Machine Learning

Machine Learning Approach based on Decision Trees

Decision Tree Learning Practical inductive inference method Same goal as Candidate-Elimination algorithm Find Boolean function of attributes Decision trees can be extended to functions with more than two output values. Widely used Robust to noise Can handle disjunctive (OR’s) expressions Completely expressive hypothesis space Easily interpretable (tree structure, if-then rules)

Training Examples Shall we play tennis today? ( Tennis 1 ) Attribute, variable, property Object, sample, example decision

Shall we play tennis today? Decision trees do classification Classifies instances into one of a discrete set of possible categories Learned function represented by tree Each node in tree is test on some attribute of an instance Branches represent values of attributes Follow the tree from root to leaves to find the output value.

The tree itself forms hypothesis Disjunction (OR’s) of conjunctions (AND’s) Each path from root to leaf forms conjunction of constraints on attributes Separate branches are disjunctions Example from PlayTennis decision tree: (Outlook=Sunny   Humidity=Normal)  (Outlook=Overcast)   (Outlook=Rain  Wind=Weak)

Types of problems decision tree learning is good for: Instances represented by attribute-value pairs For algorithm in book, attribute s take on a small number of discrete values Can be extended to real-valued attributes (numerical data) Target function has discrete output values Algorithm in book assumes Boolean functions Can be extended to multiple output values

Hypothesis space can include disjunctive expressions . In fact, hypothesis space is complete space of finite discrete-valued functions Robust to imperfect training data classification errors errors in attribute values missing attribute values Examples : Equipment diagnosis Medical diagnosis Credit card risk analysis Robot movement Pattern Recognition face recognition hexapod walking gates

ID3 Algorithm Top-down, greedy search through space of possible decision trees Remember, decision trees represent hypotheses, so this is a search through hypothesis space . What is top-down? How to start tree? What attribute should represent the root? As you proceed down tree, choose attribute for each successive node . No backtracking : So, algorithm proceeds from top to bottom

The ID3 algorithm is used to build a decision tree, given a set of non-categorical attributes C1, C2, .., Cn, the categorical attribute C, and a training set T of records. function ID3 (R: a set of non-categorical attributes, C: the categorical attribute, S: a training set) returns a decision tree; begin If S is empty, return a single node with value Failure; If every example in S has the same value for categorical attribute, return single node with that value; If R is empty, then return a single node with most frequent of the values of the categorical attribute found in examples S; [note: there will be errors, i.e., improperly classified records]; Let D be attribute with largest Gain(D,S) among R’s attributes; Let {dj| j=1,2, .., m} be the values of attribute D; Let {Sj| j=1,2, .., m} be the subsets of S consisting respectively of records with value dj for attribute D; Return a tree with root labeled D and arcs labeled d1, d2, .., dm going respectively to the trees ID3(R-{D},C,S1), ID3(R-{D},C,S2) ,.., ID3(R-{D},C,Sm); end ID3;

What is a greedy search ? At each step, make decision which makes greatest improvement in whatever you are trying optimize. Do not backtrack (unless you hit a dead end) This type of search is likely not to be a globally optimum solution, but generally works well. What are we really doing here? At each node of tree, make decision on which attribute best classifies training data at that point . Never backtrack (in ID3) Do this for each branch of tree. End result will be tree structure representing a hypothesis which works best for the training data .

Information Theory Background If there are n equally probable possible messages, then the probability p of each is 1/n Information conveyed by a message is -log(p) = log(n) Eg, if there are 16 messages, then log(16) = 4 and we need 4 bits to identify/send each message. In general, if we are given a probability distribution P = (p1, p2, .., pn) the information conveyed by distribution (aka Entropy of P) is: I(P) = -(p1*log(p1) + p2*log(p2) + .. + pn*log(pn))

Question? How do you determine which attribute best classifies data ? Answer: Entropy! Information gain : Statistical quantity measuring how well an attribute classifies the data. Calculate the information gain for each attribute . Choose attribute with greatest information gain .

