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Evaluation of learned models. Kurt Driessens again with slides stolen from Evgueni Smirnov and Hendrik Blockeel. Overview. Motivation Metrics for Classifier Evaluation Methods for Classifier Evaluation & Comparison Costs in Data Mining Cost-Sensitive Classification and Learning
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Evaluation of learned models Kurt Driessens again with slides stolen from Evgueni Smirnov and HendrikBlockeel
Overview • Motivation • Metrics for Classifier Evaluation • Methods for Classifier Evaluation & Comparison • Costs in Data Mining • Cost-Sensitive Classification and Learning • Lift Charts • ROC Curves
Motivation • It is important to evaluate classifier’s generalization performance in order to: • Determine whether to employ the classifier; (For example: when learning the effectiveness of medical treatments from a limited-size data, it is important to estimate the accuracy of the classifiers.) • Optimize the classifier. (For example: when post-pruning decision trees we must evaluate the accuracy of the decision trees on each pruning step.)
Processed data Selection Model’s Evaluation in the KDD Process Knowledge Transformed data Patterns Target data Interpretation Evaluation Data Mining data Transformation & feature selection Preprocessing & cleaning
How to evaluate the Classifier’s Generalization Performance? Assume that we test a classifier on some test set and we derive at the end the following confusion matrix: P N
Metrics for Classifier’s Evaluation Accuracy = (TP+TN)/(P+N) Error = (FP+FN)/(P+N) Precision = TP/(TP+FP) Recall/TP rate = TP/P FP Rate = FP/N P N
How to Estimate the Metrics? • We can use: • Training data; • Independent test data; • Hold-out method; • k-fold cross-validation method; • Leave-one-out method; • Bootstrap method; • And many more…
Estimation with Training Data The accuracy/error estimates on the training data are not good indicators of performance on future data. • Q: Why? • A: Because new data will probably not be exactly the same as the training data! • The accuracy/error estimates on the training data measure the degree of classifier’s overfitting. Classifier Training set Training set
Estimation with Independent Test Data Estimation with independent test data is used when we have plenty of data and there is a natural way to forming training and test data. • For example: Quinlan in 1987 reported experiments in a medical domain for which the classifiers were trained on data from 1985 and tested on data from 1986. Classifier Training set Test set
Hold-out Method The hold-out method splits the data into training data and test data (usually 2/3 for train, 1/3 for test). Then we build a classifier using the train data and test it using the test data. • used with thousands of instances, • including plenty from each class. Classifier Training set Test set Data
Data Y N Classification: Train, Validation, Test Split Results Known Model Builder + Training set + - - + Evaluate Classifier Builder Predictions + - + - Validation set + - + - Final Evaluation Final Test Set Classifier The test data can’t be used for parameter tuning!
Making the Most of the Data • 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.
Stratification • The holdout method reserves a certain amount for testing and uses the remainder for training. • Usually: one third for testing, the rest for training. • For “unbalanced” datasets, samples might not be representative. • Few or none instances of some classes. • Stratified sampling: advanced version of balancing the data. • Make sure that each class is represented with approximately equal proportions in both subsets.
Repeated Holdout Method In general, estimates can be made more reliable by repeated sampling • Each iteration, a certain proportion is randomly selected for training (possibly with stratification). • The error rates on the different iterations are averaged to yield an overall error rate. This is called the repeated holdout method.
Repeated Holdout Method, 2 Random sampling ≠ optimal • the different test sets overlap • we would like all our instances from the data to be tested at least once Can we prevent overlapping?
train train test train test train test train train k-Fold Cross-Validation • k-fold cross-validation avoids overlapping test sets: • First step: data is split into k subsets of equal size; • Second step: each subset in turn is used for testing and the remainder for training. • The subsets are stratified before the cross-validation. • The estimates are averaged to yield an overall estimate. Classifier Data
More on Cross-Validation • Standard method for evaluation: stratified 10-fold cross-validation. • Why 10? Extensive experiments have shown that this is the best choice to get an accurate estimate. • 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 is 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 sub-sampling. • Very computationally expensive.
