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Feature selection and transduction for prediction of molecular bioactivity for drug design Reporter: Yu Lun Kuo (D95922037)

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Presentation on theme: "Feature selection and transduction for prediction of molecular bioactivity for drug design Reporter: Yu Lun Kuo (D95922037)"— Presentation transcript:

1 Feature selection and transduction for prediction of molecular bioactivity for drug design Reporter: Yu Lun Kuo (D95922037) E-mail: sscc6991@gmail.comsscc6991@gmail.com Date: April 17, 2008 Bioinformatics Vol. 19 no. 6 2003 (Pages 764-771)

2 Abstract Drug discovery –Identify characteristics that separate active (binding) compounds from inactive ones. Two method for prediction of bioactivity –Feature selection method –Transductive method Improvement over using only one of the techniques 2015/4/222

3 Introduction (1/4) Discovery of a new drug –Testing many small molecules for their ability to bind to the target site –The task of determining what separate the active (binding) compounds from the inactive ones 2015/4/223

4 Introduction (2/4) Design new compounds –Not only bind –But also possess certain other properties required for a drug The task of determination can be seen in a machine learning context as one of feature selection 2015/4/224

5 Introduction (3/4) Challenging –Few positive examples Little information is given indicating positive correlation between features and the labels –Large number of features Selected from a huge collection of useful features Some features are in reality uncorrelated with the labels –Different distributions Cannot expect the data to come from a fix distribution 2015/4/225

6 Introduction (4/4) Many conventional machine learning algorithms are illequiped to deal with these Many algorithms generalize poorly –The high dimensionality of the problem –The problem size many methods are no longer computationally feasible –Most cannot deal with training and testing data coming from different distributions 2015/4/226

7 Overcome Feature selection criterion –Called unbalanced correlation score Take into account the unbalanced nature of the data Simple enough to avoid overfitting Classifier –Takes into account the different distributions in the test data compared to the training data Induction Transduction 2015/4/227

8 Overcome Induction –Builds a model based only on the distribution of the training data Transduction –Also take into account the test data inputs Combining these two techniques we obtained improved prediction accuracy 2015/4/228

9 KDD Cup Competition (1/2) We focused on a well studies data set –KDD Cup 2001 competition Knowledge Discovery and Data Mining One of the premier meetings of the data mining community –http://www.kdnuggets.com/datasets/kddcup.htmlhttp://www.kdnuggets.com/datasets/kddcup.html 2015/4/229

10 KDD Cup Competition (2/2) KDD Cup 2006 –data mining for medical diagnosis, specifically identifying pulmonary embolisms from three-dimensional computed tomography data KDD Cup 2004 – features tasks in particle physics and bioinformatics evaluated on a variety of different measures KDD Cup 2002 –focus: bioinformatics and text mining KDD Cup 2001 –focus: bioinformatics and drug discovery 2015/4/2210

11 KDD Cup 2001 (1/2) Objective –Prediction of molecular bioactivity for drug design -- binding to Thrombin Data –Training: 1909 cases (42 positive), 139,351 binary features –Test: 634 cases 2015/4/2211

12 KDD Cup 2001 (2/2) Challenge –Highly imbalanced, high-dimensional, different distribution Approach –Bayesian network predictive model –Data PreProcessor system –BN PowerPredictor system –BN PowerConstructor system 2015/4/2212

13 Data Set (1/3) Provided by DuPont Pharmaceuticals –Drug binds to a target site on thrombin, a key receptor in blood clotting Each example has a fixed length vector of 139,351 binary features in {0, 1} –Which describe three-dimensional properties of the molecule 2015/4/2213

14 Data Set (2/3) Positive examples are labeled +1 Negative examples are labeled -1 In the training set –1909 examples, 42 of which bind (rather unbalanced, positive is 2.2%) In the test set –634 additional compounds 2015/4/2214

15 Data Set (3/3) An important characteristic of the data –Very few of the feature entries are non-zero (0.68% of the 1,909 X 139,351 training matrix) 2015/4/2215

16 System Assessment Performance is evaluated according to a weighted accuracy criterion –The score of an estimate y’ of the labels y –Complete success is a score of 1 Multiply this score by 100 as the percentage weighted success rate 2015/4/2216

17 Methodology Predict the labels on the test set by using a machine learning algorithm The positively and negatively labeled training examples are split randomly into n groups –For n-fold cross validation such that as close to 1/n of the positively labeled examples are present in each group as possible Called balanced cross validation –As few positive examples 2015/4/2217

