1 / 70

Overfitting

Overfitting. Overfitting occurs when a statistical model describes random error or noise instead of the underlying relationship. Overfitting generally occurs when a model is excessively complex, such as having too many parameters relative to the number of observations.

quamar-conrad
Download Presentation

Overfitting

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Overfitting Overfitting occurs when a statistical model describes random error or noise instead of the underlying relationship. Overfitting generally occurs when a model is excessively complex, such as having too many parameters relative to the number of observations. A model which has been overfit will generally have poor predictive performance, as it can exaggerate minor fluctuations in the data.

  2. Suppose we need to solve a classification problem • We are not sure if we should us the.. • Simple linear classifier • or the • Simple quadratic classifier • How do we decide which to use? • We do cross validation and • choose the best one.

  3. 100 100 90 90 80 80 70 70 60 60 50 50 40 40 30 30 20 20 10 10 10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 80 100 100 90 90 • Simple linear classifier gets 81% accuracy • Simple quadratic classifier 99% accuracy

  4. Simple linear classifier gets 96% accuracy • Simple quadratic classifier 97% accuracy

  5. This problem is greatly exacerbated by having too little data • Simple linear classifier gets 90% accuracy • Simple quadratic classifier 95% accuracy

  6. What happens as we have more and more training examples? The accuracy for all models goes up! The chance of making a mistake goes down The cost of the mistake (if made) goes down • Simple linear 70% accuracy • Simple quadratic 90% accuracy • Simple linear 90% accuracy • Simple quadratic 95% accuracy • Simple linear 99% accuracy • Simple quadratic 99% accuracy

  7. One Solution: Charge Penalty for complex models • For example, for the simple {polynomial} classifier, we could charge 1% for every increase in the degree of the polynomial • Simple linear classifier gets 90.5% accuracy, minus 0, equals 90.5% • Simple quadratic classifier 97.0% accuracy, minus 1, equals 96.0% • Simple cubic classifier 97.05% accuracy, minus 2, equals 95.05% Accuracy = 90.5% Accuracy = 97.0% Accuracy = 97.05% 10 10 10 9 9 9 8 8 8 7 7 7 6 6 6 5 5 5 4 4 4 3 3 3 2 2 2 1 1 1 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 10 10

  8. One Solution: Charge Penalty for complex models • For example, for the simple {polynomial} classifier, we could charge 1% for every increase in the degree of the polynomial. • There are more principled ways to charge penalties • In particular, there is a technique called Minimum Description Length (MDL)

  9. Suppose you have a four feature problem, and you want to search over feature subsets. It happens to be the case that features 2 and 3, shown here Are all you need, and the other features are random

  10. 1 2 3 4 1,2 1,3 2,3 1,4 2,4 3,4 1,2,3 1,2,4 1,3,4 2,3,4 1,2,3,4 Suppose you have a four feature problem, and you want to search over feature subsets. It happens to be the case that features 2 and 3, shown here are all you need, and the other features are random 0 1 2 3 4

  11. My_Collection • We have seen that we are given features… • Suppose using these features we cannot get satisfactory accuracy results. • So far, we have two tricks • Ask for more features • Remove irrelevant or redundant features • There is another possibility…

  12. 5 2.5 10 9 8 7 6 2 5 5 4 3 2 1 5 3 1 2 3 4 5 6 7 8 10 9 2.5 3 Feature Generation • Feature generation refers to any technique to make new features from existing features • Recall pigeon problem 2, and assume we are using the linear classifier Pigeon Problem 2 Examples of class A Examples of class B Using both features works poorly, using just X works poorly, using just Y works poorly.. 4 4 5 5 6 6 3 3

  13. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 10 9 Feature Generation • Solution: Create a new feature Z Z = absolute_value(X-Y) 0 1 2 3 4 5 6 7 8 10 9 Z-axis

  14. Recall this example? It was a teaching example to show that NN could use any distance measure It would not really work very well, unless we had LOTS more data…

  15. Irish Names Japanese Names ABERCROMBIE ABERNETHY ACKART ACKERMAN ACKERS ACKLAND ACTON ADAIR ADLAM ADOLPH AFFLECK ALVIN AMMADON AIKO AIMI AINA AIRI AKANE AKEMI AKI AKIKO AKIO AKIRA AMI AOI ARATA ASUKA

  16. Z = number of vowels / word length Vowels = I O U A E Irish Names Japanese Names ABERCROMBIE 0.45 ABERNETHY 0.33 ACKART 0.33 ACKERMAN 0.375 ACKERS 0.33 ACKLAND 0.28 ACTON 0.33 AIKO 0.75 AIMI 0.75 AINA 0.75 AIRI 0.75 AKANE 0.6 AKEMI 0.6

  17. I have a box of apples.. 1 H(X) Pr(X = good) = p then Pr(X = bad) = 1 − p the entropy of X is given by 0.5 0 0 binary entropy function attains its maximum value when p = 0.5 1 All good All bad

