Steven F. Ashby Center for Applied Scientific Computing Month DD, 1997
Classification: Basic Concepts,
Decision Trees (Slides 1 to 53), and Model Evaluation
Chapter 18 in Textbook
Slides modified from Tan et al.
Machine Learning
Outline
What is the classification?
Classification Techniques
Decision Trees (slides 7 to 53)
Summary
Classification: Definition
Given a collection of records (training set )
Each record contains a set of attributes/features, one of the attributes is the class.
Find a model for class attribute as a function of the values of other attributes.
Goal: previously unseen records should be assigned a class as accurately as possible.
A test set is used to determine the accuracy of the model. Usually, the given data set is divided into training and test sets, with training set used to build the model and test set used to validate it.
What is classification?
Classification is the task of learning a target function f that maps attribute set x to one of the predefined class labels y
Illustrating Classification Task
Examples of Classification Task
Predicting tumor cells as benign or malignant
Classifying credit card transactions
as legitimate or fraudulent
Classifying secondary structures of protein
as alpha-helix, beta-sheet, or random
coil
Categorizing news stories as finance,
weather, entertainment, sports, etc
Classification Techniques
Decision Tree based Methods
Rule-based Methods
Neural Networks
Naïve Bayes and Bayesian Belief Networks
Support Vector Machines
…
Example of a Decision Tree
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Splitting Attributes
Training Data
Model: Decision Tree
categorical
categorical
continuous
class
Another Example of Decision Tree
categorical
categorical
continuous
class
MarSt
Refund
TaxInc
YES
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
There could be more than one tree that fits the same data!
NO
Decision Tree Classification Task
Decision Tree
Apply Model to Test Data
Test Data
Start from the root of tree.
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Apply Model to Test Data
Test Data
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Apply Model to Test Data
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
Apply Model to Test Data
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
Apply Model to Test Data
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
Apply Model to Test Data
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
Assign Cheat to “No”
Decision Tree Induction Algorithm
Decision Tree
Decision Tree Induction
Many Algorithms:
Hunt’s Algorithm (one of the earliest)
CART
ID3, C4.5
SLIQ,SPRINT
General Structure of Hunt’s Algorithm
Let Dt be the set of training records that reach a node t.
General Procedure:
If Dt contains records that belong the same class yt, then t is a leaf node labeled as yt
If Dt is an empty set, then t is a leaf node labeled by the default class, yd
If Dt contains records that belong to more than one class, use an attribute test to split the data into smaller subsets. Recursively apply the procedure to each subset.
Dt
?
Constructing decision trees: the Hunt’s algorithm
Xt: the set of training records for node t
y={y1,…,yc}: class labels
Step 1: If all records in Xt belong to the same class yt, then t is a leaf node labeled as yt
Step 2: If Xt contains records that belong to more than one class,
Select attribute test condition to partition the records into smaller subsets
Create a child node for each outcome of test condition
Apply algorithm recursively for each child
Example: Hunt’s Algorithm for
inducing Decision Trees
Don’t
Cheat
Refund
Don’t
Cheat
Don’t
Cheat
Yes
No
Refund
Don’t
Cheat
Yes
No
Marital
Status
Don’t
Cheat
Cheat
Single,
Divorced
Married
Taxable
Income
Don’t
Cheat
< 80K >= 80K
Refund
Don’t
Cheat
Yes
No
Marital
Status
Don’t
Cheat
Cheat
Single,
Divorced
Married
Tree Induction
Greedy strategy.
Split the records based on an attribute test that optimizes certain criterion.
Two design Issues
Determine how to split the records.
How to specify the attribute test condition?
How to determine the best split?
Determine when to stop splitting.
Tree Induction
Greedy strategy.
Split the records based on an attribute test that optimizes certain criterion.
Issues
Determine how to split the records
How to specify the attribute test condition?
How to determine the best split?
Determine when to stop splitting
How to Specify Test Condition?
Depends on attribute types
Nominal
Data are neither measured nor ordered but subjects are merely allocated to distinct categories: for example, a record of students’ course choices constitutes nominal data.
Ordinal
Ordinal data is a categorical.
Continuous
Depends on number of ways to split
2-way split
Multi-way split
Splitting Based on Nominal Attributes
Multi-way split: Use as many partitions as distinct values.
Binary split: Divides values into two subsets.
Need to find optimal partitioning.
OR
CarType
Family
Sports
Luxury
CarType
{Family,
Luxury}
{Sports}
CarType
{Sports, Luxury}
{Family}
Multi-way split: Use as many partitions as distinct values.
Binary split: Divides values into two subsets.
Need to find optimal partitioning.
What about this split?
Splitting Based on Ordinal Attributes
OR
Size
Small
Medium
Large
Size
{Medium,
Large}
{Small}
Size
{Small, Medium}
{Large}
Size
{Small, Large}
{Medium}
Splitting Based on Continuous Attributes
Different ways of handling
Discretization to form an ordinal categorical attribute
Static – discretize once at the beginning
Dynamic – ranges can be found by equal interval bucketing, equal frequency bucketing
(percentiles), or clustering.
Binary Decision: (A < v) or (A v)
consider all possible splits and finds the best cut
can be more compute intensive
Splitting Based on Continuous Attributes
Tree Induction
Greedy strategy.
Split the records based on an attribute test that optimizes certain criterion.
Issues
Determine how to split the records
How to specify the attribute test condition?
How to determine the best split?
Determine when to stop splitting
How to determine the Best Split ?