18 Methodology The method is –Trained on n-1 of groups –Tested on the remaining group –Repeated n times (different group for testing) –Final score: mean of the n scores 2015/4/2218

19 Feature Selection (1/2) Called the unbalanced correlation score –f j : the score of feature j –X: training data as a matrix X where columns are features and examples are rows Take λ very large in order to select features which have non-zero entries (λ ≧ 3) 2015/4/2219

20 Feature Selection (2/2) This score is an attempt to encode prior information that –The data is unbalanced –Large number of features –Only positive correlations are likely to be useful 2015/4/2220

21 Justification Justify the unbalanced correlation score using methods of information theory –Entropy: higher  non-regular Pi: the probability of appearance of event i 2015/4/2221

22 Entropy The probability of random appearance of a feature with an unbalanced score of N=N p -N n –N p = number of one entries associated to +1 –N n = number of one entries associated to -1 –T p = total number of positive labels in training set –T n = total number of negative labels in training set 2015/4/2222

23 Entropy Need to compute the probability that a certain N might occur randomly Finally, compute the entropy for each feature 2015/4/2223

24 Entropy and unbalanced score The entropy and unbalanced score will not reach the same feature –Because the unbalanced correlation score will no select samples with low negative In this particular problem –Reach a similar ranking of the features Due to the unbalanced nature of the data 2015/4/2224

25 Entropy and unbalanced score The first 6 features for both scores –5 out of 6 are the same ones –For 16 features, 12 coincide –Pay more attention to positive correlations 2015/4/2225

26 Multivariate unbalanced correlation The feature selection algorithm described so far is univariate –Reduces the chance of overfitting –Between the inputs and targets are too complex this assumption may be to restrictive We extend our criterion to assign a rank to a subset of feature –Rather than just a single feature 2015/4/2226

27 Multivariate unbalanced correlation By computing the logical OR of the subset of features S (as they are binary) 2015/4/2227

28 Fisher Score –μ(+): the mean of the feature values for positive –μ(-): the mean of the feature values for negative –σ(+): standard deviations –σ(-): standard deviations 2015/4/2228

29 In each case, the algorithms are evaluated for different numbers of features d –The range d = 1, …, 40 Choose a small number of features in order to render interpretability of the decision function It is anticipated that a large number of features are noisy and should not be selected 2015/4/2229

30 Classification algorithms (Inductive) The task may not simply be just to identify relevant characteristics via feature selection –But also to provide a prediction system Simplest of classifiers –We call this a logical OR classifier 2015/4/2230

31 Comparison Techniques We compared a number of rather more sophisticated classification –Support vector machines (SVM) –SVM* Make a search over all possible values of the threshold parameter in the linear model after training –K-nearest neighbors (K-NN) –K-NN* (parameter γ) –C4.5 (decision tree learner) 2015/4/2231

32 Transductive Inference One is given labeled data from which builds a general model –Then applies this model to classify previously unseen (test) data Takes into account not only the given (labeled) training set but also unlabeled data –That one wishes to classify 2015/4/2232

33 Transductive Inference Different models can be built –Trying to classify different test sets –Even if the training set is the same in all cases It is this characteristic which help to solve problem 3 –The data we are given has different distribution in the training and test sets 2015/4/2233

34 Transductive Inference Transduction is not useful in all tasks –In drug discovery in particular we believe it is useful Developers often have access to huge databases of compounds –Compounds are often generated using virtual Combinatorial Chemistry –Compound descriptors can be computed even though the compounds have not been synthesized yet 2015/4/2234

35 Transductive Inference Drug discovery is an iterative process –Machine learning method is to help choose the next test set –Step in a two-step candidate selection procedure After candidate test set has been produced Its result is the final test set 2015/4/2235

36 Transductive algorithm 2015/4/2236

37 Results (with unbalanced correlation score) C4.5 gave only 50% success rate for all 2015/4/2237 The tansductive algorithm is consistently selecting more relevant features than the inductive one the Fisher score

38 Further Results We also tested some more sophisticated multivariate feature selection methods –Not as good as using the unbalanced criterion score Using non-linear SVMs –Not improve results (50% success) SVMs as a base classifier for our transduction –Improvement over using SVMs 2015/4/2238

39 Further Results Also tried training the classifiers with larger numbers of features –Inductive methods failed to learn anything after 200 features –Transductive methods Exhibit generalization behavior up to 1000 features (TRANS-Orcub:58% success with d=1000,77% with d=200) –KDD champion Success rate 68.4% (7% of entrants higher than 60%) 2015/4/2239

40 2015/4/2240 Thanks for your attention


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