  18. Decision Tree Classifier 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Ross Quinlan Abdomen Length > 7.1? Antenna Length yes no Antenna Length > 6.0? Katydid yes no Katydid Grasshopper Abdomen Length

  19. Antennae shorter than body? Yes No 3 Tarsi? Grasshopper Yes No Foretiba has ears? Yes No Cricket Decision trees predate computers Katydids Camel Cricket

  20. Decision Tree Classification • Decision tree • A flow-chart-like tree structure • Internal node denotes a test on an attribute • Branch represents an outcome of the test • Leaf nodes represent class labels or class distribution • Decision tree generation consists of two phases • Tree construction • At start, all the training examples are at the root • Partition examples recursively based on selected attributes • Tree pruning • Identify and remove branches that reflect noise or outliers • Use of decision tree: Classifying an unknown sample • Test the attribute values of the sample against the decision tree

  21. How do we construct the decision tree? • Basic algorithm (a greedy algorithm) • Tree is constructed in a top-down recursive divide-and-conquer manner • At start, all the training examples are at the root • Attributes are categorical (if continuous-valued, they can be discretized in advance) • Examples are partitioned recursively based on selected attributes. • Test attributes are selected on the basis of a heuristic or statistical measure (e.g., information gain) • Conditions for stopping partitioning • All samples for a given node belong to the same class • There are no remaining attributes for further partitioning – majority voting is employed for classifying the leaf • There are no samples left

  22. Information Gain as A Splitting Criteria • Select the attribute with the highest information gain (information gain is the expected reduction in entropy). • Assume there are two classes, P and N • Let the set of examples S contain p elements of class P and n elements of class N • The amount of information, needed to decide if an arbitrary example in S belongs to P or N is defined as 0 log(0) is defined as0

  23. Information Gain in Decision Tree Induction • Assume that using attribute A, a current set will be partitioned into some number of child sets • The encoding information that would be gained by branching on A Note: entropy is at its minimum if the collection of objects is completely uniform

  24. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) = 0.9911 no yes Hair Length <= 5? Let us try splitting on Hair length Entropy(3F,2M) = -(3/5)log2(3/5) - (2/5)log2(2/5) = 0.9710 Entropy(1F,3M) = -(1/4)log2(1/4) - (3/4)log2(3/4) = 0.8113 Gain(Hair Length <= 5) = 0.9911 – (4/9 * 0.8113 + 5/9 * 0.9710 ) = 0.0911

  25. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) = 0.9911 no yes Weight <= 160? Let us try splitting on Weight Entropy(0F,4M) = -(0/4)log2(0/4) - (4/4)log2(4/4) = 0 Entropy(4F,1M) = -(4/5)log2(4/5) - (1/5)log2(1/5) = 0.7219 Gain(Weight <= 160) = 0.9911 – (5/9 * 0.7219 + 4/9 * 0 ) = 0.5900

  26. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) = 0.9911 no yes age <= 40? Let us try splitting on Age Entropy(1F,2M) = -(1/3)log2(1/3) - (2/3)log2(2/3) = 0.9183 Entropy(3F,3M) = -(3/6)log2(3/6) - (3/6)log2(3/6) = 1 Gain(Age <= 40) = 0.9911 – (6/9 * 1 + 3/9 * 0.9183 ) = 0.0183

  27. Of the 3 features we had, Weight was best. But while people who weigh over 160 are perfectly classified (as males), the under 160 people are not perfectly classified… So we simply recurse! no yes Weight <= 160? This time we find that we can split on Hair length, and we are done! no yes Hair Length <= 2?

  28. We need don’t need to keep the data around, just the test conditions. Weight <= 160? yes no How would these people be classified? Hair Length <= 2? Male yes no Male Female

  29. It is trivial to convert Decision Trees to rules… Weight <= 160? yes no Hair Length <= 2? Male no yes Male Female Rules to Classify Males/Females IfWeightgreater than 160, classify as Male Elseif Hair Lengthless than or equal to 2, classify as Male Else classify as Female

  30. Once we have learned the decision tree, we don’t even need a computer! This decision tree is attached to a medical machine, and is designed to help nurses make decisions about what type of doctor to call. Decision tree for a typical shared-care setting applying the system for the diagnosis of prostatic obstructions.

