CSIS 100CSIS 100 - Discussion Board Topic #1One of the object.docx

CSIS 100
CSIS 100 - Discussion Board Topic #1:
One of the objectives of this course is to enable students to
differentiate between the disciplines of Information Systems,
Information Technology, and Computer Science. Oftentimes,
these areas overlap and are difficult to distinguish – even
among professionals within the industries.
There are some distinctions that become evident, but all too
frequently, people do not understand these distinctions until
they are already deep within their programs of study.
Consequently, many decide that it is too late to pursue a
different avenue in the computing world without losing valuable
time and money spent on courses that may or may not apply to a
different major.
Given the importance of achieving effective planning from the
beginning, your first assignment in this course is to delve into
the broad areas of Information Systems, Information
Technology, and Computer Science and write about your career
choice in a discussion board post. This should be your thought
process:
· First, define each field (i.e. IS, IT, CS). Understand the
similarities and differences.
· Second, determine what jobs are available in each area.
· Third, look at the degree completion plans for each of these
programs.
· Fourth, assess your own skills (e.g. Are you good in math? Do
you like business? Do you like algorithms? Are you gifted at
problem-solving? Do you like learning about new technology?
Do you enjoy working hands-on with
equipment/hardware/wires?)
· Fifth, (and most importantly) ask God what He wants you to
pursue based on your talents, interests, and abilities.
· Sixth, based on your analysis above, what career do you hope
to obtain after graduation, and what degree will you pursue to

achieve this goal?
To facilitate your research, there are four videos in your
Reading & Study folder that will help you understand the
differences between the computing fields and become familiar
with the job opportunities in each area. Be sure to view these
videos first.
The LU Registrar’s home page has information on degree
completion plans. Here is a link to all of the currently available
ones in the university:
http://www.liberty.edu/academics/registrar/index.cfm?PID=298
1
Be sure to look at all of the ones listed for Information Systems
and Information Technology. At the time of this writing,
Computer Science is only listed under residential degree plans.
That does not mean that you should rule out Computer Science
as a potential major. You must consider all options and listen
to God’s calling upon your life. With God, all things are
possible.
Discussion Board Deliverables
Main Post:
In a minimum of 300 words, create a thread in Module 1’s
discussion board forum that describes the following:
1. Your desired career upon graduation
2. Why you chose this career
3. Your intended major
4. Your strengths, weaknesses, and interests
5. How the major supports your chosen career
6. How God has led you to reach your decision
7. A Bible verse that reminds you to trust in God to lead you in
your career/educational choices.
Note that the discussion board main thread requires a minimum
of two citations. The Holy Bible may be used in this discussion
board to satisfy these requirements.
Number your responses. Make complete sentences.
Title your main post as follows: [Your career choice] – [Your

major]
Here is an example:
Title: Network Engineer – Information Technology
(Networking/Security Cognate)
1. After graduation, I hope to obtain employment as a Network
Administrator within a medium-to-large organization and
eventually become certified as a CCNA.
2. I chose this career because I enjoy working with hardware
and troubleshooting connectivity issues. I like to help others
when it comes to setting up routers/modems in their homes and
even running the cables. There are also numerous, well-paying
jobs in this arena right now.
3. My intended major is Information Technology with a cognate
in Networking and Security.
4. My strengths are being able to see how the “big picture”
works. Looking at schematics and diagrams of networks is
fascinating, and I want to learn more about it. Math is not my
strongest area. Therefore, I have chosen a degree that requires
only a few math classes. I enjoy helping people communicate
and protecting them from hackers. Thus, I am also very
interested in the security aspect of data communications.
5. Liberty offers a degree in Information Technology with a
cognate (specialization) in data networking and security. This
degree not only prepares you to perform networking in the field,
but it also prepares you for the CCNA (Cisco Certified Network
Associate) exam, a certification that I hope to obtain after I
graduate.
6. God has directed my career choices by giving me the interest
and ability to work with networking equipment. I can see how
learning about networking and security can help protect God’s
children from cyber-attacks and persecution and, therefore, help
build His kingdom.
7. I actually have two Bible verses that I use to remind myself
of God’s control in my life (and especially my career):
Psalm 37:4 (ESV): Delight yourself in the Lord, and he will
give you the desires of your heart.

Proverbs 19:21 (ESV): Many are the plans in the mind of a man,
but it is the purpose of the lord that will stand.
Replies:
Reply to two of your classmates’ threads. Each reply should be
100 words each, include at least 1 citation, and should say
something meaningful rather than just “Great choice of career.
I really agree with you…” Use this as an opportunity to share
insights, cares, troubles, suggestions, etc. These replies should
serve to foster professional relationships and build bridges
between each other.
This discussion board should serve as an opportunity for you to
review your career and educational choices. Most importantly,
it should be a reminder let God guide you in the process and
open your heart to His will for your life. He loves you so much
and wants only the best for you. Have fun with this!
関係者外秘
関係者外秘
関係者外秘
Submission instructions: Complete the answers below the given
questions here itself. DO NOT create a new Word document.
Failure to follow instructions may lead to 10% deduction of
points from the final exam.
Q. 1. Consider the market basket transactions. When answering
the below questions, provide reasoning whenever possible. (30
points)
1. What is the maximum number of association rules that can be
extracted from this data (including rules that have zero
support)?
1. What is the maximum size of frequent itemsets that can be

extracted (assuming minsup > 0)?
1. Write an expression for the maximum number of size-3
itemsets that can be derived from this data set.
1. Find an itemset (of size 2 or larger) that has the largest
support.
1. Find a pair of items, a and b, such that the rules and have the
same confidence.
Q. 2. What is Anomaly Detection? Describe in detail the
characteristics of the Anomaly Detection Problems. Also,
describe in detail the characteristics of Anomaly Detection
Methods. (30 points)
When referring to other articles, you need to cite using the
format given here (either APA or MLA):
https://owl.purdue.edu/owl/research_and_citation/resources.htm
l
Note: Failure of citations will lead to several deduction of
points or zero points may be awarded for the answer.
Data Mining
Association Analysis: Basic Concepts

and Algorithms
Lecture Notes
Introduction to Data Mining
by
Tan, Steinbach, Kumar
© Tan,Steinbach, Kumar Introduction to Data Mining
4/18/2004 *
Association Rule MiningGiven a set of transactions, find rules
that will predict the occurrence of an item based on the
occurrences of other items in the transaction
Market-Basket transactions
Example of Association Rules
Implication means co-occurrence, not causality!
TIDItems
1
Bread, Milk
2
Bread, Diaper, Beer, Eggs

3
Milk, Diaper, Beer, Coke
4
Bread, Milk, Diaper, Beer
5
Bread, Milk, Diaper, Coke
Definition: Frequent ItemsetItemsetA collection of one or more
itemsExample: {Milk, Bread, Diaper}k-itemsetAn itemset that
of transactions that contain an itemsetE.g. s({Milk, Bread,
Diaper}) = 2/5Frequent ItemsetAn itemset whose support is
greater than or equal to a minsup threshold
TIDItems
1
Bread, Milk
2
3
4
5
Definition: Association RuleAssociation RuleAn implication
itemsetsExample:
Rule Evaluation MetricsSupport (s)Fraction of transactions that

contain both X and YConfidence (c)Measures how often items
in Y
appear in transactions that
contain X
Example:
TIDItems
1
Bread, Milk
2
3
4
5
Association Rule Mining TaskGiven a set of transactions T, the
goal of association rule mining is to find all rules having
support ≥ minsup thresholdconfidence ≥ minconf threshold
Brute-force approach:List all possible association rulesCompute
the support and confidence for each rulePrune rules that fail the
minsup and minconf thresholds
putationally prohibitive!
Mining Association Rules
Example of Rules:

=0.4, c=0.5)
Observations: All the above rules are binary partitions of the
same itemset:
{Milk, Diaper, Beer} Rules originating from the same itemset
have identical support but
can have different confidence Thus, we may decouple the
support and confidence requirements
TIDItems
1
Bread, Milk
2
3
4
5
Mining Association RulesTwo-step approach:
Frequent Itemset Generation
Rule Generation
Generate high confidence rules from each frequent itemset,
where each rule is a binary partitioning of a frequent itemset
Frequent itemset generation is still computationally expensive