But how do you measure information? Claude Shannon in 1948 at Bell Labs established the field of information theory . Mathematical function, Entropy , measures information content of random process : Takes on largest value when events are equiprobable. Takes on smallest value when only one event has non-zero probability. For two states: Positive examples and Negative examples from set S: H(S) = - p + log 2 (p + ) - p - log 2 (p - ) Entropy of set S denoted by H(S)

Entropy Largest entropy Boolean functions with the same number of ones and zeros have largest entropy

But how do you measure information? Claude Shannon in 1948 at Bell Labs established the field of information theory . Mathematical function, Entropy , measures information content of random process : Takes on largest value when events are equiprobable. Takes on smallest value when only one event has non-zero probability. For two states: Positive examples and Negative examples from set S: H(S) = - p + log 2 (p + ) - p - log 2 (p - ) Entropy = Measure of order in set S

In general: For an ensemble of random events : {A 1 ,A 2 ,...,A n } , occurring with probabilities: z = { P(A 1 ),P(A 2 ),...,P(A n )} If you consider the self-information of event, i , to be: -log 2 (P(A i )) Entropy is weighted average of information carried by each event. Does this make sense?

If an event conveys information, that means it’s a surprise. If an event always occurs , P(A i )=1, then it carries no information. -log 2 (1) = 0 If an event rarely occurs (e.g. P(A i )=0.001), it carries a lot of info. -log 2 (0.001) = 9.97 The less likely the event, the more the information it carries since, for 0  P(A i )  1, -log 2 (P(A i )) increases as P(A i ) goes from 1 to 0. (Note: ignore events with P(A i )=0 since they never occur.) Does this make sense?

What about entropy ? Is it a good measure of the information carried by an ensemble of events? If the events are equally probable, the entropy is maximum. 1) For N events, each occurring with probability 1/N . H = -  (1/N)log 2 (1/N) = -log 2 (1/N) This is the maximum value . (e.g. For N=256 (ascii characters) -log 2 (1/256) = 8 number of bits needed for characters. Base 2 logs measure information in bits.) This is a good thing since an ensemble of equally probable events is as uncertain as it gets. (Remember, information corresponds to surprise - uncertainty .)

2) H is a continuous function of the probabilities. That is always a good thing. 3) If you sub-group events into compound events, the entropy calculated for these compound groups is the same. That is good since the uncertainty is the same. It is a remarkable fact that the equation for entropy shown above (up to a multiplicative constant) is the only function which satisfies these three conditions .

Choice of base 2 log corresponds to choosing units of information. (BIT’s) Another remarkable thing: This is the same definition of entropy used in statistical mechanics for the measure of disorder. Corresponds to macroscopic thermodynamic quantity of Second Law of Thermodynamics.

The concept of a quantitative measure for information content plays an important role in many areas: For example, Data communications (channel capacity) Data compression (limits on error-free encoding) Entropy in a message corresponds to minimum number of bits needed to encode that message . In our case, for a set of training data, the entropy measures the number of bits needed to encode classification for an instance. Use probabilities found from entire set of training data. Prob(Class=Pos) = Num. of positive cases / Total case Prob(Class=Neg) = Num. of negative cases / Total cases

(Back to the story of ID3) Information gain is our metric for how well one attribute A i classifies the training data. Information gain for a particular attribute = Information about target function, given the value of that attribute. (conditional entropy) Mathematical expression for information gain : entropy Entropy for value v

ID3 algorithm (for boolean-valued function) Calculate the entropy for all training examples positive and negative cases p + = #pos/Tot p - = #neg/Tot H(S) = -p + log 2 (p + ) - p - log 2 (p - ) Determine which single attribute best classifies the training examples using information gain. For each attribute find: Use attribute with greatest information gain as a root

Using Gain Ratios The notion of Gain introduced earlier favors attributes that have a large number of values. If we have an attribute D that has a distinct value for each record, then Info(D,T) is 0, thus Gain(D,T) is maximal. To compensate for this Quinlan suggests using the following ratio instead of Gain: GainRatio(D,T) = Gain(D,T) / SplitInfo(D,T) SplitInfo(D,T) is the information due to the split of T on the basis of value of categorical attribute D. SplitInfo(D,T) = I(|T1|/|T|, |T2|/|T|, .., |Tm|/|T|) where {T1, T2, .. Tm} is the partition of T induced by value of D.

