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QV{h   fBQV{h #QV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 Z ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w tuv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? 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Y/ cFalse"   @` 8\ BЬjJ?: Y / `96"   @` 9\ BjJ?\Y: / `70"   @` :\ BXjJ?Y\/ bRain"   @` ;\ BjJ?Y bPlay"   @` <\ BjJ? Y cFalse"   @` =\ BjJ?:  Y `78"   @` >\ BhjJ?\: Y `83"   @` ?\ BjJ?\Y fOvercast  "   @` @\ BjJ? rDon t Play  "   @` A\ B\jJ?  bTrue"   @` B\ BjJ?:   `90"   @` C\ B jJ?\:  `80"   @` D\ BjJ?\ cSunny"   @` E\ B jJ? rDon t Play  "   @` F\ B)jJ?  cFalse"   @` G\ B2jJ?:   `85"   @` H\ B;jJ?\:  `85"   @` I\ BDjJ?\ cSunny"   @` J\ BNjJ? bPlay"   @` K\ BXVjJ?  cWindy"   @` L\ B`XjJ?:   fHumidity  "   @` M\ BajJ?\:  i Temperature  "   @` N\ BZjJ?\ eOutlook"   @``B O\ 0o ?ZB P\ s *1 ?ZB Q\ s *1 ?`B R\ 0o ?`B S\ 0o ?ZB T\ s *1 ?\\ZB U\ s *1 ?: : ZB V\ s *1 ?  ZB W\ s *1 ?`B X\ 0o ?ZB Y\ s *1 ?ZB Z\ s *1 ?YYZB [\ s *1 ?//ZB \\ s *1 ?ZB ]\ s *1 ?ZB ^\ s *1 ?  ZB _\ s *1 ?  ZB `\ s *1 ?] ] ZB a\ s *1 ?3 3 ZB b\ s *1 ?  ZB c\ s *1 ?  ZB d\ s *1 ?H \ 0޽h ? ̙33  0`0(  `x ` c $PP   x ` c $|4  H ` 0޽h ? ̙33  @dF(  dx d c $P    d c $  "p`PpH d 0޽h ? ̙33  Ph0(  hx h c $t{P   x h c $4P  H h 0޽h ? ̙33  ppD(  px p c $\`    p c $E   H p 0޽h ? ̙33  tP(  t~ t s *dP    t s *   H t 0޽h ? ̙33  `l0(  lx l c $XP   x l c $,  H l 0޽h ? ̙33 |\(  | | 0` 0@ j2results of: c4.5  f golf 2AR | C *Agolftree00H | 0޽h ? 33333)BH(jiTN (  R  C *AgolfdiagpF  0 0@ dc4.5  f golf decision tree represented graphically3 23AH  0޽h ? 33333)BH(jiT, l(  R  C *Agolfrule   0t zBc4.5rules  f golf: induced rules" 2"AH  0޽h ? 33333)BH(jiT  \(  \r \ S    ^B \ 6D>p@pR \ s *CkC  \ 6`  (1) b~[~[Ov:gP[Ɩy:NzS 0 (2)͑ Y (a) b_bNagĉRʑS_MRzS0 (b) NvQYO[O-N[~bĉRvOY0 (c) 1uS_MRzSTĉROYubevzS0 v0RĉRl gOY:Nbk0"|(z$~ H \ 0޽h ? ̙33y___PPT10Y+D=' = @B +  Hi(  Hr H S    ^B H 6D>p@pR H s *CkC H 6, ID3 Decision Tree Learning Algorithm ID3(Examples, Target, Attributes) Create a root node If all Examples have the same Target value, give the root this label Else if Attributes is empty label the root according to the most common value Else begin Calculate the information gain for each attribute, according to the average entropy formula Select the attribute, A, with the lowest average entropy (highest information gain) and make this the attribute tested at the root end &#     $# H H 0޽h ? ̙33y___PPT10Y+D=' = @B +?  