Frequent Itemset Generation
Given d items, there are 2d possible candidate itemsets
Frequent Itemset GenerationBrute-force approach: Each itemset
in the lattice is a candidate frequent itemsetCount the support of
each candidate by scanning the database
Match each transaction against every candidateComplexity ~
O(NMw) => Expensive since M = 2d !!!
N�
w�
M�
List of Candidates�
Computational ComplexityGiven d unique items:Total number
of itemsets = 2dTotal number of possible association rules:
If d=6, R = 602 rules
Frequent Itemset Generation StrategiesReduce the number of
candidates (M)Complete search: M=2dUse pruning techniques

to reduce M
Reduce the number of transactions (N)Reduce size of N as the
size of itemset increasesUsed by DHP and vertical-based mining
algorithms
Reduce the number of comparisons (NM)Use efficient data
structures to store the candidates or transactionsNo need to
match every candidate against every transaction
Reducing Number of CandidatesApriori principle:If an itemset
is frequent, then all of its subsets must also be frequent
Apriori principle holds due to the following property of the
support measure:
Support of an itemset never exceeds the support of its
subsetsThis is known as the anti-monotone property of support
Illustrating Apriori Principle
Found to be Infrequent
Pruned supersets
Illustrating Apriori Principle
Items (1-itemsets)
Pairs (2-itemsets)
(No need to generate
candidates involving Coke

or Eggs)
Triplets (3-itemsets)
Minimum Support = 3
If every subset is considered,
6C1 + 6C2 + 6C3 = 41
With support-based pruning,
6 + 6 + 1 = 13
ItemCount
Bread
4
Coke
2
Milk
4
Beer
3
Diaper
4
Eggs
1
ItemsetCount
{Bread,Milk}
3

{Bread,Beer}
2
{Bread,Diaper}
3
{Milk,Beer}
2
{Milk,Diaper}
3
{Beer,Diaper}
3
ItemsetCount
{Bread,Milk,Diaper}
3
Apriori AlgorithmMethod:
Let k=1Generate frequent itemsets of length 1Repeat until no
new frequent itemsets are identifiedGenerate length (k+1)
candidate itemsets from length k frequent itemsetsPrune
candidate itemsets containing subsets of length k that are
infrequent Count the support of each candidate by scanning the
DBEliminate candidates that are infrequent, leaving only those
that are frequent
Reducing Number of ComparisonsCandidate counting:Scan the
database of transactions to determine the support of each

candidate itemsetTo reduce the number of comparisons, store
the candidates in a hash structure Instead of matching each
transaction against every candidate, match it against candidates
contained in the hashed buckets
N�
k�
Buckets�
Hash Structure�
Generate Hash Tree
Suppose you have 15 candidate itemsets of length 3:
{1 4 5}, {1 2 4}, {4 5 7}, {1 2 5}, {4 5 8}, {1 5 9}, {1 3 6}, {2
3 4}, {5 6 7}, {3 4 5}, {3 5 6}, {3 5 7}, {6 8 9}, {3 6 7}, {3 6
8}
You need: Hash function Max leaf size: max number of
itemsets stored in a leaf node (if number of candidate itemsets
exceeds max leaf size, split the node)
2 3 4
5 6 7
1 4 5
1 3 6
1 2 4
4 5 7
1 2 5
4 5 8

1 5 9
3 4 5
3 5 6
3 5 7
6 8 9
3 6 7
3 6 8
1,4,7
2,5,8
3,6,9
Hash function
Association Rule Discovery: Hash tree
1,4,7
2,5,8
3,6,9
Hash Function
Candidate Hash Tree
Hash on 1, 4 or 7

1 5 9
1 4 5
1 3 6
3 4 5
3 6 7
3 6 8
3 5 6
3 5 7
6 8 9

2 3 4
5 6 7
1 2 4
4 5 7
1 2 5
4 5 8
1,4,7
2,5,8
3,6,9
Hash Function
Candidate Hash Tree
Hash on 2, 5 or 8

1 5 9
1 4 5
1 3 6
3 4 5
3 6 7
3 6 8
3 5 6
3 5 7
6 8 9
2 3 4

5 6 7
1 2 4
4 5 7
1 2 5
4 5 8
1,4,7
2,5,8
3,6,9
Hash Function
Candidate Hash Tree
Hash on 3, 6 or 9

5 6 7
1 2 4
4 5 7
1 2 5
4 5 8
Subset Operation
Given a transaction t, what are the possible subsets of size 3?
1 2 3 5 6�
Transaction, t�
2 3 5 6�
3 5 6�
2�
1�
5 6�
1 3�
3 5 6�
1 2�
6�
1 5�
5 6�
2 3�
6�

2 5�
5 6�
3�
1 2 3�1 2 5�1 2 6�
1 3 5�1 3 6�
1 5 6�
2 3 5�2 3 6�
2 5 6�
3 5 6�
Subsets of 3 items�
Level 1�
Level 2�
Level 3�
6�
3 5�
Subset Operation Using Hash Tree
transaction
1 4 5
3 6 7
3 6 8
3 5 6
3 5 7
6 8 9
2 3 4
5 6 7

1 2 4
4 5 7
1 2 5
4 5 8
1 5 9
1 3 6
3 4 5

1 2 3 5 6
1 +
2 3 5 6
3 5 6
2 +
5 6
3 +
2,5,8
3,6,9
Hash Function
1,4,7
1 5 9

1 3 6
3 4 5
transaction
1 4 5
3 6 7
3 6 8
3 5 6
3 5 7
6 8 9
2 3 4
5 6 7
1 2 4
4 5 7
1 2 5
4 5 8

2,5,8
3,6,9
Hash Function
1,4,7
1 2 3 5 6
3 5 6
1 2 +
5 6
1 3 +
6
1 5 +
3 5 6
2 +
5 6
3 +
1 +
2 3 5 6

1 5 9
1 3 6
3 4 5
transaction
Match transaction against 11 out of 15 candidates
1 4 5
3 6 7
3 6 8
3 5 6
3 5 7
6 8 9
2 3 4
5 6 7
1 2 4

4 5 7
1 2 5
4 5 8
2,5,8
3,6,9
Hash Function
1,4,7
1 2 3 5 6
3 5 6
1 2 +
5 6
1 3 +
6
1 5 +
3 5 6
2 +
5 6

3 +
1 +
2 3 5 6
Factors Affecting ComplexityChoice of minimum support
threshold lowering support threshold results in more frequent
itemsets this may increase number of candidates and max length
of frequent itemsetsDimensionality (number of items) of the
data set more space is needed to store support count of each
item if number of frequent items also increases, both
computation and I/O costs may also increaseSize of database
since Apriori makes multiple passes, run time of algorithm may
increase with number of transactionsAverage transaction width
transaction width increases with denser data setsThis may
increase max length of frequent itemsets and traversals of hash
tree (number of subsets in a transaction increases with its
width)
Compact Representation of Frequent ItemsetsSome itemsets are
redundant because they have identical support as their supersets
Number of frequent itemsets
Need a compact representation

Maximal Frequent Itemset
Border
Infrequent Itemsets
Maximal Itemsets
An itemset is maximal frequent if none of its immediate
supersets is frequent
null�
AB�
AC�
AD�
AE�
BC�
BD�
BE�
CD�
CE�
DE�
ABC�
ABD�
ABE�
ACD�
ACE�
ADE�
BCD�
BCE�
BDE�
CDE�
A�
B�
C�
D�

E�
ABCD�
ABCE�
ABDE�
ACDE�
BCDE�
ABCDE�
Closed ItemsetAn itemset is closed if none of its immediate
supersets has the same support as the itemset
Sheet1TIDItems1{A,B}2{B,C,D}3{A,B,C,D}4{A,B,D}5{A,B,C
,D}
Sheet2
Sheet3
Sheet1ItemsetSupport{A}4{B}5{C}3{D}4{A,B}4{A,C}2{A,D}
3{B,C}3{B,D}4{C,D}3
Sheet2
Sheet3
Sheet1ItemsetSupport{A,B,C}2{A,B,D}3{A,C,D}2{B,C,D}3{A
,B,C,D}2
Sheet2
Sheet3
Maximal vs Closed Itemsets
Transaction Ids
Not supported by any transactions
Sheet1TIDItems1ABC2ABCD3BCE4ACDE5DE

Maximal vs Closed Frequent Itemsets
Minimum support = 2
# Closed = 9
# Maximal = 4
Closed and maximal
Closed but not maximal
Maximal vs Closed Itemsets
Frequent Itemsets�
Closed Frequent Itemsets�
Maximal Frequent Itemsets�
Alternative Methods for Frequent Itemset GenerationTraversal
of Itemset LatticeGeneral-to-specific vs Specific-to-general
....�
Frequent itemset border�
null�
{a1,a2,...,an}�
(a) General-to-specific�
....�
null�
{a1,a2,...,an}�
(b) Specific-to-general�

....�
null�
{a1,a2,...,an}�
(c) Bidirectional�
of Itemset LatticeEquivalent Classes
null�
AB�
AC�
AD�
null�
BC�
BD�
AB�
CD�
AC�
AD�
ABC�
ABD�
BC�
ACD�
BD�
CD�
BCD�
A�
B�
C�
A�
B�

C�
D�
D�
ABC�
ABD�
ACD�
BCD�
ABCD�
(a) Prefix tree�
(b) Suffix tree�
ABCD�
of Itemset LatticeBreadth-first vs Depth-first
(a) Breadth first�
(b) Depth first�
Alternative Methods for Frequent Itemset
GenerationRepresentation of Databasehorizontal vs vertical data
layout
Horizontal Data Layout�
Vertical Data Layout�
FP-growth AlgorithmUse a compressed representation of the
database using an FP-tree

Once an FP-tree has been constructed, it uses a recursive
divide-and-conquer approach to mine the frequent itemsets
FP-tree construction
null
A:1
B:1
null
A:1
B:1
B:1
C:1
D:1
After reading TID=1:
After reading TID=2:
Sheet1TIDItems1{A,B}2{B,C,D}3{A,C,D,E}4{A,D,E}5{A,B,C
}6{A,B,C,D}7{B,C}8{A,B,C}9{A,B,D}10{B,C,E}
Sheet2
Sheet3
FP-Tree Construction

null
A:7
B:5
B:3
C:3
D:1
C:1
D:1
C:3
D:1
D:1
E:1
E:1
Pointers are used to assist frequent itemset generation
D:1
E:1
Transaction Database
Header table
Sheet2

Sheet3
Sheet1ItemPointerABCDE
Sheet2
Sheet3
FP-growth
null
A:7
B:5
B:1
C:1
D:1
C:1
D:1
C:3
D:1
D:1
Conditional Pattern base for D:
P = {(A:1,B:1,C:1),
(A:1,B:1),

(A:1,C:1),
(A:1),
(B:1,C:1)}
Recursively apply FP-growth on P
Frequent Itemsets found (with sup > 1):
AD, BD, CD, ACD, BCD
D:1
Tree Projection
Set enumeration tree:
Possible Extension: E(A) = {B,C,D,E}
Possible Extension: E(ABC) = {D,E}
Tree ProjectionItems are listed in lexicographic orderEach node
P stores the following information:Itemset for node PList of
possible lexicographic extensions of P: E(P)Pointer to projected
database of its ancestor nodeBitvector containing information
about which transactions in the projected database contain the
itemset
Projected Database
Original Database:
Projected Database for node A:

E(A)
Sheet2
Sheet3
Sheet1TIDItems1{B}2{}3{C,D,E}4{D,E}5{B,C}6{B,C,D}7{}8
{B,C}9{B,D}10{}
Sheet2
Sheet3
ECLATFor each item, store a list of transaction ids (tids)
TID-list
ECLATDetermine support of any k-itemset by intersecting tid-
lists of two of its (k-1) subsets.
3 traversal approaches: top-down, bottom-up and
hybridAdvantage: very fast support countingDisadvantage:
intermediate tid-lists may become too large for memory

Sheet1TIDItemsABCDE1A,B,E112212B,C,D423433C,E554564
A,C,D67895A,B,C,D7896A,E8107A,B98A,B,C9A,C,D10BAB11
425567788109
Sheet2
Sheet3
A,C,D67895A,B,C,D7896A,E8107A,B98A,B,C9A,C,D10BAB11
425567788109
Sheet2
Sheet3
A,C,D67895A,B,C,D7896A,E8107A,B98A,B,C9A,C,D10BABA
B111425557678788109
Sheet2
Sheet3
Rule GenerationGiven a frequent itemset L, find all non-empty
– f satisfies the minimum
confidence requirementIf {A,B,C,D} is a frequent itemset,
candidate rules:
If |L| = k, then there are 2k – 2 candidate association rules

Rule GenerationHow to efficiently generate rules from frequent
itemsets?In general, confidence does not have an anti-monotone
property
But confidence of rules generated from the same itemset has an
anti-monotone propertye.g., L = {A,B,C,D}:
D)
Confidence is anti-monotone w.r.t. number of items on the RHS
of the rule
Rule Generation for Apriori Algorithm
Lattice of rules
Low Confidence Rule
Pruned Rules
ABCD=>{ }�
BC=>AD�
BD=>AC�
CD=>AB�
AD=>BC�
AC=>BD�
AB=>CD�
D=>ABC�
C=>ABD�
B=>ACD�
A=>BCD�
ACD=>B�
ABD=>C�
ABC=>D�
BCD=>A�

Rule Generation for Apriori AlgorithmCandidate rule is
generated by merging two rules that share the same prefix
in the rule consequent
join(CD=>AB,BD=>AC)
would produce the candidate
rule D => ABC
Prune rule D=>ABC if its
subset AD=>BC does not have
high confidence
Effect of Support DistributionMany real data sets have skewed
support distribution
Support distribution of a retail data set
Effect of Support DistributionHow to set the appropriate minsup
threshold?If minsup is set too high, we could miss itemsets
involving interesting rare items (e.g., expensive products)
If minsup is set too low, it is computationally expensive and the
number of itemsets is very large
Using a single minimum support threshold may not be effective
Multiple Minimum SupportHow to apply multiple minimum

supports?MS(i): minimum support for item i e.g.:
MS(Milk)=5%, MS(Coke) = 3%,
MS(Broccoli)=0.1%, MS(Salmon)=0.5%MS({Milk,
Broccoli}) = min (MS(Milk), MS(Broccoli))
= 0.1%
Challenge: Support is no longer anti-monotone Suppose:
Support(Milk, Coke) = 1.5% and
Support(Milk, Coke, Broccoli) = 0.5%
{Milk,Coke} is infrequent but {Milk,Coke,Broccoli} is
frequent
Multiple Minimum Support
Multiple Minimum Support
Multiple Minimum Support (Liu 1999)Order the items according
to their minimum support (in ascending order)e.g.:
MS(Milk)=5%, MS(Coke) = 3%,
MS(Broccoli)=0.1%, MS(Salmon)=0.5%Ordering:
Broccoli, Salmon, Coke, Milk
Need to modify Apriori such that:L1 : set of frequent itemsF1 :
where MS(1) is mini( MS(i) )C2 : candidate itemsets of size 2 is

generated from F1
instead of L1
Multiple Minimum Support (Liu 1999)Modifications to
Apriori:In traditional Apriori, A candidate (k+1)-itemset is
generated by merging two
frequent itemsets of size k The candidate is pruned if it
contains any infrequent subsets
of size kPruning step has to be modified: Prune only if subset
contains the first item e.g.: Candidate={Broccoli, Coke, Milk}
(ordered according to
minimum support) {Broccoli, Coke} and {Broccoli, Milk}
are frequent but
{Coke, Milk} is infrequent
Candidate is not pruned because {Coke,Milk} does not contain
the first item, i.e., Broccoli.
Pattern EvaluationAssociation rule algorithms tend to produce
too many rules many of them are uninteresting or
have same support & confidence
Interestingness measures can be used to prune/rank the derived
patterns
In the original formulation of association rules, support &
confidence are the only measures used