Example: PlayTennis Four attributes used for classification: Outlook = {Sunny,Overcast,Rain} Temperature = {Hot, Mild, Cool} Humidity = {High, Normal} Wind = {Weak, Strong} One predicted (target) attribute (binary) PlayTennis = {Yes,No} Given 14 Training examples 9 positive 5 negative

Training Examples Examples, minterms, cases, objects, test cases,

Step 1: Calculate entropy for all cases: N Pos = 9 N Neg = 5 N Tot = 14 H(S) = -(9/14)*log 2 (9/14) - (5/14)*log 2 (5/14) = 0.940 14 cases 9 positive cases entropy

Step 2: Loop over all attributes, calculate gain : Attribute = Outlook Loop over values of Outlook Outlook = Sunny N Pos = 2 N Neg = 3 N Tot = 5 H(Sunny) = -(2/5)*log 2 (2/5) - (3/5)*log 2 (3/5) = 0.971 Outlook = Overcast N Pos = 4 N Neg = 0 N Tot = 4 H(Sunny) = -(4/4)*log 2 4/4) - (0/4)*log 2 (0/4) = 0.00

Outlook = Rain N Pos = 3 N Neg = 2 N Tot = 5 H(Sunny) = -(3/5)*log 2 (3/5) - (2/5)*log 2 (2/5) = 0.971 Calculate Information Gain for attribute Outlook Gain( S,Outlook ) = H(S) - N Sunny /N Tot *H(Sunny) - N Over /N Tot *H(Overcast) - N Rain /N Tot *H(Rainy) Gain( S,Outlook ) = 9.40 - (5/14)*0.971 - (4/14)*0 - (5/14)*0.971 Gain( S,Outlook ) = 0.246 Attribute = Temperature (Repeat process looping over {Hot, Mild, Cool}) Gain( S,Temperature ) = 0.029

Attribute = Humidity (Repeat process looping over {High, Normal}) Gain( S,Humidity ) = 0.029 Attribute = Wind (Repeat process looping over {Weak, Strong}) Gain( S,Wind ) = 0.048 Find attribute with greatest information gain: Gain( S,Outlook ) = 0.246, Gain( S,Temperature ) = 0.029 Gain( S,Humidity ) = 0.029, Gain( S,Wind ) = 0.048  Outlook is root node of tree

Iterate algorithm to find attributes which best classify training examples under the values of the root node Example continued Take three subsets: Outlook = Sunny (N Tot = 5) Outlook = Overcast (N Tot = 4) Outlook = Rainy (N Tot = 5) For each subset, repeat the above calculation looping over all attributes other than Outlook

For example: Outlook = Sunny (N Pos = 2, N Neg =3, N Tot = 5) H=0.971 Temp = Hot (N Pos = 0, N Neg =2, N Tot = 2) H = 0.0 Temp = Mild (N Pos = 1, N Neg =1, N Tot = 2) H = 1.0 Temp = Cool (N Pos = 1, N Neg =0, N Tot = 1) H = 0.0 Gain( S Sunny ,Temperature ) = 0.971 - (2/5)*0 - (2/5)*1 - (1/5)*0 Gain( S Sunny ,Temperature ) = 0.571 Similarly: Gain( S Sunny ,Humidity ) = 0.971 Gain( S Sunny ,Wind ) = 0.020  Humidity classifies Outlook =Sunny instances best and is placed as the node under Sunny outcome. Repeat this process for Outlook = Overcast &Rainy

Important: Attributes are excluded from consideration if they appear higher in the tree Process continues for each new leaf node until: Every attribute has already been included along path through the tree or Training examples associated with this leaf all have same target attribute value.

Note: In this example data were perfect . No contradictions Branches led to unambiguous Yes, No decisions If there are contradictions take the majority vote This handles noisy data. Another note: Attributes are eliminated when they are assigned to a node and never reconsidered . e.g. You would not go back and reconsider Outlook under Humidity ID3 uses all of the training data at once Contrast to Candidate-Elimination Can handle noisy data.

Another Example: Russell’s and Norvig’s Restaurant Domain Develop a decision tree to model the decision a patron makes when deciding whether or not to wait for a table at a restaurant. Two classes: wait, leave Ten attributes: alternative restaurant available?, bar in restaurant?, is it Friday?, are we hungry?, how full is the restaurant?, how expensive?, is it raining?,do we have a reservation?, what type of restaurant is it?, what's the purported waiting time? Training set of 12 examples ~ 7000 possible cases

A decision Tree from Introspection

ID3 A greedy algorithm for Decision Tree Construction developed by Ross Quinlan, 1987 Consider a smaller tree a better tree Top-down construction of the decision tree by recursively selecting the " best attribute " to use at the current node in the tree, based on the examples belonging to this node. Once the attribute is selected for the current node, generate children nodes, one for each possible value of the selected attribute. Partition the examples of this node using the possible values of this attribute, and assign these subsets of the examples to the appropriate child node. Repeat for each child node until all examples associated with a node are either all positive or all negative.

