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Decision Tree and Bayesian Classification
- 1. Made by :Komal Kotak
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- 3. Decision tree ▸A decisiontree is a decision support tool that uses a tree-like graph or model of decisions and their possible consequences, including chance event outcomes, resource costs, and utility. It is one way to display an algorithm. 3
- 4. Why decision tree? ▸Decisiontrees are powerful and popular tools for classification and prediction. ▸Decision trees represent rules, which can be understood by humans and used in knowledge system such as database. 4
- 5. Requirements 1 ▸Attribute-value description:object or case must be expressible in terms of a fixed collection of properties or attributes (e.g., hot, mild, cold). ▸Predefined classes (target values): the target function has discrete output values (bollean or multiclass) ▸Sufficient data: enough training cases should be provided to learn the model. 5
- 6. Easy Example 1 ADecision Tree with two choices. 6 Go to Graduate School to get my MBA. Go to Work “in the Real World”
- 7. Notation used inDecision trees ▸A box is used to show a choice that the manager has to make. ▸A circle is used to show that a probability outcome will occur. ▸Lines connect outcomes to their choice or probability outcome. 7
- 8. Easy Example 1Revisited What are some of the costs we should take into account when deciding whether or not to go to business school? • Tuition and Fees • Rent / Food / etc. • Opportunity cost of salary • Anticipated future earnings 8
- 9. Simple Decision TreeModel 9 Go to Graduate School to get my MBA. Go to Work “in the Real World” 2 Years of tuition: $55,000, 2 years of Room/Board: $20,000; 2 years of Opportunity Cost of Salary = $100,000 Total = $175,000. PLUS Anticipated 5 year salary after Business School = $600,000. NPV (business school) = $600,000 - $175,000 = $425,000 First two year salary = $100,000 (from above), minus expenses of $20,000. Final five year salary = $330,000 NPV (no b-school) = $410,000
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- 11. Easy Example 2 ▸Thefollowing decision tree is for the concept buy computer that indicates whether a customer at a company is likely to buy a computer or not. Each internal node represents a test on an attribute. Each leaf node represents a class. 11
- 12. How does atree decide where to split? ▸The decision of making strategic splits heavily affects a tree’s accuracy. The decision criteria is different for classification and regression trees. ▸Decision trees use multiple algorithms to decide to split a node in two or more sub-nodes. The creation of sub-nodes increases the homogeneity of resultant sub-nodes. In other words, we can say that purity of the node increases with respect to the target variable. Decision tree splits the nodes on all available variables and then selects the split which results in most homogeneous sub-nodes. 12
- 13. The algorithm selectionis also based on type of target variables. Let’s look at the four most commonly used algorithms in decision tree: Gini Index Chi-Square Information Gain Reduction in Variance 13
- 14. Gini Index Gini indexsays, if we select two items from a population at random then they must be of same class and probability for this is 1 if population is pure. ▸It works with categorical target variable “Success” or “Failure”. ▸It performs only Binary splits ▸Higher the value of Gini higher the homogeneity. ▸CART (Classification and Regression Tree) uses Gini method to create binary splits. 14
- 15. Gini Index Steps toCalculate Gini for a split ▸Calculate Gini for sub-nodes, using formula sum of square of probability for success and failure (p^2+q^2). ▸Calculate Gini for split using weighted Gini score of each node of that split ▸Example: –To segregate the students based on target variable ( playing cricket or not ). In the snapshot below, we split the population using two input variables Gender and Class. Now, I want to identify which split is producing more homogeneous sub-nodes using Gini index. 15
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- 17. Gini Index Split onGender: ▸Calculate, Gini for sub-node Female = (0.2)*(0.2)+(0.8)*(0.8)=0.68 ▸Gini for sub-node Male = (0.65)*(0.65)+(0.35)*(0.35)=0.55 ▸Calculate weighted Gini for Split Gender = (10/30)*0.68+(20/30)*0.55 = 0.59 Similar for Split on Class: ▸Gini for sub-node Class IX = (0.43)*(0.43)+(0.57)*(0.57)=0.51 ▸Gini for sub-node Class X = (0.56)*(0.56)+(0.44)*(0.44)=0.51 ▸Calculate weighted Gini for Split Class = (14/30)*0.51+(16/30)*0.51 = 0.51 ▸Above, you can see that Gini score for Split on Gender is higher than Split on Class, hence, the node split will take place on Gender 17
