# Data Mining Practical Machine Learning Tools and Techniques

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1 Credibility: Evaluating what s been learned Data Mining Practical Machine Learning Tools and Techniques Slides for Chapter 5 of Data Mining by I. H. Witten, E. Frank and M. A. Hall Issues: training, testing, tuning Predicting performance: confidence limits Holdout, cross validation, bootstrap Comparing schemes: the t test Predicting probabilities: loss functions Cost sensitive measures Evaluating numeric prediction The Minimum Description Length principle 2 Evaluation: the key to success Issues in evaluation How predictive is the model we learned? Error on the training data is not a good indicator of performance on future data Otherwise 1 NN would be the optimum classifier! Simple solution that can be used if lots of (labeled) data is available: Split data into training and test set However: (labeled) data is usually limited More sophisticated techniques need to be used Statistical reliability of estimated differences in performance ( significance tests) Choice of performance measure: Number of correct classifications Accuracy of probability estimates Error in numeric predictions Costs assigned to different types of errors Many practical applications involve costs 3 4

2 Training and testing I Training and testing II Natural performance measure for classification problems: error rate Success: instance s class is predicted correctly Error: instance s class is predicted incorrectly Error rate: proportion of errors made over the whole set of instances Resubstitution error: error rate obtained from training data Resubstitution error is (hopelessly) optimistic! Test set: independent instances that have played no part in formation of classifier Assumption: both training data and test data are representative samples of the underlying problem Test and training data may differ in nature Example: classifiers built using customer data from two different towns A and B To estimate performance of classifier from town A in completely new town, test it on data from B 5 6 Note on parameter tuning Making the most of the data It is important that the test data is not used in any way to create the classifier Some learning schemes operate in two stages: Stage 1: build the basic structure Stage 2: optimize parameter settings The test data can t be used for parameter tuning! Proper procedure uses three sets: training data, validation data, and test data Validation data is used to optimize parameters Once evaluation is complete, all the data can be used to build the final classifier Generally, the larger the training data the better the classifier (but returns diminish) The larger the test data the more accurate the error estimate Holdout procedure: method of splitting original data into training and test set Dilemma: ideally both training set and test set should be large! 7 8

3 Predicting performance Confidence intervals Assume the estimated error rate is 25%. How close is this to the true error rate? Depends on the amount of test data Prediction is just like tossing a (biased!) coin Head is a success, tail is an error In statistics, a succession of independent events like this is called a Bernoulli process Statistical theory provides us with confidence intervals for the true underlying proportion We can say: p lies within a certain specified interval with a certain specified confidence Example: S=750 successes in N=1000 trials Estimated success rate: 75% How close is this to true success rate p? Answer: with 80% confidence p in [73.2,76.7] Another example: S=75 and N=100 Estimated success rate: 75% With 80% confidence p in [69.1,80.1] 9 10 Mean and variance Confidence limits Mean and variance for a Bernoulli trial: p, p (1 p) Expected success rate f=s/n Mean and variance for f : p, p (1 p)/n For large enough N, f follows a Normal distribution c% confidence interval [ z X z] for random variable with 0 mean is given by: Pr [ z X z ]=c With a symmetric distribution: Pr [ z X z ]=1 2 Pr [x z ] Confidence limits for the normal distribution with 0 mean and a variance of 1: Thus: Pr[X z] 0.1% 0.5% 10% 20% 40% To use this we have to reduce our random variable f to have 0 mean and unit variance 1% 5% Pr[ 1.65 X 1.65]=90% z