Leave-One-Out Cross-Validation and Stratification • A disadvantage of Leave-One-Out-CV is that 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!
Bootstrap Method • Cross validation 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 replacementto 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.
Bootstrap Method The bootstrap method is also called the 0.632 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: • This means the training data will contain approximately 63.2% of the instances and the test data will contain approximately 36.8% of the instances.
Estimating Error with the Bootstrap Method The error estimate on the test data will be very pessimistic because the classifier is trained on approx. 63% of the instances. • Therefore, combine it with the training error: • The training error gets less weight than the error on the test data. • Repeat process several times with different replacement samples; average the results.
Confidence Intervals for Performance Assume that the error errorS(h) of the classifier h estimated by the 10-fold cross validation is 25%. • How close is the estimated error errorS(h) to the true error errorD(h) ?
Confidence intervals (2) Iftest data contain n examples, drawn independently of each other, n 30 Then with approximately N% probability, errorD(h) lies in the intervalwhere N%: 50% 68% 80% 90% 95% 98% 99%zN: 0.67 1.00 1.28 1.64 1.96 2.33 2.58
Comparison of hypotheses Given two hypotheses, which one has lower true error? • Statistical hypothesis test: • claim that both are equally good • if claim rejected, accept that 1 is better • 2 cases: • compare 2 hypotheses on possibly different test sets • compare 2 hypotheses on same test set
Different Test Sets To compare h1 and h2, estimate p1-p2from samples S1 (p’1) and S2 (p’2) • if “very likely” p1-p2 > 0 (i.e., confidence interval is entirely to the right of 0): h1 is better • similarly, < 0 : h2 is better • otherwise, no difference demonstrated Formula for confidence interval of difference:
Same Test Set When comparing hypotheses on the same data set, more powerful procedure possible • uses more information from test • possible influence of easy/difficult examples removed More informative method: • for each single example, compare h1 and h2 • how often was h1 correct and h2 wrong on the same example, vs. the other way around? • McNemar’s test
McNemar's test • Consider table: • If h1 is equally good as h2: • for each instance where h1 and h2 differ, probability 0.5 that either is correct • hence we expect B C (B+C)/2 • B and C follow binomial (+/- normal) distribution • Reject equality if B deviates too much from (B+C)/2
Example comparison Consider table below • Method with independent test sets: • 55-45 in favour of h2 (out of 100) • not very convincing • Method with same test set: • much more convincing: 10-0 in favour of h2 - h2 clearly better than h1 - might not be discovered using "conservative" comparison
Metric Evaluation Summary: • Use test sets and the hold-out method for “large” data; • Use the cross-validation method for “middle-sized” data; • Use the leave-one-out and bootstrap methods for small data; Don’t use test data for parameter tuning - use separate validation data. Comparing two classifiers to each other can use more advanced statistics: t-test, McNemar, …
Drawbacks of Accuracy • Evaluation based on accuracy is not always appropriate • Shortcomings: • can sometimes be misleading • unstable when class distribution may change • assumes symmetric misclassification costs
1: Accuracy can be misleading E.g., "99% correct prediction": is this good? • Yes, if 50% "+" and 50% "-" • No, if 1% "+" and 99% "-" always predicting "neg" gives 99% accuracy • Accuracy is a relative measure • Should be compared with "base accuracy" of always predicting the majority class base accuracy in table = max{P,N} / T • Even then, it may be misleading...