  31. PSA = serum prostate-specific antigen levels PSAD = PSA density TRUS = transrectal ultrasound  Garzotto M et al. JCO 2005;23:4322-4329

  32. The worked examples we have seen were performed on small datasets. However with small datasets there is a great danger of overfitting the data… When you have few datapoints, there are many possible splitting rules that perfectly classify the data, but will not generalize to future datasets. Yes No Wears green? Male Female For example, the rule “Wears green?” perfectly classifies the data, so does “Mothers name is Jacqueline?”, so does “Has blue shoes”…

  33. Avoid Overfitting in Classification • The generated tree may overfit the training data • Too many branches, some may reflect anomalies due to noise or outliers • Result is in poor accuracy for unseen samples • Two approaches to avoid overfitting • Prepruning: Halt tree construction early—do not split a node if this would result in the goodness measure falling below a threshold • Difficult to choose an appropriate threshold • Postpruning: Remove branches from a “fully grown” tree—get a sequence of progressively pruned trees • Use a set of data different from the training data to decide which is the “best pruned tree”

  34. 10 10 100 9 9 90 8 8 80 7 7 70 6 6 60 5 5 50 4 4 40 3 3 30 2 2 20 1 1 10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 10 10 9 9 10 20 30 40 50 60 70 80 100 90 Which of the “Pigeon Problems” can be solved by a Decision Tree? • Deep Bushy Tree • Useless • Deep Bushy Tree ? The Decision Tree has a hard time with correlated attributes

  35. Advantages/Disadvantages of Decision Trees • Advantages: • Easy to understand (Doctors love them!) • Easy to generate rules • Disadvantages: • May suffer from overfitting. • Classifies by rectangular partitioning (so does not handle correlated features very well). • Can be quite large – pruning is necessary. • Does not handle streaming data easily

  36. There now exists, perhaps tens of million of digitized pages of historical manuscripts dating back to the 12th century, that feature one or more heraldic shields The images are often stained, faded or torn

  37. Wouldn’t it be great if we could automatically hyperlink all similar shields to each other? For example, here we could link two occurrence of the Von Sax family shield. To do this, we need to consider shape, color and texture. Lets just consider shape for now… Manesse Codex an illuminated manuscript in codex form, copied and illustrated between 1304 and 1340 in Zurich

  38. Polish Spanish 3 2 1 151.7 I (Polish) 0 156.1 II (Spanish) Shaplet Dictionary 0 100 200 300 400 Shield Decision Tree I Spanish Polish French II 1 2 0 Using the entire shape is not a good idea, because the shields can have flourishes or tears Decision Tree for Shields Flourishes Tear Training data (subset) An NSF funded project (IIS 0803410) is attempting to solve this by using parts of the shapes, called shaplets* Shaplets allow you to build decision trees for shapes French Spanish Polish *Ye and Keogh (2009) Time Series Shapelets: A New Primitive for Data Mining. SIGKDD 2009

  39. Naïve Bayes Classifier Thomas Bayes 1702 - 1761 We will start off with a visual intuition, before looking at the math…

  40. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Grasshoppers Katydids Antenna Length Abdomen Length Remember this example? Let’s get lots more data…

  41. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Katydids Grasshoppers With a lot of data, we can build a histogram. Let us just build one for “Antenna Length” for now… Antenna Length

  42. We can leave the histograms as they are, or we can summarize them with two normal distributions. Let us us two normal distributions for ease of visualization in the following slides…

  43. We want to classify an insect we have found. Its antennae are 3 units long. How can we classify it? • We can just ask ourselves, give the distributions of antennae lengths we have seen, is it more probable that our insect is a Grasshopperor a Katydid. • There is a formal way to discuss the most probable classification… p(cj | d) = probability of class cj, given that we have observed d 3 Antennae length is 3

  44. p(cj | d) = probability of class cj, given that we have observed d P(Grasshopper | 3 ) = 10 / (10 +2) = 0.833 P(Katydid | 3 ) = 2 / (10 + 2) = 0.166 10 2 3 Antennae length is 3

  45. p(cj | d) = probability of class cj, given that we have observed d P(Grasshopper | 7 ) = 3 / (3 +9) = 0.250 P(Katydid | 7 ) = 9 / (3 + 9) = 0.750 9 3 7 Antennae length is 7

  46. p(cj | d) = probability of class cj, given that we have observed d P(Grasshopper | 5 ) = 6 / (6 +6) = 0.500 P(Katydid | 5 ) = 6 / (6 + 6) = 0.500 6 6 5 Antennae length is 5

  47. Bayes Classifiers • That was a visual intuition for a simple case of the Bayes classifier, also called: • Idiot Bayes • Naïve Bayes • Simple Bayes • We are about to see some of the mathematical formalisms, and more examples, but keep in mind the basic idea. • Find out the probability of the previously unseen instance belonging to each class, then simply pick the most probable class.

  48. Bayes Classifiers • Bayesian classifiers use Bayes theorem, which says p(cj | d ) = p(d | cj ) p(cj) p(d) • p(cj | d) = probability of instance d being in class cj, This is what we are trying to compute • p(d | cj) = probability of generating instance d given class cj, We can imagine that being in class cj, causes you to have feature d with some probability • p(cj) = probability of occurrence of class cj, This is just how frequent the class cj, is in our database • p(d) = probability of instance d occurring This can actually be ignored, since it is the same for all classes

More Related