Application of Interestingness Measure
Interestingness Measures
Computing Intere
information needed to compute rule interestingness can be
obtained from a contingency table
Used to define various measures support, confidence, lift, Gini,
J-measure, etc.YY Xf11f10f1+X f01f00fo+f+1f+0|T|
f11: support of X and Y

Drawback of Confidence
Coffee
CoffeeTea15520Tea755809010100
Confidence= P(Coffee|Tea) = 0.75
but P(Coffee) = 0.9 Although confidence is high, rule is
misleading P(Coffee|Tea) = 0.9375
Statistical IndependencePopulation of 1000 students600
students know how to swim (S)700 students know how to bike
(B)420 students know how to swim and bike (S,B)
Negatively correlated

Statistical-based MeasuresMeasures that take into account
statistical dependence
Example: Lift/Interest
Confidence= P(Coffee|Tea) = 0.75
but P(Coffee) = 0.9 Lift = 0.75/0.9= 0.8333 (< 1, therefore is
negatively associated)
Coffee
CoffeeTea15520Tea755809010100
Drawback of Lift & Interest
Statistical independence:
If P(X,Y)=P(X)P(Y) => Lift =
1YYX10010X090901090100YYX90090X010109010100

There are lots of measures proposed in the literature
Some measures are good for certain applications, but not for
others
What criteria should we use to determine whether a measure is
good or bad?
What about Apriori-style support based pruning? How does it
affect these measures?

Properties of A Good MeasurePiatetsky-Shapiro:
3 properties a good measure M must satisfy:M(A,B) = 0 if A
and B are statistically independent
M(A,B) increase monotonically with P(A,B) when P(A) and
P(B) remain unchanged
M(A,B) decreases monotonically with P(A) [or P(B)] when
P(A,B) and P(B) [or P(A)] remain unchanged
Comparing Different Measures
10 examples of contingency tables:
Rankings of contingency tables using various measures:
Sheet1ExampleE18123834241370100001.1581671343E2833026
221046100001.116800672E3948194127298100001.030581565E
43954308052961100001.4198707063E528861363132044311000
01.6148802655E6150020005006000100002.1428571429E74000
200010003000100001.3333333333E84000200020002000100001
.1111111111E91720712151154100001.127815235E1061248347
452100003.6889211418
Sheet2
Sheet3
Property under Variable Permutation
Does M(A,B) = M(B,A)?
Symmetric measures: support, lift, collective strength, cosine,
Jaccard, etc

Asymmetric measures: confidence, conviction, Laplace, J-
measure, etc
Property under Row/Column Scaling
Grade-Gender Example (Mosteller, 1968):
Mosteller:
Underlying association should be independent of
the relative number of male and female students
in the samples
2x
10xMaleFemaleHigh235Low1453710MaleFemaleHigh43034Lo
w2404267076

Property under Inversion Operation
Transaction 1
Transaction N
.
.
.
.
.
- -coefficient is analogous to
correlation coefficient for continuous variables
tablesYYX601070X1020307030100YYX201030X106070307010
0

Property under Null Addition
Invariant measures: support, cosine, Jaccard, etc
Non-invariant measures: correlation, Gini, mutual information,
odds ratio, etc
Different Measures have Different Properties
Sheet1SymbolMeasureRangeP1P2P3O1O2O3O3'O4FCorrelation
-1 … 0 … 1YesYesYesYesNoYesYesNolLambda0 …
1YesNoNoYesNoNo*YesNoaOdds
ratioYes*YesYesYesYesYes*YesNoQYule's Q-1 … 0 …
1YesYesYesYesYesYesYesNoYYule's Y-1 … 0 …
1YesYesYesYesYesYesYesNokCohen's-1 … 0 …
1YesYesYesYesNoNoYesNoMMutual Information0 …
1YesYesYesYesNoNo*YesNoJJ-Measure0 …
1YesNoNoNoNoNoNoNoGGini Index0 …
1YesNoNoNoNoNo*YesNosSupport0 …
1NoYesNoYesNoNoNoNocConfidence0 …
1NoYesNoYesNoNoNoYesLLaplace0 …
1NoYesNoYesNoNoNoNoVConvictionNoYesNoYes**NoNoYes
NoIInterestYes*YesYesYesNoNoNoNoISIS (cosine)0 ..
1NoYesYesYesNoNoNoYesPSPiatetsky-Shapiro's-0.25 … 0 …

0.25YesYesYesYesNoYesYesNoFCertainty factor-1 … 0 …
1YesYesYesNoNoNoYesNoAVAdded value0.5 … 1 …
1YesYesYesNoNoNoNoNoSCollective
strengthNoYesYesYesNoYes*YesNozJaccard0 ..
1NoYesYesYesNoNoNoYesKKlosgen'sYesYesYesNoNoNoNoN
owhere:O1: Symmetry under variable permutationO2:
Marginal invarianceO3: Antisymmetry under row or column
permutationO3': Inversion invarianceO4: Null
invarianceYes*: Yes if measure is normalizedYes**: Yes if
measure is symmetrized by taking max(M(A,B),M(B,A))No*:
Symmetry under row or column permutation
Sheet2
Sheet3
3
3
2
0
3
1
3
2
1
3
2
K
K
÷
÷
ø
ö
ç
ç
è
æ
-

-
÷
÷
ø
ö
ç
ç
è
æ
-
MBD01E1AF4B.unknown
Support-based PruningMost of the association rule mining
algorithms use support measure to prune rules and itemsets
Study effect of support pruning on correlation of
itemsetsGenerate 10000 random contingency tablesCompute
support and pairwise correlation for each tableApply support-
based pruning and examine the tables that are removed
Effect of Support-based Pruning
Chart1-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
All Itempairs
13
95
172
282
359
503
648

716
830
867
908
915
881
759
602
532
370
259
194
84
11
Chart2-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.9
0
0
4
7
9
7
25
35
42
86
92
75
54
28
14
16
3
5

0
0
0
Chart3-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.7
7
29
70
95
150
182
267
296
336
383
375
370
332
272
202
161
89
49
41
17
4
Chart4-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.5
13
92
164
265

327
428
542
572
640
665
664
678
645
540
406
355
237
139
118
51
5
Chart5-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.01
10
21
22
18
28
28
46
40
33
25
21
7
2
1
0

0
0
0
0
0
0
Chart6-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.03
13
71
68
66
67
84
94
93
93
77
56
30
14
5
3
0
0
0
0
0
0
Chart7-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.05
13

91
104
106
107
128
141
151
161
136
99
67
45
15
8
4
2
1
0
0
0
Sheet1CorrelationAll> 0.9> 0.7> 0.5< 0.01< 0.03< 0.05-
1130713101313-0.99502992217191-0.81724701642268104-
0.72827952651866106-0.635991503272867107-
0.550371824282884128-0.4648252675424694141-
0.3716352965724093151-0.2830423366403393161-
0.18678638366525771360908923756642156990.191575370678
730670.288154332645214450.37592827254015150.4602142024
060380.5532161613550040.63703892370020.72595491390010.
81940411180000.98401751000111045000
Sheet2
Sheet3
Effect of Support-based Pruning
Support-based pruning eliminates mostly negatively correlated

itemsets
Chart1-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
All Itempairs
13
95
172
282
359
503
648
716
830
867
908
915
881
759
602
532
370
259
194
84
11
Chart2-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.9
0
0
4
7
9

7
25
35
42
86
92
75
54
28
14
16
3
5
0
0
0
Chart3-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.7
7
29
70
95
150
182
267
296
336
383
375
370
332
272
202
161

89
49
41
17
4
Chart4-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.5
13
92
164
265
327
428
542
572
640
665
664
678
645
540
406
355
237
139
118
51
5
Chart5-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.01
10
21