Before Splitting: 10 records of class 0,
10 records of class 1
Which test condition is the best?
How to determine the Best Split ?
Greedy approach:
Nodes with homogeneous class distribution are preferred
Need a measure of node impurity:
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity
Measures of Node Impurity
Gini Index
Entropy
Misclassification error
Measures of Node Impurity
Tree Induction Using the Information Gain such as Gini Index.
How to Find the Best Split ?
B?
Yes
No
Node N3
Node N4
A?
Yes
No
Node N1
Node N2
Before Splitting:
Gain = M0 – M12 vs M0 – M34
M0
M1
M2
M3
M4
M12
M34
Measure of Impurity: GINI
Gini Index for a given node t :
(NOTE: p( j | t) is the relative frequency of class j at node t).
Maximum (1 - 1/nc) when records are equally distributed among all classes, implying least interesting information
Minimum (0.0) when all records belong to one class, implying most interesting information
Examples for computing GINI
P(C1) = 0/6 = 0 P(C2) = 6/6 = 1
Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0
P(C1) = 1/6 P(C2) = 5/6
Gini = 1 – (1/6)2 – (5/6)2 = 0.278
P(C1) = 2/6 P(C2) = 4/6
Gini = 1 – (2/6)2 – (4/6)2 = 0.444
Splitting Based on GINI
Used in CART, SLIQ, SPRINT.
When a node p is split into k partitions (children), the quality of split is computed as,
where, ni = number of records at child i,
n = number of records at node p.
Binary Attributes: Computing GINI Index
Splits into two partitions
Effect of Weighing partitions:
Larger and Purer Partitions are sought for.
B?
Yes
No
Node N1
Node N2
Gini(N1)
= 1 – (5/6)2 – (2/6)2
= 0.194
Gini(N2)
= 1 – (1/6)2 – (4/6)2
= 0.528
Gini(Children)
= 7/12 * 0.194 +
5/12 * 0.528
= 0.333
Computation Error!
7
7
5
5
Categorical Attributes: Computing Gini Index
For each distinct value, gather counts for each class in the dataset
Use the count matrix to make decisions
Multi-way split
Two-way split
(find best partition of values)
Continuous Attributes: Computing Gini Index
Use Binary Decisions based on one value
Several Choices for the splitting value
Number of possible splitting values
= Number of distinct values
Each splitting value has a count matrix associated with it
Class counts in each of the partitions, A < v and A v
Simple method to choose best v
For each v, scan the database to gather count matrix and compute its Gini index
Computationally Inefficient! Repetition of work.
Continuous Attributes: Computing Gini Index...
For efficient computation: for each attribute,
Sort the attribute on values
Linearly scan these values, each time updating the count matrix and computing gini index
Choose the split position that has the least gini index
Split Positions
Sorted Values
Alternative Splitting Criteria based on Entropy
Entropy at a given node t:
(NOTE: p( j | t) is the relative frequency of class j at node t).
Measures homogeneity of a node.
Maximum (log nc) when records are equally distributed among all classes implying least information
Minimum (0.0) when all records belong to one class, implying most information
Entropy based computations are similar to the GINI index computations
Examples for computing Entropy
P(C1) = 0/6 = 0 P(C2) = 6/6 = 1
Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0
P(C1) = 1/6 P(C2) = 5/6
Entropy = – (1/6) log2 (1/6) – (5/6) log2 (1/6) = 0.65
P(C1) = 2/6 P(C2) = 4/6
Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92
Splitting Based on Entropy
Information Gain:
Parent Node, p is split into k partitions;
ni is number of records in partition i
Measures Reduction in Entropy achieved because of the split. Choose the split that achieves most reduction (maximizes GAIN)
Used in ID3 and C4.5
Disadvantage: Tends to prefer splits that result in large number of partitions, each being small but pure.
Splitting Based on Gain Ratio
Gain Ratio:
Parent Node, p is split into k partitions
ni is the number of records in partition i
Adjusts Information Gain by the entropy of the partitioning (SplitINFO). Higher entropy partitioning (large number of small partitions) is penalized!
Used in C4.5
Designed to overcome the disadvantage of Information Gain
Splitting Criteria based on Classification Error
Classification error at a node t :
Measures misclassification error made by a node.
Maximum (1 - 1/nc) when records are equally distributed among all classes, implying least interesting information
Minimum (0.0) when all records belong to one class, implying most interesting information
Examples for Computing Error
P(C1) = 0/6 = 0 P(C2) = 6/6 = 1
Error = 1 – max (0, 1) = 1 – 1 = 0
P(C1) = 1/6 P(C2) = 5/6
Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6
P(C1) = 2/6 P(C2) = 4/6
Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
Misclassification Error vs Gini
A?
Yes
No
Node N1
Node N2
Gini(N1)
= 1 – (3/3)2 – (0/3)2
= 0
Gini(N2)
= 1 – (4/7)2 – (3/7)2
= 0.489
Gini(Children)
= 3/10 * 0
+ 7/10 * 0.489
= 0.342
Gini improves !!
Computation Error and typo!
Tree Induction
Greedy strategy.
Split the records based on an attribute test that optimizes certain criterion.
Issues
Determine how to split the records
How to specify the attribute test condition?
How to determine the best split?
Determine when to stop splitting
Stopping Criteria for Tree Induction
Stop expanding a node when all the records belong to the same class
Stop expanding a node when all the records have similar attribute values
Early termination (to be discussed later)
Decision Tree Based Classification
Advantages:
Inexpensive to construct
Extremely fast at classifying unknown records
Easy to interpret for small-sized trees
Accuracy is comparable to other classification techniques for many simple data sets
Example: C4.5
Simple depth-first construction.
