![EXAMPLE
Standard deviation
Golf players = {25, 30, 46, 45, 52,
23, 43, 35, 38, 46, 48, 52, 44, 30}
Average of golf players = (25 +
30 + 46 + 45 + 52 + 23 + 43 + 35
+ 38 + 46 + 48 + 52 + 44 + 30
)/14 = 39.78
Standard deviation of golf
players = [( (25 – 39.78)
√ 2
+ (30
– 39.78)2
+ (46 – 39.78)2
+ … + (30
– 39.78)2
)/14] = 9.32](https://image.slidesharecdn.com/21-250407132925-7fa185ed/85/21-Regression-Tree-in-machine-learning-pptx-4-320.jpg)
21. Regression Tree in machine learning.pptx
- 2. INTRODUCTION Classification Trees: When the decision tree has categorical target variable. The above tree is an example of a classification tree because we know that there are two options for the result. Regression Trees: When the decision tree has a continuous target variable. For example, a regression tree would be used for the price of a newly launched product because price can be anything depending on various constraints. Both types of decision trees fall under the Classification and Regression Tree (CART) designation.
- 4. EXAMPLE Standard deviation Golf players = {25, 30, 46, 45, 52, 23, 43, 35, 38, 46, 48, 52, 44, 30} Average of golf players = (25 + 30 + 46 + 45 + 52 + 23 + 43 + 35 + 38 + 46 + 48 + 52 + 44 + 30 )/14 = 39.78 Standard deviation of golf players = [( (25 – 39.78) √ 2 + (30 – 39.78)2 + (46 – 39.78)2 + … + (30 – 39.78)2 )/14] = 9.32
- 5. Golf players for sunny outlook = {25, 30, 35, 38, 48} Average of golf players for sunny outlook = (25+30+35+38+48)/5 = 35.2 Standard deviation of golf players for sunny outlook = (((25 – 35.2) √ 2 + (30 – 35.2)2 + … )/5) = 7.78
- 6. Golf players for overcast outlook = {46, 43, 52, 44} Average of golf players for overcast outlook = (46 + 43 + 52 + 44)/4 = 46.25 Standard deviation of golf players for overcast outlook = (((46- √ 46.25)2 +(43-46.25)2 +…)= 3.49
- 7. Golf players for overcast outlook = {45, 52, 23, 46, 30} Average of golf players for overcast outlook = (45+52+23+46+30)/5 = 39.2 Standard deviation of golf players for rainy outlook = (((45 – √ 39.2)2 +(52 – 39.2)2 +…)/5)=10.87
- 8. Weighted standard deviation for outlook = (4/14)x3.49 + (5/14)x10.87 + (5/14)x7.78 = 7.66 Standard deviation reduction for outlook = 9.32 – 7.66 = 1.66
- 9. Weighted standard deviation for humidity = (7/14)x9.36 + (7/14)x8.73 = 9.04 Standard deviation reduction for humidity = 9.32 – 9.04 = 0.27
- 11. Root Node Outlo ok !4 data - Global Std dev 5 data - Global Std dev 5 Temp Hot Mild Cool Sunny Wind Weak Strong Humidity High Normal
- 12. Golf players for sunny outlook = {25, 30, 35, 38, 48} Average of golf players for sunny outlook = (25+30+35+38+48)/5 = 35.2 Standard deviation of golf players for sunny outlook = (((25 – 35.2) √ 2 + (30 – 35.2)2 + … )/5) = 7.78 Considered as Global standard deviation for this sub data set = 7.78
- 13. Standard deviation for sunny outlook and hot temperature = 2.5 Standard deviation for sunny outlook and cool temperature = 0 Standard deviation for sunny outlook and mild temperature = 6.5
- 14. Weighted standard deviation for sunny outlook and temperature = (2/5)x2.5 + (1/5)x0 + (2/5)x6.5 = 3.6 Standard deviation reduction for sunny outlook and temperature = 7.78 – 3.6 = 4.18
- 15. Weighted standard deviations for sunny outlook and humidity = (3/5)x4.08 + (2/5)x5 = 4.45 Standard deviation reduction for sunny outlook and humidity = 7.78 – 4.45 = 3.33 Weighted standard deviations for sunny outlook and wind = (2/5)x9 + (3/5)x5.56 = 6.93 Standard deviation reduction for sunny outlook and wind = 7.78 – 6.93 = 0.85 Summarizing standard deviations for windy feature when outlook is sunny
- 17. FINAL FORM OF REGRESSION TREE https://sefiks.com/2018/08/28/a-step-by-step-regression-decision-tree-example/ Leaf Node = Golf Player 5 5 1 (2) 2
- 18. Decision Tree Entropy Information gain – Higher gain Best candidate to be selected as a node Entropy – If all the data belongs to the same class label – Entropy =0 (Pure) If the input data belongs to many class lables – Entropy = near to 1 (Impure) Nodes Input attributes (Ex: outlook) Arcs/links/edges Values of input attributes (Ex: Sunny, Rainy, Overcast) Top node – Root node Other nodes in the tree – Intermediate nodes Leaf node (last level of the tree) – Identifies the corresponding class label (Ex: Play = Yes/ No) From Decision tree Derive classification rules How many rules can be derived ? No of leaf level nodes