22
18
28
28
46
40
33
25
21
7
2
1
0
0
0
0
0
0
0
Chart6-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.03
13
71
68
66
67
84
94
93
93
77
56
30
14

5
3
0
0
0
0
0
0
Chart7-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.05
13
91
104
106
107
128
141
151
161
136
99
67
45
15
8
4
2
1
0
0
0
1130713101313-0.99502992217191-0.81724701642268104-
0.72827952651866106-0.635991503272867107-

0.550371824282884128-0.4648252675424694141-
0.3716352965724093151-0.2830423366403393161-
0.18678638366525771360908923756642156990.191575370678
730670.288154332645214450.37592827254015150.4602142024
060380.5532161613550040.63703892370020.72595491390010.
81940411180000.98401751000111045000
Sheet2
Sheet3
Chart1-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
All Itempairs
13
95
172
282
359
503
648
716
830
867
908
915
881
759
602
532
370
259
194
84
11
Chart2-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91

Correlation
Support > 0.9
0
0
4
7
9
7
25
35
42
86
92
75
54
28
14
16
3
5
0
0
0
Chart3-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.7
7
29
70
95
150
182
267
296
336

383
375
370
332
272
202
161
89
49
41
17
4
Chart4-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.5
13
92
164
265
327
428
542
572
640
665
664
678
645
540
406
355
237
139
118
51

5
Chart5-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.01
10
21
22
18
28
28
46
40
33
25
21
7
2
1
0
0
0
0
0
0
0
Chart6-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.03
13
71
68
66
67
84

94
93
93
77
56
30
14
5
3
0
0
0
0
0
0
Chart7-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.05
13
91
104
106
107
128
141
151
161
136
99
67
45
15
8
4
2

1
0
0
0
1130713101313-0.99502992217191-0.81724701642268104-
0.72827952651866106-0.635991503272867107-
0.550371824282884128-0.4648252675424694141-
0.3716352965724093151-0.2830423366403393161-
0.18678638366525771360908923756642156990.191575370678
730670.288154332645214450.37592827254015150.4602142024
060380.5532161613550040.63703892370020.72595491390010.
81940411180000.98401751000111045000
Sheet2
Sheet3
Chart1-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
All Itempairs
13
95
172
282
359
503
648
716
830
867
908
915
881
759
602
532

370
259
194
84
11
Chart2-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.9
0
0
4
7
9
7
25
35
42
86
92
75
54
28
14
16
3
5
0
0
0
Chart3-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.7
7
29

70
95
150
182
267
296
336
383
375
370
332
272
202
161
89
49
41
17
4
Chart4-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support > 0.5
13
92
164
265
327
428
542
572
640
665
664
678
645

540
406
355
237
139
118
51
5
Chart5-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.01
10
21
22
18
28
28
46
40
33
25
21
7
2
1
0
0
0
0
0
0
0
Chart6-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation

Support < 0.03
13
71
68
66
67
84
94
93
93
77
56
30
14
5
3
0
0
0
0
0
0
Chart7-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-
0.100.10.20.30.40.50.60.70.80.91
Correlation
Support < 0.05
13
91
104
106
107
128
141
151
161
136

99
67
45
15
8
4
2
1
0
0
0
1130713101313-0.99502992217191-0.81724701642268104-
0.72827952651866106-0.635991503272867107-
0.550371824282884128-0.4648252675424694141-
0.3716352965724093151-0.2830423366403393161-
0.18678638366525771360908923756642156990.191575370678
730670.288154332645214450.37592827254015150.4602142024
060380.5532161613550040.63703892370020.72595491390010.
81940411180000.98401751000111045000
Sheet2
Sheet3
Effect of Support-based PruningInvestigate how support-based
pruning affects other measures
Steps:Generate 10000 contingency tablesRank each table
according to the different measuresCompute the pair-wise
correlation between the measures
Effect of Support-based Pruning Without Support Pruning (All
Pairs) Red cells indicate correlation between

the pair of measures > 0.85 40.14% pairs have correlation >
0.85
Scatter Plot between Correlation & Jaccard Measure
Effect of Support-
pairs have correlation > 0.85
Scatter Plot between Correlation & Jaccard Measure:
Effect of Support-
pairs have correlation > 0.85
Scatter Plot between Correlation & Jaccard Measure
Subjective Interestingness MeasureObjective measure: Rank
patterns based on statistics computed from datae.g., 21
measures of association (support, confidence, Laplace, Gini,
mutual information, Jaccard, etc).
Subjective measure:Rank patterns according to user’s
interpretation A pattern is subjectively interesting if it
contradicts the
expectation of a user (Silberschatz & Tuzhilin) A pattern is
subjectively interesting if it is actionable
(Silberschatz & Tuzhilin)

Interestingness via UnexpectednessNeed to model expectation
of users (domain knowledge)
Need to combine expectation of users with evidence from data
(i.e., extracted patterns)
+
Pattern expected to be frequent
-
Pattern expected to be infrequent
Pattern found to be frequent
Pattern found to be infrequent
+
-
Expected Patterns
-
+
Unexpected Patterns
Interestingness via UnexpectednessWeb Data (Cooley et al
2001)Domain knowledge in the form of site structureGiven an
itemset F = {X1, X2, …, Xk} (Xi : Web pages) L: number of

-1) cfactor = 1
(if graph is connected), 0 (disconnected graph)Structure
Usage evidence
Use Dempster-Shafer theory to combine domain knowledge and
evidence from data
TID Items
1 Bread, Milk
2 Bread, Diaper, Beer, Eggs
3 Milk, Diaper, Beer, Coke
4 Bread, Milk, Diaper, Beer
5 Bread, Milk, Diaper, Coke
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TID Items
1 Bread, Milk

N
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k
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1 2 3 5 6
Transaction, t
2 3 5 613 5 62
5 61 33 5 61 261 55 62 362 5
5 63
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1 3 5
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Level 1
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{B}5
{C}3
{D}4
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{A,D}3
{B,C}3

{B,D}4
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ItemsetSupport
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{A,B,D}3
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I
Data Mining
Classification: Basic Concepts, Decision Trees, and Model
Evaluation
Lecture Notes for Chapter 4
by

4/18/2004 *
(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR
2002
2002
Classification: DefinitionGiven a collection of records (training
set )Each record contains a set of attributes, 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.
2002
2002
Illustrating Classification Task
2002

�
Learning algorithm�
�
�
Induction�
Deduction�
Test Set�
Model�
Training Set�
Examples of Classification TaskPredicting 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
2002
Classification TechniquesDecision Tree based MethodsRule-
based MethodsMemory based reasoningNeural NetworksNaïve
Bayes and Bayesian Belief NetworksSupport Vector Machines
2002

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
2002
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
2002
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
NoMarried
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
Decision Tree Classification Task
Decision Tree
2002
�
Tree Induction algorithm�
�
�
Induction�
Deduction�
Test Set�
Model�
Training Set�

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
2002
Refund
Marital
Status
Taxable
Income

Cheat
No
Married
80K
?
10
Test Data
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K

2002
Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
2002

Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
2002

Refund
Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
2002
Refund

Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single, Divorced
< 80K
> 80K
Test Data
Assign Cheat to “No”
2002
Refund

Marital
Status
Taxable
Income
Cheat
No
Married
80K
?
10
Decision Tree Classification Task
Decision Tree
2002
�
Tree Induction algorithm�
�
�
Induction�
Deduction�
Test Set�
Model�
Training Set�
Decision Tree InductionMany Algorithms:Hunt’s Algorithm
(one of the earliest)CARTID3, C4.5SLIQ,SPRINT

2002
General Structure of Hunt’s AlgorithmLet Dt be the set of
training records that reach a node tGeneral Procedure:If Dt
contains records that belong the same class yt, then t is a leaf
node labeled as ytIf Dt is an empty set, then t is a leaf node
labeled by the default class, ydIf 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
?
2002
Tid
Refund
Marital
Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
NoMarried
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
Hunt’s Algorithm
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
2002
Tid
Refund
Marital