4 Transforming f Examples Transformed value for f : f p p 1 p /N (i.e. subtract the mean and divide by the standard deviation) f = 75%, N = 1000, c = 80% (so that z = 1.28): p [0.732,0.767] f = 75%, N = 100, c = 80% (so that z = 1.28): Resulting equation: Solving for p : Pr [ z f p p(1 p)/ N z ]=c p [0.691,0.801] Note that normal distribution assumption is only valid for large N (i.e. N > 100) f = 75%, N = 10, c = 80% (so that z = 1.28): p= f z2 2N z f N f 2 N z2 4N 2 / 1 z2 N p [0.549,0.881] (should be taken with a grain of salt) Holdout estimation Repeated holdout method What to do if the amount of data is limited? The holdout method reserves a certain amount for testing and uses the remainder for training Usually: one third for testing, the rest for training Problem: the samples might not be representative Example: class might be missing in the test data Advanced version uses stratification Ensures that each class is represented with approximately equal proportions in both subsets Holdout estimate can be made more reliable by repeating the process with different subsamples In each iteration, a certain proportion is randomly selected for training (possibly with stratificiation) The error rates on the different iterations are averaged to yield an overall error rate This is called the repeated holdout method Still not optimum: the different test sets overlap Can we prevent overlapping? 15 16

5 Cross validation More on cross validation Cross validation avoids overlapping test sets First step: split data into k subsets of equal size Second step: use each subset in turn for Testing, the remainder for Learning (training) Called k fold cross validation Often the subsets are stratified before the cross validation is performed The error estimates are averaged to yield an overall error estimate Round S1 S2 S3 1 T L L 2 L T L 3 L L T Standard method for evaluation: stratified ten fold cross validation Why ten? Extensive experiments have shown that this is the best choice to get an accurate estimate There is also some theoretical evidence for this Stratification reduces the estimate s variance Even better: repeated stratified cross validation E.g. ten fold cross validation is repeated ten times and results are averaged (reduces the variance) Leave One Out cross validation Leave One Out CV and stratification Leave One Out: a particular form of cross validation: Set number of folds to number of training instances I.e., for n training instances, build classifier n times Makes best use of the data Involves no random subsampling Very computationally expensive (exception: NN) Disadvantage of Leave One Out CV: stratification is not possible It guarantees a non stratified sample because there is only one instance in the test set! Extreme example: random dataset split equally into two classes Best inducer predicts majority class 50% accuracy on fresh data Leave One Out CV estimate is 100% error! 19 20

6 The bootstrap CV uses sampling without replacement The same instance, once selected, can not be selected again for a particular training/test set The bootstrap uses sampling with replacement to form the training set Sample a dataset of n instances n times with replacement to form a new dataset of n instances Use this data as the training set Use the instances from the original dataset that don t occur in the new training set for testing The bootstrap Also called the bootstrap A particular instance has a probability of 1 1/n of not being picked Thus its probability of ending up in the test data is: 1 1 n n e This means the training data will contain approximately 63.2% of the instances Estimating error with the bootstrap More on the bootstrap The error estimate on the test data will be very pessimistic Trained on just ~63% of the instances Therefore, combine it with the resubstitution error: err=0.632 e test instances e training_instances The resubstitution error gets less weight than the error on the test data Repeat process several times with different replacement samples; average the results Probably the best way of estimating performance for very small datasets However, it has some problems Consider the random dataset from above A perfect memorizer will achieve 0% resubstitution error and ~50% error on test data Bootstrap estimate for this classifier: err= % %=31.6% True expected error: 50% 23 24

7 Comparing data mining schemes Comparing schemes II Frequent question: which of two learning schemes performs better? Note: this is domain dependent! Obvious way: compare 10 fold CV estimates Generally sufficient in applications (we don't loose if the chosen method is not truly better) However, what about machine learning research? Need to show convincingly that a particular method works better Want to show that scheme A is better than scheme B in a particular domain For a given amount of training data On average, across all possible training sets Let's assume we have an infinite amount of data from the domain: Sample infinitely many dataset of specified size Obtain cross validation estimate on each dataset for each scheme Check if mean accuracy for scheme A is better than mean accuracy for scheme B Paired t test Distribution of the means In practice we have limited data and a limited number of estimates for computing the mean Student s t test tells whether the means of two samples are significantly different In our case the samples are cross validation estimates for different datasets from the domain Use a paired t test because the individual samples are paired The same CV is applied twice x1 x 2 x k and y 1 y 2 y k are the 2k samples for the k different datasets mx and m y are the means With enough samples, the mean of a set of independent samples is normally distributed Estimated variances of the means are x2 /k and y2 /k William Gosset Born: 1876 in Canterbury; Died: 1937 in Beaconsfield, England Obtained a post as a chemist in the Guinness brewery in Dublin in Invented the t test to handle small samples for quality control in brewing. Wrote under the name "Student". If x and y are the true means then m x x x 2 /k are approximately normally distributed with mean 0, variance 1 m y y y 2 /k 27 28

