Classification with Decision Trees Instructor: Qiang Yang Hong Kong University of Science and Technology Qyang@cs.ust.hk Thanks: Eibe Frank and Jiawei Han Lecture 1

DECISION TREE [Quinlan93] An internal node represents a test on an attribute. A branch represents an outcome of the test, e.g., Color=red. A leaf node represents a class label or class label distribution. At each node, one attribute is chosen to split training examples into distinct classes as much as possible A new case is classified by following a matching path to a leaf node.

Training Set

Example Outlook sunny overcast rain humidity P windy high normal true false N P N P

Building Decision Tree [Q93] Top-down tree construction At start, all training examples are at the root. Partition the examples recursively by choosing one attribute each time. Bottom-up tree pruning Remove subtrees or branches, in a bottom-up manner, to improve the estimated accuracy on new cases.

Choosing the Splitting Attribute At each node, available attributes are evaluated on the basis of separating the classes of the training examples. A Goodness function is used for this purpose. Typical goodness functions: information gain (ID3/C4.5) information gain ratio gini index

Which attribute to select?

A criterion for attribute selection Which is the best attribute? The one which will result in the smallest tree Heuristic: choose the attribute that produces the “purest” nodes Popular impurity criterion: information gain Information gain increases with the average purity of the subsets that an attribute produces Strategy: choose attribute that results in greatest information gain

Computing information Information is measured in bits Given a probability distribution, the info required to predict an event is the distribution’s entropy Entropy gives the information required in bits (this can involve fractions of bits!) Formula for computing the entropy: Suppose a set S has n values: V1, V2, …Vn, where Vi has proportion pi, E.g., the weather data has 2 values: Play=P and Play=N. Thus, p1=9/14, p2=5/14.

Example: attribute “Outlook” “Outlook” = “Sunny”: “Outlook” = “Overcast”: “Outlook” = “Rainy”: Expected information for attribute: Note: this is normally not defined.

Computing the information gain Information gain: information before splitting – information after splitting Information gain for attributes from weather data:

Continuing to split

The final decision tree Note: not all leaves need to be pure; sometimes identical instances have different classes  Splitting stops when data can’t be split any further

Highly-branching attributes Problematic: attributes with a large number of values (extreme case: ID code) Subsets are more likely to be pure if there is a large number of values Information gain is biased towards choosing attributes with a large number of values This may result in overfitting (selection of an attribute that is non-optimal for prediction) Another problem: fragmentation

The gain ratio Gain ratio: a modification of the information gain that reduces its bias on high-branch attributes Gain ratio takes number and size of branches into account when choosing an attribute It corrects the information gain by taking the intrinsic information of a split into account Also called split ratio Intrinsic information: entropy of distribution of instances into branches (i.e. how much info do we need to tell which branch an instance belongs to)

Gain Ratio IntrinsicInfo should be Large when data is evenly spread over all branches Small when all data belong to one branch Gain ratio (Quinlan’86) normalizes info gain by IntrinsicInfo:

Computing the gain ratio Example: intrinsic information for ID code Importance of attribute decreases as intrinsic information gets larger Example of gain ratio: Example:

Gain ratios for weather data Outlook Temperature Info: 0.693 0.911 Gain: 0.940-0.693 0.247 Gain: 0.940-0.911 0.029 Split info: info([5,4,5]) 1.577 Split info: info([4,6,4]) 1.362 Gain ratio: 0.247/1.577 0.156 Gain ratio: 0.029/1.362 0.021 Humidity Windy Info: 0.788 0.892 Gain: 0.940-0.788 0.152 Gain: 0.940-0.892 0.048 Split info: info([7,7]) 1.000 Split info: info([8,6]) 0.985 Gain ratio: 0.152/1 Gain ratio: 0.048/0.985 0.049

More on the gain ratio “Outlook” still comes out top However: “ID code” has greater gain ratio Standard fix: ad hoc test to prevent splitting on that type of attribute Problem with gain ratio: it may overcompensate May choose an attribute just because its intrinsic information is very low Standard fix: First, only consider attributes with greater than average information gain Then, compare them on gain ratio

Gini Index If a data set T contains examples from n classes, gini index, gini(T) is defined as where pj is the relative frequency of class j in T. gini(T) is minimized if the classes in T are skewed. After splitting T into two subsets T1 and T2 with sizes N1 and N2, the gini index of the split data is defined as The attribute providing smallest ginisplit(T) is chosen to split the node.

Stopping Criteria When all cases have the same class. The leaf node is labeled by this class. When there is no available attribute. The leaf node is labeled by the majority class. When the number of cases is less than a specified threshold. The leaf node is labeled by the majority class. You can make a decision at every node in a decision tree! How?