Status
Taxable
Income
Cheat
1
Yes
Single
125K
No
2
NoMarried
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 InductionGreedy strategy.Split the records based on an
attribute test that optimizes certain criterion.
IssuesDetermine how to split the recordsHow to specify the
attribute test condition?How to determine the best
split?Determine when to stop splitting
2002
2002
How to Specify Test Condition?Depends on attribute
typesNominalOrdinalContinuous
Depends on number of ways to split2-way splitMulti-way split
2002
Splitting Based on Nominal AttributesMulti-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}
2002
Splitting Based on Ordinal AttributesMulti-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?
OR
Size

Small
Medium
Large
Size
{Medium,
Large}
{Small}
Size
{Small, Medium}
{Large}
Size
{Small, Large}
{Medium}
2002
Splitting Based on Continuous AttributesDifferent ways of
handlingDiscretization 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.
and finds the best cut can be more compute intensive

2002
Splitting Based on Continuous Attributes
2002
Taxable Income�> 80K?�
< 10K�
[10K,25K)�
Yes�
No�
[25K,50K)�
Taxable Income?�
[50K,80K)�
> 80K�
(i) Binary split�
(ii) Multi-way split�
2002
How to determine the Best Split
Before Splitting: 10 records of class 0,

10 records of class 1
Which test condition is the best?
2002
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�
How to determine the Best SplitGreedy approach: Nodes with
homogeneous class distribution are preferredNeed a measure of
node impurity:
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity
2002
C0: 5
C1: 5�
C0: 9
C1: 1�
Measures of Node ImpurityGini Index
Entropy
Misclassification error
2002
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
2002
C0
N10C1
N11
C0
N20C1
N21
C0
N30C1
N31
C0
N40C1
N41
C0

N00C1
N01
Measure of Impurity: GINIGini 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 informationMinimum
(0.0) when all records belong to one class, implying most
interesting information
2002
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
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
2002
C1
0C2
6
C1
2C2
4
C1
1C2
5

Splitting Based on GINIUsed 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.
2002
Binary Attributes: Computing GINI Index
Splits into two partitionsEffect 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
2002
ParentC1
6C2
6
Gini = 0.500
N1
N2C1
5
1C2
2
4Gini=0.333
Categorical Attributes: Computing Gini IndexFor each distinct
value, gather counts for each class in the datasetUse the count
matrix to make decisions
Multi-way split
Two-way split
(find best partition of values)
2002
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
Continuous Attributes: Computing Gini IndexUse Binary
Decisions based on one valueSeveral Choices for the splitting
valueNumber of possible splitting values
= Number of distinct valuesEach splitting value has a count
matrix associated with itClass counts in each of the partitions,
the database to gather count matrix and compute its Gini
indexComputationally Inefficient! Repetition of work.
2002
Tid
Refund
Marital
Status
Taxable

Income
Cheat
1
Yes
Single
125K
No
2
NoMarried
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�
Continuous Attributes: Computing Gini Index...For efficient
computation: for each attribute,Sort the attribute on
valuesLinearly scan these values, each time updating the count
matrix and computing gini indexChoose the split position that
has the least gini index
Split Positions
Sorted Values

2002
Alternative Splitting Criteria based on INFOEntropy 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
informationMinimum (0.0) when all records belong to one class,
implying most informationEntropy based computations are
similar to the GINI index computations
2002
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
2002
C1
0C2
6
C1
2C2
4
C1

1C2
5
Splitting Based on INFO...Information Gain:
Parent Node, p is split into k partitions;
ni is number of records in partition iMeasures Reduction in
Entropy achieved because of the split. Choose the split that
achieves most reduction (maximizes GAIN)Used in ID3 and
C4.5Disadvantage: Tends to prefer splits that result in large
number of partitions, each being small but pure.
2002
Splitting Based on INFO...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 C4.5Designed to overcome the
disadvantage of Information Gain
2002

Splitting Criteria based on Classification ErrorClassification
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 informationMinimum (0.0) when all
records belong to one class, implying most interesting
information
2002
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
2002
C1
0C2
6
C1
2C2
4
C1
1C2

5
Comparison among Splitting Criteria
For a 2-class problem:
2002
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 !!

2002
ParentC1
7C2
3
Gini = 0.42
N1
N2C1
3
4C2
0
3Gini=0.361
2002
Stopping Criteria for Tree InductionStop 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)
2002

Decision Tree Based ClassificationAdvantages:Inexpensive to
constructExtremely fast at classifying unknown recordsEasy to
interpret for small-sized treesAccuracy is comparable to other
classification techniques for many simple data sets
2002
Example: C4.5Simple depth-first construction.Uses Information
GainSorts 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
2002
Practical Issues of ClassificationUnderfitting and Overfitting
Missing Values
Costs of Classification
2002
Underfitting and Overfitting (Example)
500 circular and 500 triangular data points.
Circular points:
Triangular points:

sqrt(x12+x22) > 0.5 or
sqrt(x12+x22) < 1
2002
Underfitting and Overfitting
Overfitting
Underfitting: when model is too simple, both training and test
errors are large
2002
Overfitting due to Noise
Decision boundary is distorted by noise point
2002
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
2002
Notes on OverfittingOverfitting 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
2002
Estimating Generalization ErrorsRe-substitution errors: error on
Methods for estimating generalization errors:Optimistic
approach: e’(t) = e(t)Pessimistic approach: For each leaf
(N: number of leaf nodes) For a tree with 30 leaf nodes and 10
errors on training
(out of 1000 instances):
Training error = 10/1000 = 1%
Generalization error =
2.5%Reduced error pruning (REP): uses validation data set to
estimate generalization
error
2002
Occam’s RazorGiven 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

2002
Minimum Description Length (MDL)Cost(Model,Data) =
Cost(Data|Model) + Cost(Model)Cost is the number of bits
needed for encoding.Search for the least costly
model.Cost(Data|Model) encodes the misclassification
errors.Cost(Model) uses node encoding (number of children)
plus splitting condition encoding.
2002
Sheet1XyXy1?0?0?1?…………1?
Sheet2
Sheet3
Sheet1XyXy????…………?
Sheet2
Sheet3
How to Address OverfittingPre-Pruning (Early Stopping
Rule)Stop the algorithm before it becomes a fully-grown
treeTypical stopping conditions for a node: Stop if all instances
belong to the same class Stop if all the attribute values are the
sameMore restrictive conditions: Stop if number of instances is
less than some user-specified threshold Stop if class distribution
of instances are independent of the available features (e.g.,
improve impurity
measures (e.g., Gini or information gain).

2002
How to Address Overfitting…Post-pruningGrow decision tree to
its entiretyTrim the nodes of the decision tree in a bottom-up
fashionIf generalization error improves after trimming, replace
sub-tree by a leaf node.Class label of leaf node is determined
from majority class of instances in the sub-treeCan use MDL for
post-pruning
2002
Example of Post-Pruning
Training Error (Before splitting) = 10/30
Pessimistic error = (10 + 0.5)/30 = 10.5/30
Training Error (After splitting) = 9/30
Pessimistic error (After splitting)
PRUNE!Class = Yes20Class = No10Error = 10/30Class =
Yes8Class = No4Class = Yes3Class = No4Class = Yes4Class =
No1Class = Yes5Class = No1

2002
Examples of Post-pruningOptimistic error?
Pessimistic error?
Reduced error pruning?
Don’t prune for both cases
Don’t prune case 1, prune case 2
Case 1:
Case 2:
Depends on validation set

C0: 11
C1: 3
C0: 2
C1: 4
C0: 14
C1: 3
C0: 2
C1: 2
2002
Handling Missing Attribute ValuesMissing values affect
decision tree construction in three different ways:Affects how
impurity measures are computedAffects how to distribute
instance with missing value to child nodesAffects how a test
instance with missing value is classified
2002
Computing Impurity Measure
Split on Refund:
Entropy(Refund=Yes) = 0
Entropy(Refund=No)
= -(2/6)log(2/6) – (4/6)log(4/6) = 0.9183
Entropy(Children)