f^(  r  S (   ^B  6D>0@0R  s *CkC  6(6 d For each possible value, v, of this attribute Add a new branch below the root, corresponding to A = v Let Examples(v) be those examples with A = v If Examples(v) is empty, make the new branch a leaf node labelled with the most common value among Examples Else let the new branch be the tree created by ID3(Examples(v), Target, Attributes - {A}) end t/ /   // "H  0޽h ? ̙33y___PPT10Y+D=' = @B +  XF(  Xx X c $%P    X c $,  "p`PpH X 0޽h ? ̙33  0 F(   x   c $͌`      c $ X  "P@0 (H   0޽h ? ̙33   8J(  8~ 8 s *\P    8 c $eP0   H 8 0޽h ? ̙33  @<(  @~ @ s *jP   ~ @ s *k  H @ 0޽h ? ̙33 C(    0OP`p g/Common Decision Tree / Rule-Induction Software:0 20A<  0Б`  c4.5 DOS-based decision tree and rule induction software based on Ross Quinlan s ID3 algorithm. (Free) c5.0 / See-5 Windows-compatible version of c4.5. More flexible analysis. Replaces ID3 with a better-performing, less resource-intensive algorithm. Evaluation copy (maximum 500 cases) is free. Non-case-limited version is $475. SPSS AnswerTree (UNICEF, financial firms) Gives users the choice of several algorithms in constructing trees, depending on type of data. Impressive output, but very expensive. ($1999) 2A[AAA AANAA*AAAAH  0޽h ? 33333)BH(jiT8 x(  J  C "A see5p@L=  0p 0@ =See 5 2AI  0ߑp` See 5 is very similar to c4.5. The basic interface displays the data set currently in use, along with all the constituent files that you have created and/or See 5 has computed 2CH  0޽h ? 33333)BH(jiT  d(  dr d S 0n   ^B d 6D>0@0R d s *CkC  d 6@  1986, Schlimmer T Fisher NID4f[`N{l, /fNyX_f[`N{l0NNO9eID3{l (Wk*NSvQV{h~pR^N|Rh0k*Nh1uhQ*ghKm^\'`0@0R h s *CkC h 6v j eQ: QV{h N*N[O Q: QV{h (1) 勞[O/fckO ckOpeR1 &TR SOpeR10 (2) Yg[OhQ:NckObSO RԏVQV{h0 (3) &TR (a) {ggOo`Rpe0 (b) [O-NQsvk*N^\'`0k*N

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QV{h   fBQV{h #QV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 Z ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w uv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? 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Y/ cFalse"   @` 8\ BBjJ?: Y / `96"   @` 9\ BKjJ?\Y: / `70"   @` :\ BTjJ?Y\/ bRain"   @` ;\ B]jJ?Y bPlay"   @` <\ BfjJ? Y cFalse"   @` =\ BojJ?:  Y `78"   @` >\ BxjJ?\: Y `83"   @` ?\ BājJ?\Y fOvercast  "   @` @\ Bx|jJ? rDon t Play  "   @` A\ BjJ?  bTrue"   @` B\ B,jJ?:   `90"   @` C\ BjJ?\:  `80"   @` D\ BTjJ?\ cSunny"   @` E\ BЩjJ? rDon t Play  "   @` F\ BjJ?  cFalse"   @` G\ B伀jJ?:   `85"   @` H\ BƀjJ?\:  `85"   @` I\ BTπjJ?\ cSunny"   @` J\ B؀jJ?_U0ƼcPfNOh+'0 hp  AAI06-2ʷֲƼ71Microsoft PowerPoint@ G @\Xk@`p$@`lGg  K  y--$x bPlay"   @` K\ BjJ?  cWindy"   @` L\ BjJ?:   fHumidity  "   @` M\ BjJ?\:  i Temperature  "   @` N\ BjJ?\ eOutlook"   @``B O\ 0o ?ZB P\ s *1 ?ZB Q\ s *1 ?`B R\ 0o ?`B S\ 0o ?ZB T\ s *1 ?\\ZB U\ s *1 ?: : ZB V\ s *1 ?  ZB W\ s *1 ?`B X\ 0o ?ZB Y\ s *1 ?ZB Z\ s *1 ?YYZB [\ s *1 ?//ZB \\ s *1 ?ZB ]\ s *1 ?ZB ^\ s *1 ?  ZB _\ s *1 ?  ZB `\ s *1 ?] ] ZB a\ s *1 ?3 3 ZB b\ s *1 ?  ZB c\ s *1 ?  ZB d\ s *1 ?H \ 0޽h ? ̙33   0 0`0(  `x ` c $P   x ` c $$  H ` 0޽h ? ̙33  0 @dF(  dx d c $2P    d c $C  "p`PpH d 0޽h ? ̙33  0 Ph0(  hx h c $[P   x h c $\P  H h 0޽h ? ̙33  0 ppD(  px p c $gw    p c $Lo    H p 0޽h ? ̙33  0 tP(  t~ t s *t    t s *     H t 0޽h ? ̙33  0 `l0(  lx l c $4   x l c $"  H l 0޽h ? ̙33N 0 (  R  C *AgolfdiagpF  0 0@ dc4.5  f golf decision tree represented graphically3 23AH  0޽h ? 