8 Student s distribution Distribution of the differences With small samples (k < 100) the mean follows Student s distribution with k 1 degrees of freedom Confidence limits: Let m d = m x m y The difference of the means (m d ) also has a Student s distribution with k 1 degrees of freedom Let d 2 be the variance of the difference Assuming we have 10 estimates 9 degrees of freedom normal distribution Pr[X z] 0.1% 0.5% 1% 5% 10% 20% z Pr[X z] 0.1% 0.5% 1% 5% 10% 20% z The standardized version of m d is called the t statistic: t= m d d 2 /k We use t to perform the t test Performing the test Unpaired observations Fix a significance level If a difference is significant at the % level, there is a (100 )% chance that the true means differ Divide the significance level by two because the test is two tailed I.e. the true difference can be +ve or ve Look up the value for z that corresponds to /2 If t z or t z then the difference is significant I.e. the null hypothesis (that the difference is zero) can be rejected If the CV estimates are from different datasets, they are no longer paired (or maybe we have k estimates for one scheme, and j estimates for the other one) Then we have to use an unpaired t test with min(k, j) 1 degrees of freedom The estimate of the variance of the difference of the means becomes: x 2 2 y k j 31 32

9 Dependent estimates We assumed that we have enough data to create several datasets of the desired size Need to re use data if that's not the case E.g. running cross validations with different randomizations on the same data Samples become dependent insignificant differences can become significant A heuristic test is the corrected resampled t test: Assume we use the repeated (k fold) hold out method, with n 1 instances for training and n 2 for testing New test statistic is: t= m d 1 k n 2 n 1 d 2 Predicting probabilities Performance measure so far: success rate Also called 0 1 loss function: i { 0 if prediction is correct 1 if prediction is incorrect } Most classifiers produces class probabilities Depending on the application, we might want to check the accuracy of the probability estimates 0 1 loss is not the right thing to use in those cases Quadratic loss function Informational loss function p 1 p k are probability estimates for an instance c is the index of the instance s actual class a 1 a k = 0, except for a c which is 1 Quadratic loss is: Want to minimize j (p j a j ) 2 = j!=c p j 2 +(1 p c ) 2 E[ j p j a j 2 ] The informational loss function is log(p c ), where c is the index of the instance s actual class Number of bits required to communicate the actual class * * Let p 1 p k be the true class probabilities Then the expected value for the loss function is: p 1 log 2 p 1... p k log 2 p k Can show that this is minimized when p j = p j*, the true probabilities Justification: minimized when p j = p j * Difficulty: zero frequency problem 35 36

10 Discussion Counting the cost Which loss function to choose? Both encourage honesty Quadratic loss function takes into account all class probability estimates for an instance Informational loss focuses only on the probability estimate for the actual class Quadratic loss is bounded: 2 it can never exceed 2 1 j p j Informational loss can be infinite Informational loss is related to MDL principle [later] In practice, different types of classification errors often incur different costs Examples: Terrorist profiling Not a terrorist correct 99.99% of the time Loan decisions Oil slick detection Fault diagnosis Promotional mailing Counting the cost The confusion matrix: Aside: the kappa statistic Two confusion matrices for a 3 class problem: actual predictor (left) vs. random predictor (right) Actual class Yes No Predicted class Yes No True positive False negative False positive True negative There are many other types of cost! E.g.: cost of collecting training data Number of successes: sum of entries in diagonal (D) Kappa statistic: D observed D random D perfect D random measures relative improvement over random predictor 39 40