= 0.3 (0) + 0.6 (0.9183) = 0.551
– 0.551) = 0.3303
Missing value
Before Splitting:
Entropy(Parent)
= -0.3 log(0.3)-(0.7)log(0.7) = 0.8813
2002
Tid
Refund
Marital
Status
Taxable
Income
Class
1
Yes
Single
125K
No
2
NoMarried
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
?
Single
90K
Yes
10

Class
= Yes
Class
= NoRefund=Yes
0
3Refund=No
2
4Refund=?
1
0
Distribute Instances
Refund
Yes
No
Refund
Yes
No
Probability that Refund=Yes is 3/9
Probability that Refund=No is 6/9
Assign record to the left child with weight = 3/9 and to the right
child with weight = 6/9
2002
Tid
Refund
Marital
Status
Taxable

Income
Class
1
Yes
Single
125K
No
2
NoMarried
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
Class=Yes
0Class=No
3
Cheat=Yes
2Cheat=No
4
Tid
Refund
Marital
Status
Taxable
Income
Class
10
?
Single
90K
Yes
10

Class=Yes
2 + 6/9Class=No
4
Class=Yes
0 + 3/9Class=No
3
Classify Instances
Refund
MarSt
TaxInc
YES
NO
NO
NO
Yes
No
Married
Single,
Divorced
< 80K
> 80K
New record:
Probability that Marital Status
= Married is 3.67/6.67
Probability that Marital Status ={Single,Divorced} is
3/6.67MarriedSingleDivorcedTotalClass=No3104Class=Yes6/91
12.67Total3.67216.67

2002
Tid
Refund
Marital
Status
Taxable
Income
Class
11
No
?
85K
?
10
Other IssuesData FragmentationSearch
StrategyExpressivenessTree Replication
2002

Data FragmentationNumber of instances gets smaller as you
traverse down the tree
Number of instances at the leaf nodes could be too small to
make any statistically significant decision
2002
Search StrategyFinding an optimal decision tree is NP-hard
The algorithm presented so far uses a greedy, top-down,
recursive partitioning strategy to induce a reasonable solution
Other strategies?Bottom-upBi-directional
2002
ExpressivenessDecision tree provides expressive representation
for learning discrete-valued functionBut they do not generalize
well to certain types of Boolean functions Example: parity
function:
Class = 1 if there is an even number of Boolean attributes with
truth value = True
Class = 0 if there is an odd number of Boolean attributes with
truth value = True For accurate modeling, must have a complete
tree
Not expressive enough for modeling continuous
variablesParticularly when test condition involves only a single
attribute at-a-time
2002

Decision Boundary Border line between two neighboring
regions of different classes is known as decision boundary
Decision boundary is parallel to axes because test condition
involves a single attribute at-a-time
2002
y < 0.33?�
: 0
: 3�
: 4
: 0�
y < 0.47?�
: 4
: 0�
: 0
: 4�
x < 0.43?�
Yes�
Yes�
No�
No�
Yes�
No�
0�
0.1�
0.2�
0.3�
0.4�
0.5�
0.6�
0.7�
0.8�
0.9�
1�

0�
0.1�
0.2�
0.3�
0.4�
0.5�
0.6�
0.7�
0.8�
0.9�
1�
x�
y�
Oblique Decision Trees Test condition may involve multiple
attributes More expressive representation Finding optimal test
condition is computationally expensive
x + y < 1
Class = +
Class =
2002
Tree Replication Same subtree appears in multiple branches
2002

Model EvaluationMetrics for Performance EvaluationHow to
evaluate the performance of a model?
Methods for Performance EvaluationHow to obtain reliable
estimates?
Methods for Model ComparisonHow to compare the relative
performance among competing models?
2002
estimates?
2002
Metrics for Performance EvaluationFocus on the predictive
capability of a modelRather 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
CLASSClass=YesClass=NoClass=YesabClass=Nocd

2002
Metrics for Performance Evaluation…
Most widely-used metric:PREDICTED CLASS
ACTUAL
CLASSClass=YesClass=NoClass=Yesa
(TP)b
(FN)Class=Noc
(FP)d
(TN)

2002
Limitation of AccuracyConsider a 2-class problemNumber of
Class 0 examples = 9990Number 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
2002
Cost Matrix
C(i|j): Cost of misclassifying class j example as class i
PREDICTED CLASS
ACTUAL
CLASSC(i|j)Class=YesClass=NoClass=YesC(Yes|Yes)C(No|Yes
)Class=NoC(Yes|No)C(No|No)
2002

Computing Cost of Classification
Accuracy = 80%
Cost = 3910
Accuracy = 90%
Cost = 4255Cost MatrixPREDICTED CLASS
ACTUAL
CLASSC(i|j)+-+-1100-10Model M1PREDICTED CLASS
ACTUAL
CLASS+-+15040-60250Model M2PREDICTED CLASS
ACTUAL
CLASS+-+25045-5200

2002
Cost vs AccuracyCountPREDICTED CLASS
ACTUAL
CLASSClass=YesClass=NoClass=YesabClass=NocdCostPREDI
CTED CLASS
ACTUAL
CLASSClass=YesClass=NoClass=YespqClass=Noqp

estimates?
2002
Methods for Performance EvaluationHow to obtain a reliable
estimate of performance?
Performance of a model may depend on other factors besides the
learning algorithm:Class distributionCost of
misclassificationSize of training and test sets
2002
Learning CurveLearning curve shows how accuracy changes
with varying sample sizeRequires 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 estimateVariance of
estimate
2002
Methods of EstimationHoldoutReserve 2/3 for training and 1/3

for testing Random subsamplingRepeated holdoutCross
validationPartition data into k disjoint subsetsk-fold: train on k-
1 partitions, test on the remaining oneLeave-one-out:
k=nStratified sampling oversampling vs
undersamplingBootstrapSampling with replacement
2002
estimates?
2002
ROC (Receiver Operating Characteristic)Developed in 1950s for
signal detection theory to analyze noisy signals Characterize the
trade-off between positive hits and false alarmsROC curve plots
TP (on the y-axis) against FP (on the x-axis)Performance of
each classifier represented as a point on the ROC curvechanging
the threshold of algorithm, sample distribution or cost matrix
changes the location of the point
2002
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
2002
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 guessingBelow diagonal line: prediction
is opposite of the true class
2002
Using ROC for Model ComparisonNo model consistently
outperform the otherM1 is better for small FPRM2 is better for
large FPR
Area Under the ROC curveIdeal: Area = 1Random guess: Area
= 0.5
2002
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)InstanceP(+|A)True
Class10.95+20.93+30.87-40.85-50.85-60.85+70.76-
80.53+90.43-100.25+
2002
How to construct an ROC curve
Threshold >=
ROC Curve:
2002
Class
+

-
+
-
-
-
+
-
+
+P
0.25
0.43
0.53
0.76
0.85
0.85
0.85
0.87
0.93

0.95
1.00TP
5
4
4
3
3
3
3
2
2
1
0
FP
5
5
4
4
3
2
1
1
0
0
0TN
0
0
1
1
2
3
4
4
5
5

5
FN
0
1
1
2
2
2
2
3
3
4
5TPR
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

Test of SignificanceGiven two models:Model M1: accuracy =
85%, tested on 30 instancesModel 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?
2002
Confidence Interval for AccuracyPrediction can be regarded as
a Bernoulli trialA Bernoulli trial has 2 possible
outcomesPossible outcomes for prediction: correct or
wrongCollection of Bernoulli trials has a Binomial distribution:
ictions e.g: Toss a
fair coin 50 times, how many heads would turn up?
Given x (# of correct predictions) or equivalently, acc=x/N, and
N (# of test instances),
Can we predict p (true accuracy of model)?
2002
Confidence Interval for AccuracyFor 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 -
Z1-
2002
Confidence Interval for AccuracyConsider a model that
produces an accuracy of 80% when evaluated on 100 test
instances:N=100, acc = 0.8Let 1-
confidence)From probability table, Z -
wer)0.6700.7110.7630.7740.789p(upper)0.8880.8660.8330.8240
.811