- 18. Chi-Square It is analgorithm to find out the statistical significance between the differences between sub-nodes and parent node. We measure it by sum of squares of standardized differences between observed and expected frequencies of target variable. ▸It works with categorical target variable “Success” or “Failure”. ▸It can perform two or more splits. ▸Higher the value of Chi-Square higher the statistical significance of differences between sub-node and Parent node. ▸Chi-Square of each node is calculated using formula, ▸Chi-square = ((Actual – Expected)^2 / Expected)^1/2 ▸It generates tree called CHAID (Chi-square Automatic Interaction Detector) 18
- 19. Chi-Square Steps to CalculateChi-square for a split: ▸Calculate Chi-square for individual node by calculating the deviation for Success and Failure both ▸Calculated Chi-square of Split using Sum of all Chi-square of success and Failure of each node of the split Example: Let’s work with above example that we have used to calculate Gini. 19
- 20. Chi-Square Split on Gender: ▸Firstwe are populating for node Female, Populate the actual value for “Play Cricket” and “Not Play Cricket”, here these are 2 and 8 respectively. ▸Calculate expected value for “Play Cricket” and “Not Play Cricket”, here it would be 5 for both because parent node has probability of 50% and we have applied same probability on Female count(10). ▸Calculate deviations by using formula, Actual – Expected. It is for “Play Cricket” (2 – 5 = -3) and for “Not play cricket” ( 8 – 5 = 3). ▸Calculate Chi-square of node for “Play Cricket” and “Not Play Cricket” using formula with formula, = ((Actual – Expected)^2 / Expected)^1/2. You can refer below table for calculation. ▸Follow similar steps for calculating Chi-square value for Male node. ▸Now add all Chi-square values to calculate Chi-square for split Gender. 20
- 21. Chi-Square Split on Class: Performsimilar steps of calculation for split on Class and you will come up with below table. Above, you can see that Chi-square also identify the Gender split is more significant compare to Class. 21
- 22. Information Gain Look atthe image below and think which node can be described easily. I am sure, your answer is C because it requires less information as all values are similar. On the other hand, B requires more information to describe it and A requires the maximum information. In other words, we can say that C is a Pure node, B is less Impure and A is more impure. Now, we can build a conclusion that less impure node requires less information to describe it. And, more impure node requires more information. Information theory is a measure to define this degree of disorganization in a system known as Entropy. If the sample is completely homogeneous, then the entropy is zero and if the sample is an equally divided (50% – 50%), it has entropy of one. 22
- 23. Information Gain Entropy canbe calculated using formula:- ▸Here p and q is probability of success and failure respectively in that node. Entropy is also used with categorical target variable. It chooses the split which has lowest entropy compared to parent node and other splits. The lesser the entropy, the better it is. Steps to calculate entropy for a split: ▸Calculate entropy of parent node ▸Calculate entropy of each individual node of split and calculate weighted average of all sub-nodes available in split. 23
- 24. Information Gain Example: Let’suse this method to identify best split for student example. ▸Entropy for parent node = -(15/30) log2 (15/30) – (15/30) log2 (15/30) = 1. Here 1 shows that it is a impure node. ▸Entropy for Female node = -(2/10) log2 (2/10) – (8/10) log2 (8/10) = 0.72 and for male node, -(13/20) log2 (13/20) – (7/20) log2 (7/20) = 0.93 ▸Entropy for split Gender = Weighted entropy of sub-nodes = (10/30)*0.72 + (20/30)*0.93 = 0.86 ▸Entropy for Class IX node, -(6/14) log2 (6/14) – (8/14) log2 (8/14) = 0.99 and for Class X node, -(9/16) log2 (9/16) – (7/16) log2 (7/16) = 0.99. ▸Entropy for split Class = (14/30)*0.99 + (16/30)*0.99 = 0.99 Above, you can see that entropy for Split on Gender is the lowest among all, so the tree will split on Gender. We can derive information gain from entropy as 1- Entropy. 24
- 25. Reduction in Variance ▸Tillnow, we have discussed the algorithms for categorical target variable. Reduction in variance is an algorithm used for continuous target variables (regression problems). This algorithm uses the standard formula of variance to choose the best split. The split with lower variance is selected as the criteria to split the population: ▸Above X-bar is mean of the values, X is actual and n is number of values. ▸Steps to calculate Variance: ▸Calculate variance for each node. ▸Calculate variance for each split as weighted average of each node variance. 25
- 26. Reduction in Variance ▸Example:- Let’s assign numerical value 1 for play cricket and 0 for not playing cricket. Now follow the steps to identify the right split: ▸ Variance for Root node, here mean value is (15*1 + 15*0)/30 = 0.5 and we have 15 one and 15 zero. Now variance would be ((1-0.5)^2+(1-0.5)^2+….15 times+(0-0.5)^2+(0-0.5)^2+…15 times) / 30, this can be written as (15*(1-0.5)^2+15*(0-0.5)^2) / 30 = 0.25 ▸Mean of Female node = (2*1+8*0)/10=0.2 and Variance = (2*(1-0.2)^2+8*(0-0.2)^2) / 10 = 0.16 ▸Mean of Male Node = (13*1+7*0)/20=0.65 and Variance = (13*(1-0.65)^2+7*(0-0.65)^2) / 20 = 0.23 ▸Variance for Split Gender = Weighted Variance of Sub-nodes = (10/30)*0.16 + (20/30) *0.23 = 0.21 ▸Mean of Class IX node = (6*1+8*0)/14=0.43 and Variance = (6*(1-0.43)^2+8*(0-0.43)^2) / 14= 0.24 ▸Mean of Class X node = (9*1+7*0)/16=0.56 and Variance = (9*(1-0.56)^2+7*(0-0.56)^2) / 16 = 0.25 ▸Variance for Split Gender = (14/30)*0.24 + (16/30) *0.25 = 0.25 Above, you can see that Gender split has lower variance compare to parent node, so the split would take place on Gender variable. 26