Uses Information Gain
Sorts Continuous Attributes at each node.
Needs entire data to fit in memory.
Unsuitable for Large Datasets.
Needs out-of-core sorting.
You can download the software from:
http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
Practical Issues of Classification
Underfitting and Overfitting
Missing Values
Costs of Classification
Underfitting and Overfitting (Example)
500 circular and 500 triangular data points.
Circular points:
0.5 sqrt(x12+x22) 1
Triangular points:
sqrt(x12+x22) > 0.5 or
sqrt(x12+x22) < 1
Underfitting and Overfitting
Overfitting
Underfitting: when model is too simple, both training and test errors are large
Overfitting due to Noise
Decision boundary is distorted by noise point
Overfitting due to Insufficient Examples
Lack of data points in the lower half of the diagram makes it difficult to predict correctly the class labels of that region
- Insufficient number of training records in the region causes the decision tree to predict the test examples using other training records that are irrelevant to the classification task
Notes on Overfitting
Overfitting results in decision trees that are more complex than necessary
Training error no longer provides a good estimate of how well the tree will perform on previously unseen records
Need new ways for estimating errors
Notes on Overfitting
Example from pattern recognition
Generalization
The ultimate goal is to do as well as possible on new, unseen data (a test set), but we only have access to labels (“ground truth”) for the training set
What makes generalization possible?
*
Generalization
Inductive bias: set of assumptions a learner uses to predict the target value for previously unseen inputs
This is the same as modeling or choosing a target hypothesis class
Types of inductive bias
Occam’s razor
Similarity/continuity bias: similar inputs should have similar outputs
…
*
Achieving good generalization
Consideration 1: Bias
How well does your model fit the observed data?
It may be a good idea to accept some fitting error, because it may be due to noise or other “accidental” characteristics of one particular training set
*
Achieving good generalization
Consideration 2: Variance
How robust is the model to the selection of a particular training set?
To put it differently, if we learn models on two different training sets, how consistent will the models be?
*
Bias/variance tradeoff
Models with too many parameters may fit the training data well (low bias), but are sensitive to choice of training set (high variance)
*
Bias/variance tradeoff
Models with too many parameters may fit the training data well (low bias), but are sensitive to choice of training set (high variance)
Models with too few parameters may not fit the data well (high bias) but are consistent across different training sets (low variance)
2
*
Bias/variance tradeoff
Models with too many parameters may fit the training data well (low bias), but are sensitive to choice of training set (high variance)
Generalization error is due to overfitting
Models with too few parameters may not fit the data well (high bias) but are consistent across different training sets (low variance)
Generalization error is due to underfitting
2
*
Underfitting and overfitting
How to recognize underfitting?
High training error and high test error
How to deal with underfitting?
Find a more complex model
*
Underfitting and overfitting
How to recognize overfitting?
Low training error, but high test error
How to deal with overfitting?
Get more training data
Decrease the number of parameters in your model
Regularization: penalize certain parts of the parameter space or introduce additional constraints to deal with a potentially ill-posed problem
*
Occam’s Razor
Given two models of similar generalization errors, one should prefer the simpler model over the more complex model
For complex models, there is a greater chance that it was fitted accidentally by errors in data
Therefore, one should include model complexity when evaluating a model
*
Criterion for attribute selection
Which is the best attribute?
The one which will result in the smallest tree
Heuristic: choose the attribute that produces the “purest” nodes
Need a good measure of purity!
Maximal when?
Minimal when?
Handling Missing Attribute Values
Missing values affect decision tree construction in three different ways:
Affects how impurity measures are computed
Affects how to distribute instance with missing value to child nodes
Affects how a test instance with missing value is classified
Other Issues
Data Fragmentation
Search Strategy
Expressiveness
Tree Replication
Model Evaluation
Metrics for Performance Evaluation
How to evaluate the performance of a model?
Methods for Performance Evaluation
How to obtain reliable estimates?
Methods for Model Comparison
How to compare the relative performance among competing models?
Model Evaluation
Metrics for Performance Evaluation
How to evaluate the performance of a model?
Methods for Performance Evaluation
How to obtain reliable estimates?
Methods for Model Comparison
How to compare the relative performance among competing models?
Metrics for Performance Evaluation
Focus on the predictive capability of a model
Rather than how fast it takes to classify or build models, scalability, etc.
