Tutorial 2. Fall /21. CPSC 340: Machine Learning and Data Mining
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1 1/21 Tutorial 2 CPSC 340: Machine Learning and Data Mining Fall 2016
2 Overview 2/21 1 Decision Tree Decision Stump Decision Tree 2 Training, Testing, and Validation Set 3 Naive Bayes Classifier
3 Decision Stump 3/21 Decision stump: simple decision tree with 1 splitting rule based on 1 feature. Binary example: Assigns a label to each leaf based on the most frequent label. Most intuitive score: classification accuracy.
4 The newsgroups.mat Dataset 4/21 The newsgroups.mat Matlab file contains the following objects: 1 X: A sparse binary matrix. Each row corresponds to a post, and each column corresponds to a word from the word list. A value of 1 means that the word occurred in the post. 2 y: A vector with values 1 through 4, with the value corresponding to the newsgroup that the post came from. 3 Xtest and ytest: the word lists and newsgroup labels for additional newsgroup posts. 4 groupnames: The names of four newsgroups. 5 wordlist: A list of words that occur in posts to these newsgroups.
5 Example: Binary Decision Stump for newsgroups.mat 5/21
6 Decision Tree 6/21 Decision stumps have only 1 rule based on only 1 feature. Very limited class of models: usually not very accurate for most tasks. Decision trees allow sequences of splits based on multiple features. Very general class of models: can get very high accuracy. However, it s computationally infeasible to find the best decision tree. Most common decision tree learning algorithm in practice: Greedy recursive splitting.
7 Problem 1: Decision Tree for newsgroups.mat 7/21 For a maximum depth of 2, 1) draw the learned decision tree. and 2) re-write the function as a simple program using if/else statements.
8 Solution: Decision Tree for newsgroups.mat 8/21
9 Solution: Decision Tree for newsgroups.mat 9/21 Decision tree: If-else statement:
10 Training, Testing, and Validation Set 10/21 Given training data, we would like to learn a model to minimize error on the testing data How do we decide decision tree depth? We care about test error. But we can t look at test data. So what do we do????? One answer: Use part of your train data to approximate test error. Split training objects into training set and validation set: Train model on the training data. Test model on the validation data.
11 Cross-Validation 11/21 Isn t it wasteful to only use part of your data? k-fold cross-validation: Train on k-1 folds of the data, validate on the other fold. Repeat this k times with different splits, and average the score. Figure 1: Adapted from Wikipedia. Note: if examples are ordered, split should be random.
12 12/21 Problem 2: 2-Fold Cross Validation for newsgroups.mat Modify the code below to compute the 2-fold cross-validation scores on the training data alone. Find the depth that would be chosen by cross-validation.
13 Solution: 2-Fold Cross Validation for newsgroups.mat 13/21
14 Naive Bayes Classifier 14/21 Naive Bayes is a probabilistic classifier. Based on Bayes theorem. Strong independence assumption between features.
15 Naive Bayes Classifier 14/21 Naive Bayes is a probabilistic classifier. Based on Bayes theorem. Strong independence assumption between features. In the rest of this tutorial, We use y i for the label of object i (element i of y). We use x i for the features of object i (row i of X). We use x ij for feature j of object i. We use d for the number of features in object i.
16 Naive Bayes Classifier 15/21 Bayes rule
17 Naive Bayes Classifier 15/21 Bayes rule Since the denominator does not depend on y i, we are only interested in the numerator: p(y i x i ) p(x i y i )p(y i )
18 Naive Bayes Classifier 16/21 The numerator is equal to the joint probability: p(x i y i )p(y i ) = p(x i, y i ) = p(x i1,..., x id, y i )
19 Naive Bayes Classifier 16/21 The numerator is equal to the joint probability: p(x i y i )p(y i ) = p(x i, y i ) = p(x i1,..., x id, y i ) Chain rule: p(x i1,..., x id, y i ) = p(x i1 x i2,..., x id, y i )p(x i2,..., x id, y i ) =... = p(x i1 x i2,..., x id, y i )p(x i2 x i3,..., x id, y i )... p(x id y i )p(y i )
20 Naive Bayes Classifier 16/21 The numerator is equal to the joint probability: p(x i y i )p(y i ) = p(x i, y i ) = p(x i1,..., x id, y i ) Chain rule: p(x i1,..., x id, y i ) = p(x i1 x i2,..., x id, y i )p(x i2,..., x id, y i ) =... = p(x i1 x i2,..., x id, y i )p(x i2 x i3,..., x id, y i )... p(x id y i )p(y i ) Each feature in x i is independent of the others given y i : p(x ij x ij+1,..., x id, y i ) = p(x ij y i )
21 Naive Bayes Classifier The numerator is equal to the joint probability: p(x i y i )p(y i ) = p(x i, y i ) = p(x i1,..., x id, y i ) Chain rule: p(x i1,..., x id, y i ) = p(x i1 x i2,..., x id, y i )p(x i2,..., x id, y i ) =... = p(x i1 x i2,..., x id, y i )p(x i2 x i3,..., x id, y i )... p(x id y i )p(y i ) Each feature in x i is independent of the others given y i : p(x ij x ij+1,..., x id, y i ) = p(x ij y i ) Therefore: d p(y i, x i ) p(y i ) p(x ij y i ) j=1 16/21
22 Problem 4: Naive Bayes Classifier 17/21
23 Problem 4: Naive Bayes Classifier 17/21
24 Solution: Naive Bayes Classifier 18/21 We need
25 Solution: Naive Bayes Classifier 18/21 We need p(flu = N headache = Y, runny nose = N, fever = Y ) p(headache = Y flu = N)p(runny nose = N flu = N)p(fever = Y flu = N)p(flu = N) = =
26 Solution: Naive Bayes Classifier 18/21 We need p(flu = N headache = Y, runny nose = N, fever = Y ) p(headache = Y flu = N)p(runny nose = N flu = N)p(fever = Y flu = N)p(flu = N) = = p(flu = Y headache = Y, runny nose = N, fever = Y ) p(headache = Y flu = Y )p(runny nose = N flu = Y )p(fever = Y flu = Y )p(flu = Y ) = =
27 Solution: Naive Bayes Classifier 18/21 We need p(flu = N headache = Y, runny nose = N, fever = Y ) p(headache = Y flu = N)p(runny nose = N flu = N)p(fever = Y flu = N)p(flu = N) = = p(flu = Y headache = Y, runny nose = N, fever = Y ) p(headache = Y flu = Y )p(runny nose = N flu = Y )p(fever = Y flu = Y )p(flu = Y ) = =
28 Bayes Theorem 19/21 Bayes Theorem enables us to reverse probabilities:
29 Problem 3: Prosecutor s fallacy 20/21 A crime has been committed in a large city and footprints are found at the scene of the crime. The guilty person matches the footprints, p(f G) = 1. Out of the innocent people, 1% match the footprints by chance, p(f G) = A person is interviewed at random and his/her footprints are found to match those at the crime scene. Determine the probability that the person is guilty, or explain why this is not possible, p(g F ) =? Let F be the event that the footprints match. Let G be the event that the person is guilty G be the event that the person is innocent.
30 Solution: Prosecutor s fallacy 21/21 p(g F ) = p(f G)p(G) p(f ) = p(f G)p(G) p(f G)p(G) + p(f G)p( G)
31 Solution: Prosecutor s fallacy 21/21 p(g F ) = p(f G)p(G) p(f ) = p(f G)p(G) p(f G)p(G) + p(f G)p( G) p(g) =? Impossible!
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