11 Classification with costs Cost sensitive classification Two cost matrices: Success rate is replaced by average cost per prediction Cost is given by appropriate entry in the cost matrix Can take costs into account when making predictions Basic idea: only predict high cost class when very confident about prediction Given: predicted class probabilities Normally we just predict the most likely class Here, we should make the prediction that minimizes the expected cost Expected cost: dot product of vector of class probabilities and appropriate column in cost matrix Choose column (class) that minimizes expected cost Cost sensitive classification: Example Cost sensitive learning Cost Matrix Spam predicted class actual class Spam 0 1 Let p s =Pr(Spam) Ham Ham 10 0 Regard expected costs for each class (dot product of vector of class probabilities and appropriate column in cost matrix) EC(Spam)=p s *0 + (1 p s )*10 EC(Ham)=p s *1 + (1 p s )*0 Decide for Spam iff EC(Spam) < EC(Ham): p s > 10/11 So far we haven't taken costs into account at training time Most learning schemes do not perform costsensitive learning They generate the same classifier no matter what costs are assigned to the different classes Example: standard decision tree learner Simple methods for cost sensitive learning: Resampling of instances according to costs Weighting of instances according to costs Some schemes can take costs into account by varying a parameter, e.g. naïve Bayes 43 44

12 Lift charts Generating a lift chart In practice, costs are rarely known Decisions are usually made by comparing possible scenarios Example: promotional mailout to 1,000,000 households Mail to all; 0.1% respond (1000) Data mining tool identifies subset of 100,000 most promising, 0.4% of these respond (400) 40% of responses for 10% of cost may pay off Identify subset of 400,000 most promising, 0.2% respond (800) A lift chart allows a visual comparison Sort instances according to predicted probability of being positive: Predicted probability x axis is sample size y axis is number of true positives Actual class Yes Yes No Yes A hypothetical lift chart ROC curves 40% of responses for 10% of cost 80% of responses for 40% of cost ROC curves are similar to lift charts Stands for receiver operating characteristic Used in signal detection to show tradeoff between hit rate and false alarm rate over noisy channel Differences to lift chart: y axis shows percentage of true positives in sample rather than absolute number x axis shows percentage of false positives in sample rather than sample size 47 48

13 A sample ROC curve Cross validation and ROC curves Simple method of getting a ROC curve using cross validation: Collect probabilities for instances in test folds Sort instances according to probabilities This method is implemented in WEKA However, this is just one possibility Another possibility is to generate an ROC curve for each fold and average them Jagged curve one set of test data Smooth curve use cross validation ROC curves for two schemes The convex hull Given two learning schemes we can achieve any point on the convex hull! TP and FP rates for scheme 1: t 1 and f 1 For a small, focused sample, use method A For a larger one, use method B In between, choose between A and B with appropriate probabilities TP and FP rates for scheme 2: t 2 and f 2 If scheme 1 is used to predict 100 q % of the cases and scheme 2 for the rest, then TP rate for combined scheme: q t 1 + (1 q) t 2 FP rate for combined scheme: q f 1 +(1 q) f

14 More measures... Summary of some measures Percentage of retrieved documents that are relevant: precision=tp/(tp+fp) Percentage of relevant documents that are returned: recall =TP/(TP+FN) Precision/recall curves have hyperbolic shape Summary measures: average precision at 20%, 50% and 80% recall (three point average recall) F measure=(2 recall precision)/(recall+precision) sensitivity specificity = (TP / (TP + FN)) (TN / (FP + TN)) Area under the ROC curve (AUC): probability that randomly chosen positive instance is ranked above randomly chosen negative one Lift chart ROC curve Recallprecision curve Domain Marketing Communicatio ns Information retrieval Plot TP Subset size TP rate FP rate Recall Precisio n Explanation TP (TP+FP)/ (TP+FP+TN+FN) TP/(TP+FN) FP/(FP+TN) TP/(TP+FN) TP/(TP+FP) Cost curves Cost curves plot expected costs directly Example for case with uniform costs (i.e. error): Cost curves: example with costs Probability cost functionp c [+]= p[+]c[+ -] p[+]c[+ -] p[-]c[- +] Normalized expected cost=fn p c [+] fp 1 p c [+] 55 56