- 27. TREE PRUNING ▸Pruning isa technique in machine learning that reduces the size of decision trees by removing sections of the tree that provide little power to classify instances. Pruning reduces the complexity of the final classifier, and hence improves predictive accuracy by the reduction of overfitting. 27
- 28. Techniques of TREEPRUNING ▸Pruning can occur in a top down or bottom up fashion. A top down pruning will traverse nodes and trim sub trees starting at the root, while a bottom up pruning will start at the leaf nodes. Below are several popular pruning algorithms. ▸Reduced error pruning ▸One of the simplest forms of pruning is reduced error pruning. Starting at the leaves, each node is replaced with its most popular class. If the prediction accuracy is not affected then the change is kept. While somewhat naive, reduced error pruning has the advantage of simplicity and speed. 28
- 29. Techniques of TREEPRUNING ▸Cost complexity pruning ▸Cost complexity pruning generates a series of trees T0 . . . Tm where T0 is the initial tree and Tm is the root alone. At step i the tree is created by removing a sub tree from tree i-1 and replacing it with a leaf node with value chosen as in the tree building algorithm. The sub tree that is removed is chosen as follows. Define the error rate of tree T over data set S as err(T,S). The sub tree that minimizes is chosen for removal. The function prune(T,t) defines the tree gotten by pruning the sub trees t from the tree T. Once the series of trees has been created, the best tree is chosen by generalized accuracy as measured by a training set or cross-validation. 29
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- 31. Bayesian Classification: Why? ▸Probabilisticlearning: Calculate explicit probabilities for hypothesis, among the most practical approaches to certain types of learning problems ▸Incremental: Each training example can incrementally increase/decrease the probability that a hypothesis is correct. Prior knowledge can be combined with observed data. ▸Probabilistic prediction: Predict multiple hypotheses, weighted by their probabilities ▸Standard: Even when Bayesian methods are computationally intractable, they can provide a standard of optimal decision making against which other methods can be measured 31
- 32. Bayesian Theorem ▸Given trainingdata D, posteriori probability of a hypothesis h, P(h|D) follows the Bayes theorem ▸MAP (maximum posteriori) hypothesis ▸Practical difficulty: require initial knowledge of many probabilities, significant computational cost. 32 )( )()|()|( DP hPhDPDhP .)()|(maxarg)|(maxarg hPhDP Hh DhP HhMAP h
- 33. Naïve Bayes Classifier ▸Asimplified assumption: attributes are conditionally independent: ▸Greatly reduces the computation cost, only count the class distribution. 33 n i jijj CdPCPDCP 1 )|()()|(
- 34. Naïve Bayes Classifier ▸Ifi-th attribute is categorical: P(di|C) is estimated as the relative freq of samples having value di as i-th attribute in class C ▸If i-th attribute is continuous: P(di|C) is estimated thru a Gaussian density function ▸Computationally easy in both cases 34
- 35. Play-tennis example: estimatingP(xi|C) 35 Outlook Temperature Humidity Windy Class sunny hot high false N sunny hot high true N overcast hot high false P rain mild high false P rain cool normal false P rain cool normal true N overcast cool normal true P sunny mild high false N sunny cool normal false P rain mild normal false P sunny mild normal true P overcast mild high true P overcast hot normal false P rain mild high true N P(true|n) = 3/5P(true|p) = 3/9 P(false|n) = 2/5P(false|p) = 6/9 P(high|n) = 4/5P(high|p) = 3/9 P(normal|n) = 2/5P(normal|p) = 6/9 P(hot|n) = 2/5P(hot|p) = 2/9 P(mild|n) = 2/5P(mild|p) = 4/9 P(cool|n) = 1/5P(cool|p) = 3/9 P(rain|n) = 2/5P(rain|p) = 3/9 P(overcast|n) = 0P(overcast|p) = 4/9 P(sunny|n) = 3/5P(sunny|p) = 2/9 windy humidity temperature outlook P(n) = 5/14 P(p) = 9/14
- 36. Naive Bayesian Classifier ▸Givena training set, we can compute the probabilities 36 O utlook P N H um idity P N sunny 2/9 3/5 high 3/9 4/5 overcast 4/9 0 norm al 6/9 1/5 rain 3/9 2/5 Tem preature W indy hot 2/9 2/5 true 3/9 3/5 m ild 4/9 2/5 false 6/9 2/5 cool 3/9 1/5
- 37. Play-tennis example: classifyingX ▸An unseen sample X = <rain, hot, high, false> ▸P(X|p)·P(p) = P(rain|p)·P(hot|p)·P(high|p)·P(false|p)·P(p) = 3/9·2/9·3/9·6/9·9/14 = 0.010582 ▸P(X|n)·P(n) = P(rain|n)·P(hot|n)·P(high|n)·P(false|n)·P(n) = 2/5·2/5·4/5·2/5·5/14 = 0.018286 ▸Sample X is classified in class n (don’t play) 37
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