33333)BH(jiT, 0 l(  R  C *Agolfrule   0, zBc4.5rules  f golf: induced rules" 2"AH  0޽h ? 33333)BH(jiT  0 \(  \r \ S    ^B \ 6D>p@pR \ s *CkC  \ 6d{`  (1) b~[~[Ov:gP[Ɩy:NzS 0 (2)͑ Y (a) b_bNagĉRʑS_MRzS0 (b) NvQYO[O-N[~bĉRvOY0 (c) 1uS_MRzSTĉROYubevzS0 v0RĉRl gOY:Nbk0"|(z$~ H \ 0޽h ? ̙33y___PPT10Y+D=' = @B +  0 Hi(  Hr H S 4Ɓ   ^B H 6D>p@pR H s *CkC H 6Ɂ, ID3 Decision Tree Learning Algorithm ID3(Examples, Target, Attributes) Create a root node If all Examples have the same Target value, give the root this label Else if Attributes is empty label the root according to the most common value Else begin Calculate the information gain for each attribute, according to the average entropy formula Select the attribute, A, with the lowest average entropy (highest information gain) and make this the attribute tested at the root end &#     $# H H 0޽h ? ̙33y___PPT10Y+D=' = @B +?  0 f^(  r  S    ^B  6D>0@0R  s *CkC  6 6 d For each possible value, v, of this attribute Add a new branch below the root, corresponding to A = v Let Examples(v) be those examples with A = v If Examples(v) is empty, make the new branch a leaf node labelled with the most common value among Examples Else let the new branch be the tree created by ID3(Examples(v), Target, Attributes - {A}) end t/ /   // "H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0 XF(  Xx X c $P    X c $  "p`PpH X 0޽h ? ̙33  0 0 F(   x   c $`      c $@3  "P@0 (H   0޽h ? ̙33   0 8J(  8~ 8 s *<P    8 c $0EP0   H 8 0޽h ? ̙33  0 @<(  @~ @ s *lIP   ~ @ s *O  H @ 0޽h ? ̙338 0 x(  J  C "A see5p@L=  0 0@ =See 5 2AI  0 "p` See 5 is very similar to c4.5. 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Դǹ2Y5cBmGFA0 F] dfϟj [+m`y{}#;u 3xӡ  "$&(+-+/?1S3g5{79;=?ACFH/JCLWNkPRTVXZ\^ ac3eGg[iokmoqsuwy|#~7K_(s1Y( @  5 )Equation Equation.30 Microsoft lQ_ 3.0t/ 0DTimes New RomanTT2ܖ 0ܖD[SOes New RomanTTx--'@Times New Roman-.  2 q2009."System(-@Times New Roman-.  2 q-.-@Times New Roman-.  2 q4.-@Times New Roman-.  2 q-.-@Times New Roman-.  2 q1.--. 2 q@ʷֲ.--. 2 qM ߼˹.-@Times New Roman-.  2 q1.-κ-. 332  .-κ-. 332 Eѧϰ .-κ-. 33 2 u(2)9.--. 2 EFʷֲ.--. 2 U= пԺ.---%--'՜.+,0    XĻʾܿѧʵU33 ?Times New Roman _GB2312 Courier NewGulimSymbol WingdingsArial WeQV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 Z ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w uv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? 