Pruning Pruning simplifies a decision tree to prevent overfitting to noise in the data Two main pruning strategies: Postpruning: takes a fully-grown decision tree and discards unreliable parts Prepruning: stops growing a branch when information becomes unreliable Postpruning preferred in practice because of early stopping in prepruning

Prepruning Usually based on statistical significance test Stops growing the tree when there is no statistically significant association between any attribute and the class at a particular node Most popular test: chi-squared test ID3 used chi-squared test in addition to information gain Only statistically significant attributes where allowed to be selected by information gain procedure

Postpruning Builds full tree first and prunes it afterwards Attribute interactions are visible in fully-grown tree Problem: identification of subtrees and nodes that are due to chance effects Two main pruning operations: Subtree replacement Subtree raising Possible strategies: error estimation, significance testing, MDL principle

Subtree replacement Bottom-up: tree is considered for replacement once all its subtrees have been considered

Estimating error rates Pruning operation is performed if this does not increase the estimated error Of course, error on the training data is not a useful estimator (would result in almost no pruning) One possibility: using hold-out set for pruning (reduced-error pruning) C4.5’s method: using upper limit of 25% confidence interval derived from the training data Standard Bernoulli-process-based method

Post-pruning in C4.5 Bottom-up pruning: at each non-leaf node v, if merging the subtree at v into a leaf node improves accuracy, perform the merging. Method 1: compute accuracy using examples not seen by the algorithm. Method 2: estimate accuracy using the training examples: Consider classifying E examples incorrectly out of N examples as observing E events in N trials in the binomial distribution. For a given confidence level CF, the upper limit on the error rate over the whole population is with CF% confidence.

Pessimistic Estimate Usage in Statistics: Sampling error estimation Example: population: 1,000,000 people, population mean: percentage of the left handed people sample: 100 people sample mean: 6 left-handed How to estimate the REAL population mean? Possibility(%) 25% confidence interval 2 6 10 L0.25(100,6) U0.25(100,6)

C4.5’s method Error estimate for subtree is weighted sum of error estimates for all its leaves Error estimate for a node: If c = 25% then z = 0.69 (from normal distribution) f is the error on the training data N is the number of instances covered by the leaf

Example Outlook ? sunny cloudy yes overcast yes yes no

Example cont. Consider a subtree rooted at Outlook with 3 leaf nodes: Sunny: Play = yes : (0 error, 6 instances) Overcast: Play= yes: (0 error, 9 instances) Cloudy: Play = no (0 error, 1 instance) The estimated error for this subtree is 6*0.074+9*0.050+1*0.323=1.217 If the subtree is replaced with the leaf “yes”, the estimated error is So no pruning is performed

Another Example If we split, we can compute average error using If we have a single leaf node If we split, we can compute average error using ratios 6:2:6 this gives 0.51 f=5/14 e=0.46 f=0.33 e=0.47 f=0.5 e=0.72 f=0.33 e=0.47

Other Trees Classification Trees Current node Children nodes (L, R): Decision Trees GINI index used in CART (STD= )

Efforts on Scalability Most algorithms assume data can fit in memory. Recent efforts focus on disk-resident implementation for decision trees. Random sampling Partitioning Examples SLIQ (EDBT’96 -- [MAR96]) SPRINT (VLDB96 -- [SAM96]) PUBLIC (VLDB98 -- [RS98]) RainForest (VLDB98 -- [GRG98])

Questions Consider the following variations of decision trees

1. Apply KNN to each leaf node Instead of choosing a class label as the majority class label, use KNN to choose a class label

2. Apply Naïve Bayesian at each leaf node For each leave node, use all the available information we know about the test case to make decisions Instead of using the majority rule, use probability/likelihood to make decisions

3. Use error rates instead of entropy If a node has N1 positive class labels P, and N2 negative class labels N, If N1> N2, then choose P The error rate = N2/(N1+N2) at this node The expected error at a parent node can be calculated as weighted sum of the error rates at each child node The weights are the proportion of training data in each child

Cost Sensitive Decision Trees When the FP and FN have different costs, the leaf node label is different depending on the costs: If growing a tree has a smaller total cost, then choose an attribute with minimal total cost. Otherwise, stop and form a leaf. Label leaf according to minimal total cost: Suppose the leaf have P positive examples and N negative examples FP denotes the cost of a false positive example and FN false negative If (P×FN  N×FP) THEN label = positive ELSE label = negative

5. When there is missing value in the test data, we allow tests to be done Attribute selection criterion: minimal total cost (Ctotal = Cmc + Ctest) instead of minimal entropy in C4.5 Typically, if there are missing values, then to obtain a value for a missing attribute (say Temperature) will incur new cost But may increase accuracy of prediction, thus reducing the miss classification costs In general, there is a balance between the two costs We care about the total cost

Stream Data Suppose that you have built a decision tree from a set of training data Now suppose that some additional training data comes at a regular interval, in fast speed How do you adapt the existing tree to fit the new data? Suggest an ‘online’ algorithm. Hint: find a paper online on this topic