2002
Comparing Performance of 2 ModelsGiven two models, say M1
and M2, which is better?M1 is tested on D1 (size=n1), found
error rate = e1M2 is tested on D2 (size=n2), found error rate =
e2Assume D1 and D2 are independentIf n1 and n2 are
sufficiently large, then
Approximate:
2002
Comparing Performance of 2 ModelsTo test if performance
difference is statistically significant: d = e1 –
where dt is the true differenceSince D1 and D2 are independent,
their variance adds up:

At (1-
2002
An Illustrative ExampleGiven: M1: n1 = 30, e1 = 0.15
M2: n2 = 5000, e2 = 0.25d = |e2 – e1| = 0.1 (2-sided test)
=> Interval contains 0 => difference may not be
statistically significant
2002
Comparing Performance of 2 AlgorithmsEach learning
algorithm may produce k models:L1 may produce M11 , M12,
…, M1kL2 may produce M21 , M22, …, M2kIf models are
generated on the same test sets D1,D2, …, Dk (e.g., via cross-
validation)For each set: compute dj = e1j – e2jdj has mean dt
2002

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
C0: 9
C1: 1
C0: 5
C1: 5
)
|
(
max
1
)
(
t
i

P
t
Error
i
-
=
N1
N2
C1
5
1
C2
2
4
Gini=0.3
33
Parent
C1
7
C2

3
Gini = 0.
42
N1
N2
C1
3
4
C2
0
3
Gini=0.361
å
=
-
=
k
i
i
i

n
n
n
n
SplitINFO
1
log
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
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
å
=
=
k
i
i
split
i
GINI
n
n
GINI
1
)
(

SplitINFO
GAIN
GainRATIO
Split
split
=
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
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
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid
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
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
Taxable
Income
> 80K?

YesNo
Taxable
Income?
(i) Binary split(ii) Multi-way split
< 10K
[10K,25K)[25K,50K)[50K,80K)
> 80K
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid
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
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
CarType
{Sports,
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{Family}
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3
1
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2
4
Gini
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C1
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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
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87
92
97
110
122
172
230
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3
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3
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3
0
3
0
3
0
3
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0
7
1
6
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5
3
4
3
4
3
4
3
4
4
3
5
2
6
1
7
0
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0.420
0.400
0.375
0.343
0.417
0.400
0.300
0.343
0.375
0.400
0.420
å
-
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10
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Income
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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
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?
...
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No
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1
c
10
c
20
C0: 0
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...
c
11
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C1 N31
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C1
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Data Mining
Cluster Analysis: Basic Concepts
and Algorithms
Lecture Notes for Chapter 8
by
4/18/2004 *

2002
2002
What is Cluster Analysis?Finding groups of objects such that
the objects in a group will be similar (or related) to one another
and different from (or unrelated to) the objects in other groups

Inter-cluster distances are maximized
Intra-cluster distances are minimized
2002
Applications of Cluster AnalysisUnderstandingGroup related
documents for browsing, group genes and proteins that have
similar functionality, or group stocks with similar price
fluctuations
SummarizationReduce the size of large data sets
Clustering precipitation in Australia
2002
Discovered Clusters
Industry Group
1
Applied-Matl-DOWN,Bay-Network-Down,3-COM-DOWN,
Cabletron-Sys-DOWN,CISCO-DOWN,HP-DOWN,
DSC-Comm-DOWN,INTEL-DOWN,LSI-Logic-DOWN,
Micron-Tech-DOWN,Texas-Inst-Down,Tellabs-Inc-Down,
Natl-Semiconduct-DOWN,Oracl-DOWN,SGI-DOWN,

Sun-DOWNTechnology1-DOWN
2
Apple-Comp-DOWN,Autodesk-DOWN,DEC-DOWN,
ADV-Micro-Device-DOWN,Andrew-Corp-DOWN,
Computer-Assoc-DOWN,Circuit-City-DOWN,
Compaq-DOWN, EMC-Corp-DOWN, Gen-Inst-DOWN,
Motorola-DOWN,Microsoft-DOWN,Scientific-Atl-DOWN
Technology2-DOWN
3
Fannie-Mae-DOWN,Fed-Home-Loan-DOWN,
MBNA-Corp-DOWN,Morgan-Stanley-DOWN
Financial-DOWN
4
Baker-Hughes-UP,Dresser-Inds-UP,Halliburton-HLD-UP,
Louisiana-Land-UP,Phillips-Petro-UP,Unocal-UP,
Schlumberger-UP
Oil-UP
What is not Cluster Analysis?Supervised classificationHave
class label information
Simple segmentationDividing students into different registration
groups alphabetically, by last name
Results of a queryGroupings are a result of an external
specification
Graph partitioningSome mutual relevance and synergy, but
areas are not identical

2002
Notion of a Cluster can be Ambiguous
How many clusters?

Six Clusters
2002
Types of ClusteringsA clustering is a set of clusters
Important distinction between hierarchical and partitional sets
of clusters
Partitional ClusteringA division data objects into non-
overlapping subsets (clusters) such that each data object is in
exactly one subset
Hierarchical clusteringA set of nested clusters organized as a
hierarchical tree
2002

Partitional Clustering
Original Points
A Partitional Clustering
2002
Hierarchical Clustering
Traditional Hierarchical Clustering
Non-traditional Hierarchical Clustering
Non-traditional Dendrogram
Traditional Dendrogram
2002
Other Distinctions Between Sets of ClustersExclusive versus
non-exclusiveIn non-exclusive clusterings, points may belong to
multiple clusters.Can represent multiple classes or ‘border’
pointsFuzzy versus non-fuzzyIn fuzzy clustering, a point
belongs to every cluster with some weight between 0 and
1Weights must sum to 1Probabilistic clustering has similar
characteristicsPartial versus completeIn some cases, we only
want to cluster some of the dataHeterogeneous versus
homogeneousCluster of widely different sizes, shapes, and
densities
2002

Types of Clusters Well-separated clusters
Center-based clusters
Contiguous clusters
Density-based clusters
Property or Conceptual
Described by an Objective Function
2002
Types of Clusters: Well-SeparatedWell-Separated Clusters: A
cluster is a set of points such that any point in a cluster is closer
(or more similar) to every other point in the cluster than to any
point not in the cluster.
3 well-separated clusters
2002
Types of Clusters: Center-BasedCenter-based A cluster is a set
of objects such that an object in a cluster is closer (more
similar) to the “center” of a cluster, than to the center of any
other cluster The center of a cluster is often a centroid, the
average of all the points in the cluster, or a medoid, the most
“representative” point of a cluster
4 center-based clusters
2002

Types of Clusters: Contiguity-BasedContiguous Cluster
(Nearest neighbor or Transitive)A cluster is a set of points such
that a point in a cluster is closer (or more similar) to one or
more other points in the cluster than to any point not in the
cluster.
8 contiguous clusters
2002
Types of Clusters: Density-BasedDensity-basedA cluster is a
dense region of points, which is separated by low-density
regions, from other regions of high density. Used when the
clusters are irregular or intertwined, and when noise and
outliers are present.
6 density-based clusters

2002
Types of Clusters: Conceptual ClustersShared Property or
Conceptual ClustersFinds clusters that share some common
property or represent a particular concept.
.
2 Overlapping Circles
2002
Types of Clusters: Objective FunctionClusters Defined by an
Objective FunctionFinds clusters that minimize or maximize an
objective function. Enumerate all possible ways of dividing the
points into clusters and evaluate the `goodness' of each
potential set of clusters by using the given objective function.
(NP Hard) Can have global or local objectives. Hierarchical
clustering algorithms typically have local objectives Partitional
algorithms typically have global objectivesA variation of the
global objective function approach is to fit the data to a
parameterized model. Parameters for the model are determined
from the data. Mixture models assume that the data is a
‘mixture' of a number of statistical distributions.
2002
Types of Clusters: Objective Function …Map the clustering
problem to a different domain and solve a related problem in
that domainProximity matrix defines a weighted graph, where
the nodes are the points being clustered, and the weighted edges

CSIS 100CSIS 100 - Discussion Board Topic #1One of the object.docx

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