15 Evaluating numeric prediction Same strategies: independent test set, cross validation, significance tests, etc. Difference: error measures Actual target values: a 1 a 2 a n Predicted target values: p 1 p 2 p n Most popular measure: mean squared error p 1 a p n a n 2 Easy to manipulate mathematically n Other measures The root mean squared error : p 1 a p n a n 2 n The mean absolute error is less sensitive to outliers than the mean squared error: p 1 a 1... p n a n n Sometimes relative error values are more appropriate (e.g. 10% for an error of 50 when predicting 500) Improvement on the mean Correlation coefficient How much does the scheme improve on simply predicting the average? Measures the statistical correlation between the predicted values and the actual values The relative squared error is: p 1 a p n a n 2 a a a a n 2 The relative absolute error is: p 1 a 1... p n a n a a 1... a a n S PA = i p i p a i a n 1 Scale independent, between 1 and +1 Good performance leads to large values! S PA S P S A S P = i p i p 2 n 1 S A = i a i a 2 n

16 Which measure? The MDL principle Best to look at all of them Often it doesn t matter Example: A Root mean-squared error 67.8 Mean absolute error 41.3 Root rel squared error 42.2% Relative absolute error 43.1% Correlation coefficient 0.88 B % 40.1% 0.88 C % 34.8% 0.89 D % 30.4% 0.91 MDL stands for minimum description length The description length is defined as: space required to describe a theory + space required to describe the theory s mistakes In our case the theory is the classifier and the mistakes are the errors on the training data Aim: we seek a classifier with minimal DL MDL principle is a model selection criterion D best C second-best A, B arguable Model selection criteria Elegance vs. errors Model selection criteria attempt to find a good compromise between: The complexity of a model Its prediction accuracy on the training data Reasoning: a good model is a simple model that achieves high accuracy on the given data Also known as Occam s Razor : the best theory is the smallest one that describes all the facts William of Ockham, born in the village of Ockham in Surrey (England) about 1285, was the most influential philosopher of the 14th century and a controversial theologian. Theory 1: very simple, elegant theory that explains the data almost perfectly Theory 2: significantly more complex theory that reproduces the data without mistakes Theory 1 is probably preferable Classical example: Kepler s three laws on planetary motion Less accurate than Copernicus s latest refinement of the Ptolemaic theory of epicycles 63 64

17 MDL and compression MDL and Bayes s theorem MDL principle relates to data compression: The best theory is the one that compresses the data the most I.e. to compress a dataset we generate a model and then store the model and its mistakes We need to compute (a) size of the model, and (b) space needed to encode the errors (b) easy: use the informational loss function (a) need a method to encode the model 65 L[T]= length of the theory L[E T]=training set encoded wrt the theory Description length= L[T] + L[E T] Bayes s theorem gives a posteriori probability of a theory given the data: Pr [E T]Pr [T] Pr [T E]= Pr [E] Equivalent to: logpr [T E]= logpr [E T] logpr[t] logpr[e] constant 66 MDL and MAP Discussion of MDL principle MAP stands for maximum a posteriori probability Finding the MAP theory corresponds to finding the MDL theory Difficult bit in applying the MAP principle: determining the prior probability Pr[T] of the theory Corresponds to difficult part in applying the MDL principle: coding scheme for the theory I.e. if we know a priori that a particular theory is more likely we need fewer bits to encode it Advantage: makes full use of the training data when selecting a model Disadvantage 1: appropriate coding scheme/prior probabilities for theories are crucial Disadvantage 2: no guarantee that the MDL theory is the one which minimizes the expected error Note: Occam s Razor is an axiom! Epicurus s principle of multiple explanations: keep all theories that are consistent with the data 67 68

18 MDL and clustering Description length of theory: bits needed to encode the clusters e.g. cluster centers Description length of data given theory: encode cluster membership and position relative to cluster e.g. distance to cluster center Works if coding scheme uses less code space for small numbers than for large ones With nominal attributes, must communicate probability distributions for each cluster 69

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