Assume all examples -, except blue region (+) • Which of these classifiers is best? Classifier 1 Classifier 2 + + IF green area THEN pos: 92% correct IF false THEN pos 96% correct
An alternative measure: correlation • e.g., correlation = (ad-bc) / TposTminT+T- • close to 1: high correlation predictions - classes • close to 0: no correlation • (close to -1: predicting the opposite) • Avoids the unintuitive results just mentioned actual value note: +/- are actual values pos/neg are predictions prediction
2: Accuracy is sensitive to class distributions If class distribution in test set differs from that in training set, accuracy will also differ E.g.: • Suppose a classifier has TP = 0.8, TN = 0.6 • Tested on test set with T+/T = 0.5, T-/T = 0.5: • Acc = 0.7 • Employed in environment with T+/T = 0.3, T-/T = 0.7: • Acc = 0.66
3: Accuracy ignores misclassification costs Accuracy ignores possibility of different misclassification costs • sometimes, incorrectly predicting "pos" costs more/less than incorrectly predicting "neg” E.g.: • not treating an ill patient vs. treating a healthy patient • refusing credit to client who would have paid back vs. assigning credit to client who won't pay back Need to distinguish probability of making different types of errors
Misclassification Costs Solution: distinguish “predictive accuracy” for different classes • Acc: probability that some instance is classified correctly • Decomposed into • TP: “true positive” rate, (estimated) probability that a positive instance is classified correctly • TN: “true negative” rate, (estimated) probability that a negative instance is classified correctly • We also define • FP = 1-TN: “false positive rate”: estimated probability that a negative is classified as positive • analogously FN = 1-TP
Misclassification Costs (2) Consider costs CFP and CFN = cost of false positive resp. false negative Expected cost of a single prediction: C = CFPP(pos|-)P(-) + CFNP(neg|+)P(+) • estimated by C = CFPFPT-/T + CFNFNT+ /T Note : • Acc is weighted average of TP and TN Acc = TP T+/T + TN T-/T • C is not computable from Acc alone
Cost Sensitive Learning Simple methods for cost sensitive learning: • Resampling of instances according to costs • Weighting of instances according to costs In Weka Cost Sensitive Classification and Learning can be applied for any classifier using the meta scheme: CostSensitiveClassifier.
Lift Charts • In practice, decisions are usually made by comparing possible scenarios taking into account different costs. • E.g. • Promotional mailout to 1,000,000 households. If we mail to all households, we get 0.1% respond (1000). • Data mining tool identifies (a) subset of 100,000 householdswith 0.4% respond (400); or (b) subset of 400,000 householdswith 0.2% respond (800); • Depending on the costs we can make final decision using lift charts! • A lift chart allows a visual comparison.
Generating a Lift Chart Instances are sorted according to their predicted probability of being a true positive:Rank Predicted probability Actual class 1 0.95 Pos 2 0.93 Pos 3 0.93 Neg 4 0.88 Pos … … … In the lift chart, x axis is sample size and y axis is number of true positives.
ROC diagrams • ROC = "Receiver operating characteristic" • Allows to see • how well a classifier will perform given certain misclassification costs and class distribution • in which environments one classifier is better than another • Explicitly aims at solving problems 2 and 3 mentioned before
ROC diagram (2) • ROC diagram plots TP-rate versus FP-rate • From confusion matrix: • TP = a/(a+c) = a/T+ • FP = b/(b+d) = b/T- actual value prediction
Classifier in ROC diagram 1 classifier = 1 point on ROC diagram no positives forgotten TP perfect prediction 1 if false then pos no negatives returned random prediction if true then pos 0 1 FP
Dominance in the ROC Space Classifier A dominates classifier B if and only if TPrA > TPrB and FPrA < FPrB.
ROC Convex Hull (ROCCH) Determined by the dominant classifiers Classifiers below ROCCH are always sub-optimal. Any point of the line segment connecting two classifiers can be achieved by randomly choosing between them; The classifiers on ROCCH can be combined to form a hybrid.
Rank classifiers Rank classifiers assign a rank to their predictions some predictions are more certain than others -> higher rank E.g. (1) decision trees: • use purity of leaf used for prediction to rank it • E.g. leaf with 90% positives is ranked higher than leaf with 80% positives (2) neural nets: • criterion: <0.5 = neg, >=0.5 = pos • but 0.9 is more certainly positive than 0.51 • raise/lower threshold of 0.5:TP and FP go down or up
Rank classifiers yield aROCcurve each specific threshold = 1 point on that curve TP 1 Ranker Ranker with low threshold: better than Blue Ranker with high threshold: worse than Red 0 1 FP
ROC for one Classifier Reasonable separation between the classes, mostly convex. Fairly poor separation between the classes, mostly convex. Good separation between the classes, convex curve. Poor separation between the classes, large and small concavities. Random performance.