Choosing the Best Attribute The key problem is choosing which attribute to split a given set of examples. Some possibilities are: Random: Select any attribute at random Least-Values: Choose the attribute with the smallest number of possible values ( fewer branches ) Most-Values: Choose the attribute with the largest number of possible values ( smaller subsets ) Max-Gain: Choose the attribute that has the largest expected information gain , i.e. select attribute that will result in the smallest expected size of the subtrees rooted at its children. The ID3 algorithm uses the Max-Gain method of selecting the best attribute.

Splitting Examples by Testing Attributes

Another example : Tennis 2 (simplified former example)

The entropy is the average number of bits/message needed to represent a stream of messages. Examples: if P is (0.5, 0.5) then I(P) is 1 if P is (0.67, 0.33) then I(P) is 0.92, if P is (1, 0) then I(P) is 0. The more uniform is the probability distribution, the greater is its information gain/entropy.

What is the hypothesis space for decision tree learning? Search through space of all possible decision trees from simple to more complex guided by a heuristic: information gain The space searched is complete space of finite, discrete-valued functions. Includes disjunctive and conjunctive expressions Method only maintains one current hypothesis In contrast to Candidate-Elimination Not necessarily global optimum attributes eliminated when assigned to a node No backtracking Different trees are possible

Inductive Bias: (restriction vs. preference) ID3 searches complete hypothesis space But, incomplete search through this space looking for simplest tree This is called a preference (or search) bias Candidate-Elimination Searches an incomplete hypothesis space But, does a complete search finding all valid hypotheses This is called a restriction (or language) bias Typically, preference bias is better since you do not limit your search up-front by restricting hypothesis space considered.

How well does it work? Many case studies have shown that decision trees are at least as accurate as human experts. A study for diagnosing breast cancer: humans correctly classifying the examples 65% of the time, the decision tree classified 72% correct. British Petroleum designed a decision tree for gas-oil separation for offshore oil platforms/ It replaced an earlier rule-based expert system. Cessna designed an airplane flight controller using 90,000 examples and 20 attributes per example.

Extensions of the Decision Tree Learning Algorithm Using gain ratios Real-valued data Noisy data and Overfitting Generation of rules Setting Parameters Cross-Validation for Experimental Validation of Performance Incremental learning

Algorithms used: ID3 Quinlan (1986) C4.5 Quinlan(1993) C5.0 Quinlan Cubist Quinlan CART Classification and regression trees Breiman (1984) ASSISTANT Kononenco (1984) & Cestnik (1987) ID3 is algorithm discussed in textbook Simple, but representative Source code publicly available Entropy first time was used C4.5 (and C5.0) is an extension of ID3 that accounts for unavailable values, continuous attribute value ranges, pruning of decision trees, rule derivation, and so on.

Real-valued data Select a set of thresholds defining intervals; each interval becomes a discrete value of the attribute We can use some simple heuristics always divide into quartiles We can use domain knowledge divide age into infant (0-2), toddler (3 - 5), and school aged (5-8) or treat this as another learning problem try a range of ways to discretize the continuous variable Find out which yield “better results” with respect to some metric.

Noisy data and Overfitting Many kinds of " noise " that could occur in the examples: Two examples have same attribute/value pairs , but different classifications Some values of attributes are incorrect because of: Errors in the data acquisition process Errors in the preprocessing phase The classification is wrong (e.g., + instead of -) because of some error Some attributes are irrelevant to the decision-making process, e.g., color of a die is irrelevant to its outcome. Irrelevant attributes can result in overfitting the training data.