Confusion Matrix:
a: TP (true positive)
b: FN (false negative)
c: FP (false positive)
d: TN (true negative)
PREDICTED CLASS
ACTUAL
CLASS Class=Yes Class=No
Class=Yes a b
Class=No c d
Metrics for Performance Evaluation…
Most widely-used metric:
PREDICTED CLASS
ACTUAL
CLASS Class=Yes Class=No
Class=Yes a
(TP) b
(FN)
Class=No c
(FP) d
(TN)
Limitation of Accuracy
Consider a 2-class problem
Number of Class 0 examples = 9990
Number of Class 1 examples = 10
If model predicts everything to be class 0, accuracy is 9990/10000 = 99.9 %
Accuracy is misleading because model does not detect any class 1 example
Cost Matrix
C(i|j): Cost of misclassifying class j example as class i
PREDICTED CLASS
ACTUAL
CLASS C(i|j) Class=Yes Class=No
Class=Yes C(Yes|Yes) C(No|Yes)
Class=No C(Yes|No) C(No|No)
Computing Cost of Classification
Accuracy = 80%
Cost = 3910
Accuracy = 90%
Cost = 4255
Cost Matrix PREDICTED CLASS
ACTUAL
CLASS C(i|j) + -
+ -1 100
- 1 0
Model M1 PREDICTED CLASS
ACTUAL
CLASS + -
+ 150 40
- 60 250
Model M2 PREDICTED CLASS
ACTUAL
CLASS + -
+ 250 45
- 5 200
Cost vs Accuracy
Count PREDICTED CLASS
ACTUAL
CLASS Class=Yes Class=No
Class=Yes a b
Class=No c d
Cost PREDICTED CLASS
ACTUAL
CLASS Class=Yes Class=No
Class=Yes p q
Class=No q p
N = a + b + c + d
Accuracy = (a + d)/N
Cost = p (a + d) + q (b + c)
= p (a + d) + q (N – a – d)
= q N – (q – p)(a + d)
= N [q – (q-p) Accuracy]
Accuracy is proportional to cost if
1. C(Yes|No)=C(No|Yes) = q
2. C(Yes|Yes)=C(No|No) = p
Cost-Sensitive Measures
Precision and Recall are two widely used metrics employed in applications where successful detection of one of the classes is considered more important than detection of the other classes.
Cost-Sensitive Measures
Precision is biased towards C(Yes|Yes) & C(Yes|No)
Recall is biased towards C(Yes|Yes) & C(No|Yes)
F-measure is biased towards all except C(No|No)
ROC (Receiver Operating Characteristic)
A good classification model should be located as close as possible to the upper left corner of the diagram.
Performance of each classifier represented as a point on the ROC curve
changing the threshold of algorithm, sample distribution or cost matrix changes the location of the point
ROC curve
Receiver Operating Characteristic (ROC)
Graphical approach for displaying the tradeoff between true positive rate(TPR) and false positive rate (FPR) of a classifier
TPR = positives correctly classified/total positives
FPR = negatives incorrectly classified/total negatives
TPR on y-axis and FPR on x-axis
ROC Curve
- 1-dimensional data set containing 2 classes (positive and negative)
- any points located at x > t is classified as positive
At threshold t:
TP=0.5, FN=0.5, FP=0.12, FN=0.88
ROC curve
ROC curve
Points of interests (TP, FP)
(0, 0): everything is negative
(1, 1): everything is positive
(1, 0): perfect (ideal)
Diagonal line
Random guessing (50%)
Area Under Curve (AUC)
Measurement how good the model on the average
Good to compare with other methods
ROC Curve
(TP,FP):
(0,0): declare everything
to be negative class
(1,1): declare everything
to be positive class
(1,0): ideal
Diagonal line:
Random guessing
Below diagonal line:
prediction is opposite of the true class
ROC Curve
A model that is strictly better than another would have a larger area under the ROC curve.
If the model is perfect, then its area under the ROC curve would equal to 1.
If the model simply performs random guessing, then its area under the ROC curve would equal 0.5.
Model Evaluation
Metrics for Performance Evaluation
How to evaluate the performance of a model?
Methods for Performance Evaluation
How to obtain reliable estimates?
Methods for Model Comparison
How to compare the relative performance among competing models?
Methods for Performance Evaluation
How to obtain a reliable estimate of performance?
Performance of a model may depend on other factors besides the learning algorithm:
Class distribution
Cost of misclassification
Size of training and test sets
Learning Curve
Learning curve shows how accuracy changes with varying sample size
Requires a sampling schedule for creating learning curve:
Arithmetic sampling
(Langley, et al)
Geometric sampling
(Provost et al)
Effect of small sample size:
Bias in the estimate
Variance of estimate
Methods of Estimation
Holdout
Reserve 2/3 for training and 1/3 for testing
Random subsampling
Repeated holdout
Cross validation
Partition data into k disjoint subsets
k-fold: train on k-1 partitions, test on the remaining one
Leave-one-out: k = n
Methods of Estimation
Stratified sampling: Stratified sampling is a probability sampling technique wherein the researcher divides the entire population into different subgroups or strata, then randomly selects the final subjects proportionally from the different strata
oversampling vs undersampling
Bootstrap
Sampling with replacement
Model Evaluation
Metrics for Performance Evaluation
How to evaluate the performance of a model?
Methods for Performance Evaluation
How to obtain reliable estimates?
Methods for Model Comparison
How to compare the relative performance among competing models?
ROC (Receiver Operating Characteristic)
A good classification model should be located as close as possible to the upper left corner of the diagram.
Performance of each classifier represented as a point on the ROC curve
changing the threshold of algorithm, sample distribution or cost matrix changes the location of the point
ROC Curve
– 1-dimensional data set containing 2 classes (positive and negative)
– any points located at x > t is classified as positive
At threshold t:
TP=0.5, FN=0.5, FP=0.12, FN=0.88
ROC Curve
(TP,FP):
(0,0): declare everything
to be negative class
(1,1): declare everything
to be positive class
(1,0): ideal
Diagonal line:
Random guessing
Below diagonal line:
prediction is opposite of the true class
ROC Curve
A model that is strictly better than another would have a larger area under the ROC curve.
If the model is perfect, then its area under the ROC curve would equal to 1.
If the model simply performs random guessing, then its area under the ROC curve would equal 0.5.
