação Entropia Informação Mútua Divergência e Entropia Cruzada
3 Ganho de Informação H(X,Y) = H(X) + H(Y X) H(X,Y) = H(Y) + H(X Y) H(X) + H(Y X) = H(Y) + H(X Y) H(X) H(X Y) = H(Y) H(Y X) I(X,Y) = H(Y) H(Y X) IG(X) = H(Y) H(Y X)
4 Seleção de Atributos (X 1,,X k,y) IG(X i ) = H(Y) H(Y X i ) i=1,,k melhor atributo IG(X j ) = max i=1,,k {H(Y) H(Y X i )} IG(X j ) = H(Y) min i=1,,k {H(Y X i )}
5 wait for a table? 1. Alternate alternative restaurant nearby? 2. Bar comfortable bar area to wait in? 3. Fri/Sat Friday or Saturday? 4. Hungry are we hungry? 5. Patrons people in (None, Some, Full) 6. Price ($, $$, $$$) 7. Raining raining outside? 8. Reservation have we made a reservation? 9. Type (French, Italian, Thai, Burger) 10. WaitEstimate (0-10, 10-30, 30-60, >60)
6 Learning data Examples described by attribute values (Boolean, discrete, continuous) E.g., situations where I will/won't wait for a table: Classification of examples is positive (T) or negative (F)
7 Learning data NO attribute is known 6 positive (T) and 6 negative (F) H(Y) = H(6/12) = H(1/2) = 1
8 Learning data PATRONS attribute is known 6 positive (T) and 6 negative (F) H(Y Patron) =?
9 Learning data PATRONS = none 0 positive (T) and 2 negative (F) H(Y none) = H(0/2) = H(0) = 0
10 Learning data PATRONS = some 4 positive (T) and 0 negative (F) H(Y some) = H(4/4) = H(1) = 0
11 Learning data PATRONS = full 2 positive (T) and 4 negative (F) H(Y full) = H(1/3) = (1/3).lg(3) + (2/3).lg(3/2) = lg(3) 2/3 =.92
12 Learning data PATRONS attribute is known H(Y Patron) = (2/12).H(Y none) +(4/12).H(Y some) +(6/12).H(Y full)
13 Learning data PATRONS attribute is known H(Y Patron) = (2/12).0(Y none) +(4/12).0(Y some) +(6/12)..92ull)
14 Learning data PATRONS attribute is known H(Y Patron) =.46).0(Y none) +(4/12).0(Y some) +(6/12)..92ull)
15 Choosing an attribute A good attribute splits the examples into subsets that are (ideally) "all positive" or "all negative" IG(Patrons) = =.54 IG(Type) = 1-1 =.0
16 Learned tree Decision tree learned from the 12 examples Substantially simpler than true tree---a more complex hypothesis isn t justified by small amount of data
17 Decision tree learning Aim: find a small tree consistent with the training examples Idea: (recursively) choose "most significant" attribute as root of (sub)tree
18 Worst Case Complexity n examples k attributes complete tree of height k n.(k-l) total work per tree level l n.[k+(k-1)+ +1] total work O(n.k 2 )
19 Learning subset PATRONS = full 2 positive (T) and 4 negative (F) H(Y full) = H(1/3) = (1/3).lg(3) + (2/3).lg(3/2) = lg(3) 2/3 =.92
20 Learning subset PATRONS = full TYPE is known 2 positive (T) and 4 negative (F) H(Y full,type) = (2/6).H(1/2) + (1/6).H(0) + (2/6).H(1/2) + (1/6).H(0)
21 Learning subset PATRONS = full TYPE is known 2 positive (T) and 4 negative (F) H(Y full,type) = (2/6).(1) + (1/6).(0) + (2/6).(1) + (1/6).(0) =.66
22 Naive-DT H(Y full,tai) = -P(T full,tai). lg( P(T full,tai) ) -P(F full,tai). lg( P(F full,tai) ) P(T full,tai) = P(tai full,t). P(T full) / P(tai full) P(F full,tai) = P(tai full,f). P(F full) / P(tai full) P(T full,tai) P(tai full,t). P(T full) P(F full,tai) P(tai full,f). P(F full) P(tai full) = P(tai full,t). P(T full) + P(tai full,f). P(F full)
23 Naive-DT Naive Bayes Hypotesis P(tai full,t) = P(tai T) = P(tai,T)/ P(T) P(tai full,f) = P(tai F) = P(tai,F)/ P(F) Initial Set Counts Subset Counts P(T full,tai) P(tai T). P(T full) = (2/6).(4/6) = 8/36 P(F full,tai) P(tai F). P(F full) = (2/6).(2/6) = 4/36 P(tai full) = (8/36) + (4/36) = 12/36 = 1/3 H(Y full,tai) = H(8/12) = H(2/3) = H(1/3) =.92 H(Y full,type) = P(tai full).h(y full,tai) +... H(Y full,type) = (1/3).(.92) +...
24 Naive-DT Initial Set Counts Naive Bayes Hypotesis P(french full,t) = P(french T) = P(french,T)/ P(T) P(french full,f) = P(french F) = P(french,F)/ P(F) Subset Counts P(T full, french) P(french T). P(T full) = (1/6).(4/6) = 4/36 P(F full, french) P(french F). P(F full) = (1/6).(2/6) = 2/36 P(french full) = (4/36) + (2/36) = 6/36 = 1/6 H(Y full, french) = H(4/6) = H(2/3) = H(1/3) =.92 H(Y full,type) = P(tai full).h(y full,tai) + P(french full).h(y full,french) +... H(Y full,type) = (1/3).(.92) + (1/6).(.92) +...
25 Initial Counting small for classe=1..c total[classe] = 0 small for feature=1..k for valor=1..v[k] count[feature, valor, classe] = 0 for i=1..n total[y[i]] ++ for feature=1..k count[feature, x[i,feature], y[i]] ++
26 Subset counting
27 Decision trees the UNKNOWN true tree
28 Expressiveness Decision trees can express any function of the input attributes. E.g., for Boolean functions, truth table row path to leaf: Trivially, there is a consistent decision tree for any training set with one path to leaf for each example (unless f nondeterministic in x) but it probably won't generalize to new examples Prefer to find more compact decision trees
29 Hypothesis spaces How many distinct decision trees with n Boolean attributes? = number of Boolean functions = number of distinct truth tables with 2 n rows = 2 2n E.g., with 6 Boolean attributes, there are 18,446,744,073,709,551,616 trees
30 Hypothesis spaces How many distinct decision trees with n Boolean attributes? = number of Boolean functions = number of distinct truth tables with 2 n rows = 2 2n E.g., with 6 Boolean attributes, there are 18,446,744,073,709,551,616 trees How many purely conjunctive hypotheses (e.g., Hungry Rain)? Each attribute can be in (positive), in (negative), or out 3 n distinct conjunctive hypotheses More expressive hypothesis space increases chance that target function can be expressed increases number of hypotheses consistent with training set may get worse predictions
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