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QV{h   fBQV{h #QV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 Z ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w uv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? " "*"*  d"Discrete vs. Continuous Attributes##  # TSuppose A is an attribute with continuous values e.g., Humidity in Golf example A1, A2, .., Am are the values in the training data, in increasing order For each Aj TL = Partition of training data whose A values are <= Aj TH = Partition of training data whose A values are > Aj Calculate Gain(T,A) or GainRatio(T,A) Select the best partitionp1T1"""*"*"*D"*"*5"*"*4"*" " "p  7  6     f6Decision Trees  Advantages   oComprehensibility Fast classification Mature technology There are many systems C4.5 [Quinlan, 1993] C5.0(See-5)*888 8  c   wx.QV{hID4 $ /QV{hID4 $ 0QV{hID4 $ 1QV{hID5 $ 2 ID5{l $ 3ID5^\'`cGS{l $  }QV{h $ 4R_~f[`Nv{t $   5Goldf[`Nt $  6Goldf[`Nt $  8Goldf[`Nt $  9Goldf[`Nt $  :Goldf[`Nt $  ; g>NƋ{l( < g>NƋ{l( = !jWct|~ > !jWct|~ ? !jWct|~ 7Valiantf[`Nt $  AValiantf[`Nt $  BValiantf[`Nt $  CValiantf[`Nt $  /y      !"#$%&'()*+,-./0123O 2 0 vn(  r  S />P  > ^B  6D>P@PR  s *CkC  6D< .GP X/f[Ozz N*Ni_/f XvN*NP[Ɩ0Yg[O(Wi_-NR:NckO &TR:NSO0i_h:y/fNyi_vc i_{|/fN~i_h:y0f[`N!jW/fi_{|v gHevSf[`N'`0Valiant f[`NtNBl[vhi_v_}Yя [FIKLMNOPQSTUVZXa Z\Y "bcfikqrestuvwx{|}~ /2$n}cތ(L$$$$b$˲ƚ[Z&ډG g$b$ {0)r'Wt4^r$$$$$$$$R$2c ]}@MYe$$$$$$b$:fT@Jzl$$$$$$$$$R$卢Lo+(| 0AA@ g4BdBd<- 0ppp@ <4!d!d w 0T-<4BdBd w 0T-uʚ;2Nʚ;<4dddd  x 0 V0___PPT10 pp___PPT9 4 ,H?.+ S_ i ؚ~N]zfO =:,{mQz R_~f[`N(2)$ !  QV{h   fBQV{h #QV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 ion.30 Microsoft lQ_ 3.0/ 0DTimes New RomanTT41ܖ 0ܖD[SOes New RomanTT41ܖ 0ܖ DN[_GB2312RomanTT41ܖ 0ܖ10DCourier NewmanTT41ܖ 0ܖ1@DGulimr NewmanTT41ܖ 0ܖ"PDSymbol NewmanTTZ ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w uv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? " "*"*  d"Discrete vs. Continuous Attributes##  # TSuppose A is an attribute with continuous values e.g., Humidity in Golf example A1, A2, .., Am are the values in the training data, in increasing order For each Aj TL = Partition of training data whose A values are <= Aj TH = Partition of training data whose A values are > Aj Calculate Gain(T,A) or GainRatio(T,A) Select the best partitionp1T1"""*"*"*D"*"*5"*"*4"*" " "p  7  6     f6Decision Trees  Advantages   oComprehensibility Fast classification Mature technology There are many systems C4.5 [Quinlan, 1993] C5.0(See-5)*888 8  c   wx.QV{hID4 $ /QV{hID4 $ 0QV{hID4 $ 1QV{hID5 $ 2 ID5{l $ 3ID5^\'`cGS{l $  }QV{h $ 4R_~f[`Nv{t $   5Goldf[`Nt $  6Goldf[`Nt $  8Goldf[`Nt $  9Goldf[`Nt $  :Goldf[`Nt $  ; g>NƋ{l( < g>NƋ{l( = !jWct|~ > !jWct|~ ? !jWct|~ 7Valiantf[`Nt $  AValiantf[`Nt $  BValiantf[`Nt $  CValiantf[`Nt $  /y      !"#$%&'()*+,-./0123r,CC1EZ( @  7 )Equation Equat41ܖ 0ܖ`DWingdingswmanTT41ܖ 0ܖpDArialngswmanTT41ܖ 0ܖ"DWebdingswmanTT41ܖ 0ܖDNSeeOngswmanTT41ܖ 0ܖ@ .  @n?" dd@  @@``  qi     \               %f<<@ B ;Y=> [FIKLMNOPQSTUVZXa Z\Y "bcfikqrestuvwx{|}~ /2$n}cތ(L$$$$b$˲ƚ[Z&ډG g$b$ {0)r'Wt4^r$$$$$$$$R$2c ]}@MYe$$$$$$b$:fT@Jzl$$$$$$$$$R$卢Lo+(| 0AA@ g4BdBdU1 0 ppp@ <4!d!d w 0T1<4BdBd w 0T1uʚ;2Nʚ;<4dddd  x 0 V0___PPT10 pp___PPT9 4 ,H?.