*
Decision Trees: Summary
Representation = decision trees
Bias = preference for small decision trees
Heuristic function = information gain or
information content or others
Overfitting and pruning
Advantage is simplicity and easy conversion to rules.
Questions & Suggestions?
The End
Appendix
ROC (Receiver Operating Characteristic)
from Wikipedia
In signal detection theory, a receiver operating characteristic (ROC), or simply ROC curve, is a graphical plot of the sensitivity, or true positive rate, vs. false positive rate (1 − specificity or 1 − true negative rate), for a binary classifier system as its discrimination threshold is varied. The ROC can also be represented equivalently by plotting the fraction of true positives out of the positives (TPR = true positive rate) vs. the fraction of false positives out of the negatives (FPR = false positive rate). Also known as a Relative Operating Characteristic curve, because it is a comparison of two operating characteristics (TPR & FPR) as the criterion changes.[1]
ROC analysis provides tools to select possibly optimal models and to discard suboptimal ones independently from (and prior to specifying) the cost context or the class distribution. ROC analysis is related in a direct and natural way to cost/benefit analysis of diagnostic decision making. The ROC curve was first developed by electrical engineers and radar engineers during World War II for detecting enemy objects in battle fields, also known as the signal detection theory, and was soon introduced in psychology to account for perceptual detection of stimuli. ROC analysis since then has been used in medicine, radiology, and other areas for many decades, and it has been introduced relatively recently in other areas like machine learning and data mining.
ROC (Receiver Operating Characteristic)
Developed in 1950s for signal detection theory to analyze noisy signals
Characterize the trade-off between positive hits and false alarms
ROC curve plots TP (on the y-axis) against FP (on the x-axis)
ROC (Receiver Operating Characteristic)
A good classification model should be located as close as possible to the upper left corner of the diagram.
Performance of each classifier represented as a point on the ROC curve
changing the threshold of algorithm, sample distribution or cost matrix changes the location of the point
ROC Curve
– 1-dimensional data set containing 2 classes (positive and negative)
– any points located at x > t is classified as positive
At threshold t:
TP=0.5, FN=0.5, FP=0.12, FN=0.88
ROC Curve
(TP,FP):
(0,0): declare everything
to be negative class
(1,1): declare everything
to be positive class
(1,0): ideal
Diagonal line:
Random guessing
Below diagonal line:
prediction is opposite of the true class
ROC Curve
A model that is strictly better than another would have a larger area under the ROC curve.
If the model is perfect, then its area under the ROC curve would equal to 1.
If the model simply performs random guessing, then its area under the ROC curve would equal 0.5.
Using ROC for Model Comparison
No model consistently outperform the other
M1 is better for small FPR
M2 is better for large FPR
Area Under the ROC curve
Ideal:
Area = 1
Random guess:
Area = 0.5
How to Construct an ROC curve
Use classifier that produces posterior probability for each test instance P(+|A)
Sort the instances according to P(+|A) in decreasing order
Apply threshold at each unique value of P(+|A)
Count the number of TP, FP,
TN, FN at each threshold
TP rate, TPR = TP/(TP+FN)
FP rate, FPR = FP/(FP + TN)
Instance P(+|A) True Class
1 0.95 +
2 0.93 +
3 0.87 –
4 0.85 –
5 0.85 –
6 0.85 +
7 0.76 –
8 0.53 +
9 0.43 –
10 0.25 +
How to Construct an ROC curve: Algorithm
Step 1: assume that the continuous-valued outputs are defined for the positive class, sort the test records in increasing order of their output values.
Step 2: select the lowest ranked test record. Assign the selected record and those ranked above it to the positive class.
How to Construct an ROC curve: Algorithm
Step 3: select the next test record from the sorted list. Classify the selected record and those ranked above it as positive, while those ranked below it as negative. Update the counts of TP and FP by examining the actual class label of the previously selected record. If the previously selected record is a positive class, the TP count is decremented amd the FP count remains the same as before. If the previously selected record is a negative class, the FP count is decremented and TP count remains the same as before.
Step 4: Repeat step 3 and update the TP and FP counts accordingly until the highest ranked test record is selected.
Step 5: Plot the TPR against FPR of the classifier.
How to construct an ROC curve (Exercise)
Threshold >=
ROC Curve:
Test of Significance (skip)
Given two models:
Model M1: accuracy = 85%, tested on 30 instances
Model M2: accuracy = 75%, tested on 5000 instances
Can we say M1 is better than M2?
How much confidence can we place on accuracy of M1 and M2?
Can the difference in performance measure be explained as a result of random fluctuations in the test set?
Confidence Interval for Accuracy (skip)
Prediction can be regarded as a Bernoulli trial
A Bernoulli trial has 2 possible outcomes
Possible outcomes for prediction: correct or wrong
Collection of Bernoulli trials has a Binomial distribution:
x Bin(N, p) x: number of correct predictions
e.g: Toss a fair coin 50 times, how many heads would turn up?
Expected number of heads = Np = 50 0.5 = 25
Given x (# of correct predictions) or equivalently, acc=x/N, and N (# of test instances),
Can we predict p (true accuracy of model)?
Confidence Interval for Accuracy (skip)
For large test sets (N > 30),
acc has a normal distribution
with mean p and variance
p(1-p)/N
Confidence Interval for p:
Area = 1 –
Z/2
Z1- /2
Confidence Interval for Accuracy (skip)
Consider a model that produces an accuracy of 80% when evaluated on 100 test instances:
N=100, acc = 0.8
Let 1- = 0.95 (95% confidence)
From probability table, Z/2=1.96
1- Z
0.99 2.58
0.98 2.33
0.95 1.96
0.90 1.65
N 50 100 500 1000 5000
p(lower) 0.670 0.711 0.763 0.774 0.789
p(upper) 0.888 0.866 0.833 0.824 0.811
Comparing Performance of 2 Models (skip)
Given two models, say M1 and M2, which is better?