Fix overfitting/overlearning problem By cross validation (see later) By pruning lower nodes in the decision tree. For example, if Gain of the best attribute at a node is below a threshold, stop and make this node a leaf rather than generating children nodes. Overfitting: learning result fits data (training examples) well but does not hold for unseen data This means, the algorithm has poor generalization Often need to compromise fitness to data and generalization power Overfitting is a problem common to all methods that learn from data (b and (c): better fit for data, poor generalization (d): not fit for the outlier (possibly due to noise), but better generalization

Pruning Decision Trees Pruning of the decision tree is done by replacing a whole subtree by a leaf node. The replacement takes place if a decision rule establishes that the expected error rate in the subtree is greater than in the single leaf. E.g., Training: eg, one training red success and one training blue Failures Test: three red failures and one blue success Consider replacing this subtree by a single Failure node. After replacement we will have only two errors instead of five failures. Color 1 success 0 failure 0 success 1 failure red blue Color 1 success 3 failure 1 success 1 failure red blue 2 success 4 failure FAILURE

Incremental Learning Incremental learning Change can be made with each training example Non-incremental learning is also called batch learning Good for adaptive system ( learning while experiencing ) when environment undergoes changes Often with Higher computational cost Lower quality of learning results ITI (by U. Mass): incremental DT learning package

Evaluation Methodology Standard methodology: cross validation Collect a large set of examples (all with correct classifications!). Randomly divide collection into two disjoint sets: training and test . 3. Apply learning algorithm to training set giving hypothesis H 4. Measure performance of H w.r.t. test set Important: keep the training and test sets disjoint! Learning is not to minimize training error (wrt data) but the error for test/cross-validation: a way to fix overfitting To study the efficiency and robustness of an algorithm, repeat steps 2-4 for different training sets and sizes of training sets. If you improve your algorithm, start again with step 1 to avoid evolving the algorithm to work well on just this collection.

Restaurant Example Learning Curve

Decision Trees to Rules It is easy to derive a rule set from a decision tree: write a rule for each path in the decision tree from the root to a leaf. In that rule the left-hand side is easily built from the label of the nodes and the labels of the arcs. The resulting rules set can be simplified: Let LHS be the left hand side of a rule. Let LHS' be obtained from LHS by eliminating some conditions. We can certainly replace LHS by LHS' in this rule if the subsets of the training set that satisfy respectively LHS and LHS' are equal. A rule may be eliminated by using metaconditions such as "if no other rule applies".

C4.5 C4.5 is an extension of ID3 that accounts for unavailable values, continuous attribute value ranges, pruning of decision trees, rule derivation, and so on. C4.5: Programs for Machine Learning J. Ross Quinlan, The Morgan Kaufmann Series in Machine Learning, Pat Langley, Series Editor. 1993. 302 pages. paperback book & 3.5" Sun disk. $77.95. ISBN 1-55860-240-2

Summary of DT Learning Inducing decision trees is one of the most widely used learning methods in practice Can out-perform human experts in many problems Strengths include Fast simple to implement can convert result to a set of easily interpretable rules empirically valid in many commercial products handles noisy data Weaknesses include: "Univariate" splits/partitioning using only one attribute at a time so limits types of possible trees large decision trees may be hard to understand requires fixed-length feature vectors

Summary of ID3 Inductive Bias Short trees are preferred over long trees It accepts the first tree it finds Information gain heuristic Places high information gain attributes near root Greedy search method is an approximation to finding the shortest tree Why would short trees be preferred? Example of Occam’s Razor: Prefer simplest hypothesis consistent with the data. (Like Copernican vs. Ptolemic view of Earth’s motion)

Homework Assignment Tom Mitchell’s software See: http://www.cs.cmu.edu/afs/cs.cmu.edu/project/theo-3/www/ml.html Assignment #2 (on decision trees) Software is at: http://www.cs.cmu.edu/afs/cs/project/theo-3/mlc/hw2/ Compiles with gcc compiler Unfortunately, README is not there, but it’s easy to figure out: After compiling, to run: dt [-s <random seed> ] <train %> <prune %> <test %> <SSV-format data file> %train, %prune, & %test are percent of data to be used for training, pruning & testing. These are given as decimal fractions. To train on all data, use 1.0 0.0 0.0 Data sets for PlayTennis and Vote are include with code. Also try the Restaurant example from Russell & Norvig Also look at www.kdnuggets.com/ (Data Sets) Machine Learning Database Repository at UC Irvine - (try “zoo” for fun)

1. Think how the method of finding best variable order for decision trees that we discussed here be adopted for: ordering variables in binary and multi-valued decision diagrams finding the bound set of variables for Ashenhurst and other functional decompositions 2. Find a more precise method for variable ordering in trees, that takes into account special function patterns recognized in data 3. Write a Lisp program for creating decision trees with entropy based variable selection. Questions and Problems

Sources Tom Mitchell Machine Learning, Mc Graw Hill 1997 Allan Moser Tim Finin, Marie desJardins Chuck Dyer

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