+ S_ i ؚ~N]zfO =:,{Nz R_~f[`N(2)2 ' 6' 6  QV{h   fBQV{h #QV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 Z ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w uv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? " "*"*  d"Discrete vs. Continuous Attributes##  # TSuppose A is an attribute with continuous values e.g., Humidity in Golf example A1, A2, .., Am are the values in the training data, in increasing order For each Aj TL = Partition of training data whose A values are <= Aj TH = Partition of training data whose A values are > Aj Calculate Gain(T,A) or GainRatio(T,A) Select the best partitionp1T1"""*"*"*D"*"*5"*"*4"*" " "p  7  6     f6Decision Trees  Advantages   oComprehensibility Fast classification Mature technology There are many systems C4.5 [Quinlan, 1993] C5.0(See-5)*888 8  c   wx.QV{hID4 $ /QV{hID4 $ 0QV{hID4 $ 1QV{hID5 $ 2 ID5{l $ 3ID5^\'`cGS{l $  }QV{h $ 4R_~f[`Nv{t $   5Goldf[`Nt $  6Goldf[`Nt $  8Goldf[`Nt $  9Goldf[`Nt $  :Goldf[`Nt $  ; g>NƋ{l( < g>NƋ{l( = !jWct|~ > !jWct|~ ? !jWct|~ 7Valiantf[`Nt $  AValiantf[`Nt $  BValiantf[`Nt $  CValiantf[`Nt $  /y      !"#$%&'()*+,-./0123S  0 zr0  (   l  C PpPH  0޽h ? ̙33y___PPT10Y+D=' = @B +rS/1EZ( @  7 )Equation Equation.30 Microsoft lQ_ 3.0/ 0DTimes New RomanTT41ܖ 0ܖD[SOes New RomanTT41ܖ 0ܖ DN[_GB2312RomanTT41ܖ 0ܖ10DCourier NewmanTT41ܖ 0ܖ1@DGulimr NewmanTT41ܖ 0ܖ"PDSymbol NewmanTT41ܖ 0ܖ`DWingdingswmanTT41ܖ 0ܖpDArialngswmanTT41ܖ 0ܖ"DWebdingswmanTT41ܖ 0ܖDNSeeOngswmanTT41ܖ 0ܖ@ .  @n?" dd@  @@``  qi     \               %f<<@ B ;Y=> [FIKLMNOPQSTUVZXa Z\Y "bcfikqrestuvwx{|}~ /2$n}cތ(L$$$$b$˲ƚ[Z&ډG g$b$ {0)r'Wt4^r$$$$$$$$R$2c ]}@MYe$$$$$$b$:fT@Jzl$$$$$$$$$R$卢Lo+(| 0AA@ g4BdBdU1 0 ppp@ <4!d!d w 0T1<4BdBd w 0T1uʚ;2Nʚ;<4dddd  x 0 V0___PPT10 pp___PPT9 4 ,H?.+ S_ i ؚ~N]zfO =:,{Nz R_~f[`N(2)2 ' 6' 6  QV{h   fBQV{h #QV{h$ ]Decision Trees   Tree structure Node = query on attribute Link = attribute value Leaf = class Recursively separate data into sub-populations Prediction: Traverse path, yield most probable class<>d > d   $ CLS{l$ QV{h $ %QV{h $ 'QV{h $ (QV{h $ +QV{h $ jGolf Example (1)    m Definitions  rInformative Establishes Good decision trees Entropy Measure how informative is a node Definition: P=(p1,p2,& ,pn) Then Entropy of P is : I(P)= -(p1*log(p1) + p2*log(p2) +& + pn*log(pn) )  .W .%2fm  =      n Definitions (Cont.)  Entropy For example: If P=(0.5,0.5) I(P)=1 If P=(2/3,1/3) I(P)=0.92 If P= (1,0) I(P)=0 What do u see ? J Entropy of previous golf example I(P)=I(9/14,5/14)=-(9/14*log(9/14) + 5/14*log(5/14)) = 0.94 Z ZM" ZZ!ZXZ > ! W    o Definitions (Cont.)  