M1 is tested on D1 (size=n1), found error rate = e1
M2 is tested on D2 (size=n2), found error rate = e2
Assume D1 and D2 are independent
If n1 and n2 are sufficiently large, then
Approximate:
Comparing Performance of 2 Models (skip)
To test if performance difference is statistically significant: d = e1 – e2
d ~ N(dt,t) where dt is the true difference
Since D1 and D2 are independent, their variance adds up:
At (1-) confidence level,
An Illustrative Example (skip)
Given: M1: n1 = 30, e1 = 0.15
M2: n2 = 5000, e2 = 0.25
d = |e2 – e1| = 0.1 (2-sided test)
At 95% confidence level, Z/2=1.96
=> Interval contains 0 => difference may not be
statistically significant
Comparing Performance of 2 Algorithms (skip)
Each learning algorithm may produce k models:
L1 may produce M11 , M12, …, M1k
L2 may produce M21 , M22, …, M2k
If models are generated on the same test sets D1,D2, …, Dk (e.g., via cross-validation)
For each set: compute dj = e1j – e2j
dj has mean dt and variance t
Estimate:
ROC Analysis
Thanks to Jeremy Wyatt
Thanks to Gavin Brown
Evaluating vision algorithms
You have designed a new edge detection technique.
You give it to me, and I try it on my image dataset where the task is to predict whether the scene contains a chair or not.
I tell you that it achieved 95% accuracy on my data.
Is your technique a success?
*
Types of errors
But suppose that
The 95% is the correctly classified pixels
Only 5% of the pixels are actually edges
It misses all the edge pixels
How do we count the effect of different types of error?
Types of errors
Prediction
Edge Not edge
Ground Truth
Not Edge Edge
True
Positive (TP) False
Negative (FN)
False
Positive (FP) True Negative (TN)
True Positive
Two parts to each: whether you got it correct or not, and what you guessed. For example for a particular pixel, our guess might be labelled…
Did we get it correct? True, we did get it correct.
False Negative
Did we get it correct? False, we did not get it correct.
or maybe it was labelled as one of the others, maybe…
What did we say? We said ‘positive’, i.e. edge.
What did we say? We said ‘negative, i.e. not edge.
Sensitivity and Specificity
Count up the total number of each label (TP, FP, TN, FN) over a large dataset. In ROC analysis, we use two statistics:
Sensitivity =
TP
TP+FN
Specificity =
TN
TN+FP
Can be thought of as the likelihood of spotting a positive case when presented with one.
Or… the proportion of edges we find.
Can be thought of as the likelihood of spotting a negative case when presented with one.
Or… the proportion of non-edges that we find
Sensitivity = = ?
TP
TP+FN
Specificity = = ?
TN
TN+FP
Prediction
Ground Truth
1
1
0
0
60
30
20
80
80+20 = 100 cases in the dataset were class 0 (non-edge)
60+30 = 90 cases in the dataset were class 1 (edge)
90+100 = 190 examples (pixels) in the data overall
The ROC space
1 – Specificity
Sensitivity
This is edge detector B
This is edge detector A
1.0
0.0
1.0
Note
The ROC Curve
Draw a ‘convex hull’ around many points:
1 – Specificity
Sensitivity
This point is not on the convex hull.
ROC Analysis
1 – specificity
sensitivity
All the optimal detectors lie on the convex hull.
Which of these is best depends on the ratio of edges to non-edges, and the different cost of misclassification
Any detector on this side can lead to a better detector by flipping its output.
Take-home point : You should always quote sensitivity and specificity for your algorithm, if possible plotting an ROC graph. Remember also though, any statistic you quote should be an average over a suitable range of tests for your algorithm.
