( D) I(2,3) I(4,0) I(3,2) weighted avg. of entropies
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1 Decision Tree Induction using Information Gain Let I(x,y) as the entropy in a dataset with x number of class 1(i.e., play ) and y number of class (i.e., don t play outcomes. The entropy at the root, i.e., before partitioning the data set, is: 9 9 Info( D) I(9,) log ( ) log ( ) We will consider the four attributes in order to partition the data set and start building a decision tree. 1) Outlook (as the candidate attribute) at the root 4 Info ( D) I(,3) I(4,0) I(3,) weighted avg. of entropies log log (0) sunny overcast rain 3 3 log log 4 (0.9710) (0) (0.9710) /3 4/0 3/ Information Gain ()= =0.468 ) Temperature at the root Infotemperature( D) I(,) I(4,) I(3,1) (1) log log log log (1) (0.9183) (0.8113) Information Gain (temperature) ) Humidity at the root Info humidity (D) = 7 I(3,4) + 7 I(6,1) = 7 (0.98) + 7 (0.917) = temperature hot mild cool / 4/ 3/1 humidity high normal 3/4 6/1 IG(humidity) = =0.118
2 4) Windy at the root windy Info windy (D) = 6 I(3,3) + 8 I(6,) true false = 6 (1) + 8 (0.8113) = /3 6/ IG(windy) = = Now, is selected to partition the dataset because it has the highest information gain (0.468). Let s grow the tree at the sunny branch first. Note that no splitting occurs at the overcast branch because its leaves are pure. Splitting Sunny branch whose entropy is I(,3) = ) Temperature at =sunny Info temperature (D sunny ) = I(0,) + I(1,1) + 1 I(1,0) = 0.4 IG(temperature) = = sunny temperature hot mild cool 0/ 1/1 1/0 ) Humidity at =sunny Info humidity (D sunny ) = 3 I(0,3) + I(,0) = 0 IG(humidity) = = sunny humidity high normal 0/3 /0 3) Windy at =sunny Info windy (D sunny ) = I(1,1) + 3 I(1,) = IG(windy) = =0.0 sunny windy true false 0/ 1/0 Humidity has the highest information gain (0.9710). Therefore, humidity is selected to split the tree. The leaves are pure at the humidity branch.
3 Let us then grow the tree at the rain branch. Splitting Rain branch whose entropy is I(3,) = ) Temperature at =rain Info temperature (D rain ) = 0 I(0,0) + 3 I(,1) + I(1,1) = IG(temperature) = = 0.0 ) Humidity at =rain Info humidity (D rain ) = I(1,1) + 3 I(,1) = IG(humidity) = = 0.0 rain rain temperature hot mild cool 0/0 /1 1/1 humidity high normal 1/1 /1 3) Windy at =rain Info windy (D rain ) = 3 I(0,3) + I(,0) = 0 IG(windy) = = rain windy true false 0/3 /0 Windy has the highest information gain (0.9710). Therefore, windy is selected to split the tree. The leaves are pure at the windy branch. The decision tree is now complete. sunny humidity high normal overcast 4/0 rain windy true false 0/3 /0 0/3 /0
4 Decision Tree Induction using Gain Ratio The entropy at the root, i.e., before partitioning the data set, is: 9 9 Info( D) I(9,) log ( ) log ( ) ) Outlook (as the candidate attribute) at the root 4 Info ( D) I(,3) I(4,0) I(3,) Information Gain ()= =0.468 SplitInfo (D) = log 4 log 4 log = Gain Ratio() = 0.468/1.774=0.164 ) Temperature at the root Infotemperature ( D) I(,) I(4,) I(3,1) Information Gain (temperature) SplitInfo temperature (D) = 4 log 4 6 log 6 4 log 4 = 1.67 Gain Ratio(temperature) = 0.09/1.67= ) Humidity at the root Info humidity (D) = 7 I(3,4) + 7 I(6,1) = IG(humidity) = =0.118 SplitInfo humidity (D) = 7 log 7 7 log 7 = 1 Gain Ratio(humidity) = 0.118/1= ) Windy at the root Info windy (D) = 6 I(3,3) + 8 I(6,) = 0.89 IG(windy) = = SplitInfo windy (D) = 6 log 6 8 log 8 = 0.98 Gain Ratio(windy) = /0.98=0.0488
5 Now,, which has the highest gain ratio(0.164), is selected to partition the dataset. Let s grow the tree at the sunny branch. Splitting Sunny branch whose entropy is ) Temperature at =sunny Info temperature (D sunny ) = I(0,) + I(1,1) + 1 I(1,0) = 0.4 IG(temperature) = = SplitInfo temperature (D sunny ) = log log Gain Ratio(windy) = 0.710/1.19 = log 1 = 1.19 ) Humidity at =sunny Info humidity (D sunny ) = 3 I(0,3) + I(,0) = 0 IG(humidity) = = SplitInfo humidity (D sunny ) = 3 log 3 log Gain Ratio(windy) = /0.9710=1 = ) Windy at =sunny Info windy (D sunny ) = I(1,1) + 3 I(1,) = IG(windy) = =0.000 SplitInfo windy (D sunny ) = log 3 log Gain Ratio(windy) = 0.000/0.9710= = Humidity has the highest gain ratio (1). Therefore, humidity is selected to split the tree. The leaves are pure at the humidity branch. Let us then grow the tree at the rain branch. Splitting Outlook=Rain branch whose entropy is ) Temperature at =rain
6 Info temperature (D rain ) = 0 I(0,0) + 3 I(,1) + I(1,1) = IG(temperature) = = SplitInfo temperature (D rain ) = 0 3 log 3 log Gain Ratio(temperature) = 0.000/0.9710=0.006 = ) Humidity at =rain Info humidity (D rain ) = I(1,1) + 3 I(,1) = IG(humidity) = = SplitInfo humidity (D rain ) = log 3 log Gain Ratio(humidity) = 0.000/0.9710= = ) Windy at =rain Info windy (D rain ) = 3 I(0,3) + I(,0) = 0 IG(windy) = = SplitInfo windy (D rain ) = 3 log 3 log Gain Ratio(windy) = /0.9710=1 = Windy has the highest gain ratio (1). Therefore, windy is selected to split the tree. The leaves are pure at the windy branch. The decision tree is now complete.