Entropy of an attribute For example: I(Outlook,T)= 5/14 * I(2/5,3/5) + 4/14 * I(4/4,0) + 5/14 * I)(3/5,2/5) = 0.694 While I(windy,T)=0.892 P    p Information Gain (2)   ZInformation Gain Gain(T, A) = I(T)  I(T,A) I(T) = expected information for distinguishing classes = -(p/(p+n)log2(p/(p+n))+n/(p+n)log2(n/(p+n))) I(T, A) = expected information of tree with A as root = Si(pi+ni)/(p+n)*I(Ti) p, n: number of positive/negative training data pi, ni: number of positive/negative training data in the training data Ti partitioned by the attribute value Ai Select an attribute with highest information gain Prefer an attribute A with smallest I(T,A) i.e., Prefer the attribute that makes non-uniform distribution* R76"2" j""M C" 1"*"*B"*%"*2"j"@   <   q Golf Example (2)    The initial information I(T) = I(9/14, 5/14) = 0.94 The information for Outlook attribute Sunny, Overcast, Rain I(T,Outlook) = 5/14*I(2/5,3/5) + 4/14*I(4/4,0/4) + 5/14*I(3/5,2/5) = 0.694 Gain(T,Outlook) = 0.94-0.694 = 0.246 The information for Windy attribute True, False I(T,Windy) = 6/14*I(3/6,3/6) + 8/14*I(6/8,2/8) = 0.892 Gain(T,Windy) = 0.94-0.892 = 0.048 select Outlook attribute for the first partitionPP&PP$PfP1P""&""$"f"1"  r Definitions (Cont.)  Entropy of an attribute Definition of Gain() Gain(X,T) = I(T)  I(X,T) Gain(Outlook,T) = I(T)  I(Outlook,T) = 0.94  0.694 = 0.246 Gain(Windy,T) = I(T)  I(Windy,T) = 0.94  0.892 = 0.048 We say, Outlook is more informative than Windy, Why? JZZZZ0Z Z  9 w uv,ID3 {l ( &QV{h $ QV{h $ kC4.5 Extensions  C4.5 is an extensions of ID3 accounts for Depth-first strategy is used Unavailable values Ex: only given Outlook to be Sunny Continuous attribute value ranges Ex: humidity is greater than 75 Pruning of decision trees Rule derivation *Z0Z#Z"Z Z+Z*/" +  ^Decision Trees: Training   [C4.5, Quinlan, 1993] Generate_tree(R, C, T) // R: set of non-categorical attribute // C: categorical attribute, T: training data if T has the same categorical value then return a single node with the value if R={} then return a single node with most frequent categorical value in T A = attribute with highest information gain, Gain(T,A) among R Let A1, A2, .., Am the attribute values of A Let T1, T2, .., Tm the subsets of T partitioned by of Ai return a node with A and m links as follows for i = 1 to m do Generate_tree(R-{A}, C, Ti)U;} " ""U" "  "  "  *  "  *  "  *  "  *  "  *  "  * % "  * ? " "*"*  d"Discrete vs. Continuous Attributes##  # TSuppose A is an attribute with continuous values e.g., Humidity in Golf example A1, A2, .., Am are the values in the training data, in increasing order For each Aj TL = Partition of training data whose A values are <= Aj TH = Partition of training data whose A values are > Aj Calculate Gain(T,A) or GainRatio(T,A) Select the best partitionp1T1"""*"*"*D"*"*5"*"*4"*" " "p  7  6     f6Decision Trees  Advantages   oComprehensibility Fast classification Mature technology There are many systems C4.5 [Quinlan, 1993] C5.0(See-5)*888 8  c   wx.QV{hID4 $ /QV{hID4 $ 0QV{hID4 $ 1QV{hID5 $ 2 ID5{l $ 3ID5^\'`cGS{l $  }QV{h $ 4R_~f[`Nv{t $   5Goldf[`Nt $  6Goldf[`Nt $  8Goldf[`Nt $  9Goldf[`Nt $  :Goldf[`Nt $  ; g>NƋ{l( < g>NƋ{l( = !jWct|~ > !jWct|~ ? !jWct|~ 7Valiantf[`Nt $  AValiantf[`Nt $  BValiantf[`Nt $  CValiantf[`Nt $  /y      !"#$%&'()*+,-./0123r7U1