�
Learning algorithm�
�
�
Induction�
Deduction�
Test Set�
Model�
Training Set�
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
No
Married
100K
No
3
No
Single
70K
No
4
Yes
Married
120K
No
5
No
Divorced
95K
Yes
6
No
Married
60K
No
7
Yes
Divorced
220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
�
Tree Induction algorithm�
�
�
Induction�
Deduction�
Test Set�
Model�
Training Set�
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
�
Tree Induction algorithm�
�
�
Induction�
Deduction�
Test Set�
Model�
Training Set�
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
No
Married
100K
No
3
No
Single
70K
No
4
Yes
Married
120K
No
5
No
Divorced
95K
Yes
6
No
Married
60K
No
7
Yes
Divorced
220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
No
Married
100K
No
3
No
Single
70K
No
4
Yes
Married
120K
No
5
No
Divorced
95K
Yes
6
No
Married
60K
No
7
Yes
Divorced
220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
Taxable Income�> 80K?�
< 10K�
[10K,25K)�
Yes�
No�
[25K,50K)�
Taxable Income?�
[50K,80K)�
> 80K�
(i) Binary split�
(ii) Multi-way split�
Own Car?�
C0: 6
C1: 4�
C0: 4
C1: 6�
Car Type?�
C0: 1
C1: 3�
C0: 8
C1: 0�
C0: 1
C1: 7�
C0: 1
C1: 0�
C0: 1
C1: 0�
C0: 0
C1: 1�
Student ID?�
…�
Yes�
No�
Family�
Sports�
Luxury�
c1�
c10�
c20�
C0: 0
C1: 1�
…�
c11�
C0: 5
C1: 5�
C0: 9
C1: 1�
C0
N10
C1
N11
C0
N20
C1
N21
C0
N30
C1
N31
C0
N40
C1
N41
C0
N00
C1
N01
C1
0
C2
6
Gini=0.000
C1
2
C2
4
Gini=0.444
C1
3
C2
3
Gini=0.500
C1
1
C2
5
Gini=0.278
C1
0
C2
6
C1
2
C2
4
C1
1
C2
5
Parent
C1
6
C2
6
Gini = 0.500
N1
N2
C1
5
1
C2
2
4
Gini=0.333
CarType
{Sports, Luxury}
{Family}
C1
3
1
C2
2
4
Gini
0.400
CarType
{Sports}
{Family,Luxury}
C1
2
2
C2
1
5
Gini
0.419
CarType
Family
Sports
Luxury
C1
1
2
1
C2
4
1
1
Gini
0.393
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
No
Married
100K
No
3
No
Single
70K
No
4
Yes
Married
120K
No
5
No
Divorced
95K
Yes
6
No
Married
60K
No
7
Yes
Divorced
220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
Taxable Income�> 80K?�
Yes�
No�
C1
0
C2
6
C1
2
C2
4
C1
1
C2
5
C1
0
C2
6
C1
2
C2
4
C1
1
C2
5
Parent
C1
7
C2
3
Gini = 0.42
N1
N2
C1
3
4
C2
0
3
Gini=0.361
Class
+
–
+
–
–
–
+
–
+
+
P
0.25
0.43
0.53
0.76
0.85
0.85
0.85
0.87
0.93
0.95
1.00
TP
5
4
4
3
3
3
3
2
2
1
0
FP
5
5
4
4
3
2
1
1
0
0
0
TN
0
0
1
1
2
3
4
4
5
5
5
FN
0
1
1
2
2
2
2
3
3
4
5
TPR
1
0.8
0.8
0.6
0.6
0.6
0.6
0.4
0.4
0.2
0
FPR
1
1
0.8
0.8
0.6
0.4
0.2
0.2
0
0
0
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid
Attrib1 Attrib2 Attrib3 Class
1 Yes Large 125K
No
2 No Medium 100K
No
3 No Small 70K
No
4 Yes Medium 120K
No
5 No Large 95K
Yes
6 No Medium 60K
No
7 Yes Large 220K
No
8 No Small 85K
Yes
9 No Medium 75K
No
10 No Small 90K
Yes
10
Tid
Attrib1 Attrib2 Attrib3 Class
11 No Small 55K
?
12 Yes Medium 80K
?
13 Yes Large 110K
?
14 No Small 95K
?
15 No Large 67K
?
10
Test Set
Learning
algorithm
Training Set
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
No
Married
100K
No
3
No
Single
70K
No
4
Yes
Married
120K
No
5
No
Divorced
95K
Yes
6
No
Married
60K
No
7
Yes
Divorced
220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid
Attrib1 Attrib2 Attrib3 Class
1 Yes Large 125K
No
2 No Medium 100K
No
3 No Small 70K
No
4 Yes Medium 120K
No
5 No Large 95K
Yes
6 No Medium 60K
No
7 Yes Large 220K
No
8 No Small 85K
Yes
9 No Medium 75K
No
10 No Small 90K
Yes
10
Tid
Attrib1 Attrib2 Attrib3 Class
11 No Small 55K
?
12 Yes Medium 80K
?
13 Yes Large 110K
?
14 No Small 95K
?
15 No Large 67K
?
10
Test Set
Tree
Induction
algorithm
Training Set
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid
Attrib1 Attrib2 Attrib3 Class
1 Yes Large 125K
No
2 No Medium 100K
No
3 No Small 70K
No
4 Yes Medium 120K
No
5 No Large 95K
Yes
6 No Medium 60K
No
7 Yes Large 220K
No
8 No Small 85K
Yes
9 No Medium 75K
No
10 No Small 90K
Yes
10
Tid
Attrib1 Attrib2 Attrib3 Class
11 No Small 55K
?
12 Yes Medium 80K
?
13 Yes Large 110K
?
14 No Small 95K
?
15 No Large 67K
?
10
Test Set
Tree
Induction
algorithm
Training Set
Tid Refund Marital
Status
Taxable
Income
Cheat
1 Yes Single 125K
No
2 No Married 100K
No
3 No Single 70K
No
4 Yes Married 120K
No
5 No Divorced 95K
Yes
6 No Married 60K
No
7 Yes Divorced 220K
No
8 No Single 85K
Yes
9 No Married 75K
No
10 No Single 90K
Yes
10
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
No
Married
100K
No
3
No
Single
70K
No
4
Yes
Married
120K
No
5
No
Divorced
95K
Yes
6
No
Married
60K
No
7
Yes
Divorced
220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
Taxable
Income
> 80K?
YesNo
Taxable
Income?
(i) Binary split(ii) Multi-way split
< 10K
[10K,25K)[25K,50K)[50K,80K)
> 80K
Own
Car?
C0: 6
C1: 4
C0: 4
C1: 6
C0: 1
C1: 3
C0: 8
C1: 0
C0: 1
C1: 7
Car
Type?
C0: 1
C1: 0
C0: 1
C1: 0
C0: 0
C1: 1
Student
ID?