7 Decision Tree Induction using Gini Index The Gini index at the root, i.e., before partitioning the data set, is: n gini(d) = 1 p j = 1 ( 9 ) ( ) = 0.49 j=1 1) Outlook (as the candidate attribute) at the root gini (D) = ( ) gini(d sunny) + ( 4 ) gini(d overcast) + ( ) gini(d rain) = ( ) [1 ( ) ( 3 ) ] + ( 4 ) (0) + ( ) [1 (3 ) ( ) ] = ( ) (0.48) + ( 4 ) (0) + ( ) (0.48) = Reduction in impurity = Δgini() = = ) Temperature at the root gini temperature (D) = ( 4 ) gini(d hot) + ( 6 ) gini(d mild) + ( 4 ) gini(d cool) = ( 4 ) [1 ( 4 ) ( 4 ) ] + ( 6 ) [1 (4 6 ) ( 6 ) ] + ( 4 ) [1 (3 4 ) ( 1 4 ) ] = ( 4 ) (0.) + ( 6 ) (0.4444) + ( 4 ) (0.37) = Reduction in impurity = Δgini(temperature) = = ) Humidity at the root gini humidity (D) = ( 7 ) gini(d high) + ( 7 ) gini(d normal) = ( 7 ) [1 (3 7 ) ( 4 7 ) ] + ( 7 ) [1 (6 7 ) ( 1 7 ) ] = ( 7 ) (0.4898) + ( 7 ) (0.449) = Reduction in impurity = Δgini(humidity) = = ) Windy at the root gini windy (D) = ( 6 ) gini(d true) + ( 8 ) gini(d false) = ( 6 ) [1 (3 6 ) ( 3 6 ) ] + ( 8 ) [1 (6 8 ) ( 8 ) ] = ( 6 ) (0.) + ( 8 ) (0.37) = Reduction in impurity = Δgini(windy) = =
8 Now,, which has the highest Gini index reduction, is selected to partition the dataset. Let s grow the tree at the sunny branch. Splitting Sunny branch whose Gini index is ) Temperature at =sunny gini temperature (D sunny ) = ( ) gini(d hot) + ( ) gini(d mild) + ( 1 ) gini(d cool) = ( ) (0) + ( ) [1 (1 ) ( 1 ] ) + ( 1 ) (0) = 0. Reduction in impurity = Δgini(temperature) = = 0.8 ) Humidity at =sunny gini humidity (D sunny ) = ( 3 ) gini(d high) + ( ) gini(d normal) = ( 3 ) (0) + ( ) (0) = 0 Reduction in impurity = Δgini(humidity) = = ) Windy at =sunny gini windy (D sunny ) = ( ) gini(d true) + ( 3 ) gini(d false) = ( 3 ) (0.) + ( ) (0.4444) = Reduction in impurity = Δgini(windy) = = Humidity has the highest gini index reduction. Therefore, humidity is selected to split the tree. The leaves are pure at the humidity branch. Let us then grow the tree at the rain branch. Splitting Outlook=Rain branch whose Gini index is ) Temperature at =rain
9 gini temperature (D rain ) = ( 0 ) gini(d hot) + ( 3 ) gini(d mild) + ( ) gini(d cool) = 0 + ( 3 ) [1 ( 3 ) = ( 1 3 ) ] + ( ) [1 (1 ) ( 1 ) ] Reduction in impurity = Δgini(temperature) = = ) Humidity at =rain gini humidity (D rain ) = ( ) gini(d high) + ( 3 ) gini(d normal) = ( ) (0.) + (3 ) (0.4444) = Reduction in impurity = Δgini(humidity) = = ) Windy at =rain gini windy (D rain ) = ( 3 ) gini(d true) + ( ) gini(d false) = ( 3 ) (0) + ( ) (0) = 0 Reduction in impurity = Δgini(windy) = = 0.48 Windy has the highest Gini index reduction. Therefore, windy is selected to split the tree. The leaves are pure at the windy branch. The decision tree is now complete.
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