…
Yes
No
Family
Sports
Luxuryc
1
c
10
c
20
C0: 0
C1: 1
…
c
11
C0: 5
C1: 5
C0: 9
C1: 1
C0 N10
C1 N11
C0 N20
C1 N21
C0 N30
C1 N31
C0 N40
C1 N41
C0 N00
C1 N01
å
–
=
j
t
j
p
t
GINI
2
)]
|
(
[
1
)
(
C1
0
C2
6
Gini=0.000
C1
2
C2
4
Gini=0.444
C1
3
C2
3
Gini=0.500
C1
1
C2
5
Gini=0.278
C1
0
C2
6
C1
2
C2
4
C1
1
C2
5
å
=
=
k
i
i
split
i
GINI
n
n
GINI
1
)
(
Parent
C1
6
C2
6
Gini = 0.500
N1
N2
C1
5
1
C2
2
4
Gini=0.3
33
CarType
{Sports,
Luxury}
{Family}
C1
3
1
C2
2
4
Gini
0.400
CarType
{Sports}
{
Family,
Luxury}
C1
2
2
C2
1
5
Gini
0.419
CarType
Family
Sports
Luxury
C1
1
2
1
C2
4
1
1
Gini
0.393
Tid Refund Marital
Status
Taxable
Income
Cheat
1 Yes Single 125K
No
2 No Married 100K
No
3 No Single 70K
No
4 Yes Married 120K
No
5 No Divorced 95K
Yes
6 No Married 60K
No
7 Yes Divorced 220K
No
8 No Single 85K
Yes
9 No Married 75K
No
10 No Single 90K
Yes
10
Taxable
Income
> 80K?
YesNo
Cheat
No
No
No
Yes
Yes
Yes
No
No
No
No
Taxable Income
60
70
75
85
90
95
100
120
125
220
55
65
72
80
87
92
97
110
122
172
230
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
Yes
0
3
0
3
0
3
0
3
1
2
2
1
3
0
3
0
3
0
3
0
3
0
No
0
7
1
6
2
5
3
4
3
4
3
4
3
4
4
3
5
2
6
1
7
0
Gini
0.420
0.400
0.375
0.343
0.417
0.400
0.300
0.343
0.375
0.400
0.420
å
–
=
j
t
j
p
t
j
p
t
Entropy
)
|
(
log
)
|
(
)
(
å
–
=
j
t
j
p
t
j
p
t
Entropy
)
|
(
log
)
|
(
)
(
2
÷
ø
ö
ç
è
æ
–
=
å
=
k
i
i
split
i
Entropy
n
n
p
Entropy
GAIN
1
)
(
)
(
SplitINFO
GAIN
GainRATIO
Split
split
=
å
=
–
=
k
i
i
i
n
n
n
n
SplitINFO
1
log
)
|
(
max
1
)
(
t
i
P
t
Error
i
–
=
Parent
C1
7
C2
3
Gini = 0.
42
N1
N2
C1
3
4
C2
0
3
Gini=0.361
FN
FP
TN
TP
TN
TP
d
c
b
a
d
a
+
+
+
+
=
+
+
+
+
=
Accuracy
c
b
a
a
p
r
rp
b
a
a
c
a
a
+
+
=
+
=
+
=
+
=
2
2
2
(F)
measure
–
F
(r)
Recall
(p)
Precision
d
w
c
w
b
w
a
w
d
w
a
w
4
3
2
1
4
1
Accuracy
Weighted
+
+
+
+
=
Class
+
–
+
–
–
–
+
–
+
+
P
0.25
0.43
0.53
0.76
0.85
0.85
0.85
0.87
0.93
0.95
1.00
TP
5
4
4
3
3
3
3
2
2
1
0
FP
5
5
4
4
3
2
1
1
0
0
0
TN
0
0
1
1
2
3
4
4
5
5
5
FN
0
1
1
2
2
2
2
3
3
4
5
TPR
1
0.8
0.8
0.6
0.6
0.6
0.6
0.4
0.4
0.2
0
FPR
1
1
0.8
0.8
0.6
0
.4
0.2
0.2
0
0
0
a
a
a
–
=
<
-
-
<
-
1
)
/
)
1
(
(
2
/
1
2
/
Z
N
p
p
p
acc
Z
P
)
(
2
4
4
2
2
2
/
2
2
2
/
2
2
/
a
a
a
Z
N
acc
N
acc
N
Z
Z
acc
N
p
+
´
´
-
´
´
+
±
+
´
´
=
(
)
(
)
2
2
2
1
1
1
,
~
,
~
s
m
s
m
N
e
N
e
i
i
i
i
n
e
e
)
1
(
ˆ
-
=
s
2
)
2
1
(
2
1
)
1
1
(
1
ˆ
ˆ
2
2
2
1
2
2
2
1
2
n
e
e
n
e
e
t
-
+
-
=
+
@
+
=
s
s
s
s
s
t
t
Z
d
d
s
a
ˆ
2
/
±
=
0043
.
0
5000
)
25
.
0
1
(
25
.
0
30
)
15
.
0
1
(
15
.
0
ˆ
=
-
+
-
=
d
s
128
.
0
100
.
0
0043
.
0
96
.
1
100
.
0
±
=
´
±
=
t
d
t
k
t
k
j
j
t
t
d
d
k
k
d
d
s
s
a
ˆ
)
1
(
)
(
ˆ
1
,
1
1
2
2
-
-
=
±
=
-
-
=
å