Discuss association analysis and the advanced concepts (in Chapter six).? After reviewing the PPT material ?and also research on other resources and answer the following questions:
- What are the techniques in handling categorical attributes?
- How do continuous attributes differ from categorical attributes?
- What is a concept hierarchy?
- Note the major patterns of data and how they work.
Data Mining
Association Rules: Advanced Concepts and Algorithms
Lecture Notes for Chapter 6
Introduction to Data Mining
by
Tan, Steinbach, Kumar
? Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 *
Continuous and Categorical Attributes
Example of Association Rule:
{Number of Pages ?[5,10) ? (Browser=Mozilla)} ? {Buy = No}
How to apply association analysis formulation to non-asymmetric binary variables?
|
Session Id |
Country |
Session Length (sec) |
Number of Web Pages viewed |
Gender |
Browser Type |
Buy |
|
1 |
USA |
982 |
8 |
Male |
IE |
No |
|
2 |
China |
811 |
10 |
Female |
Netscape |
No |
|
3 |
USA |
2125 |
45 |
Female |
Mozilla |
Yes |
|
4 |
Germany |
596 |
4 |
Male |
IE |
Yes |
|
5 |
Australia |
123 |
9 |
Male |
Mozilla |
No |
|
? |
? |
? |
? |
?
|
?
|
?
|
10
Handling Categorical Attributes
- Transform categorical attribute into asymmetric binary variables
- Introduce a new ?item? for each distinct attribute-value pair
- Example: replace Browser Type attribute with
- Browser Type = Internet Explorer
- Browser Type = Mozilla
- Browser Type = Mozilla
Handling Categorical Attributes
- Potential Issues
- What if attribute has many possible values
- Example: attribute country has more than 200 possible values
- Many of the attribute values may have very low support
Potential solution: Aggregate the low-support attribute values
- What if distribution of attribute values is highly skewed
- Example: 95% of the visitors have Buy = No
- Most of the items will be associated with (Buy=No) item
Potential solution: drop the highly frequent items
Handling Continuous Attributes
- Different kinds of rules:
- Age?[21,35) ? Salary?[70k,120k) ? Buy
- Salary?[70k,120k) ? Buy ? Age: ?=28, ?=4
- Different methods:
- Discretization-based
- Statistics-based
- Non-discretization based
- minApriori
Discretization Issues
- Size of the discretized intervals affect support & confidence
- If intervals too small
- may not have enough support
- If intervals too large
- may not have enough confidence
- Potential solution: use all possible intervals
{Refund = No, (Income = $51,250)} ? {Cheat = No}
{Refund = No, (60K ? Income ? 80K)} ? {Cheat = No}
{Refund = No, (0K ? Income ? 1B)} ? {Cheat = No}
Approach by Srikant & Agrawal
- Preprocess the data
- Discretize attribute using equi-depth partitioning
- Use partial completeness measure to determine number of partitions
- Merge adjacent intervals as long as support is less than max-support
- Apply existing association rule mining algorithms
- Determine interesting rules in the output
Approach by Srikant & Agrawal
- Discretization will lose information
- Use partial completeness measure to determine how much information is lost
C: frequent itemsets obtained by considering all ranges of attribute values
P: frequent itemsets obtained by considering all ranges over the partitions
P is K-complete w.r.t C if P ? C,and ?X ? C, ? X? ? P such that:
1. X? is a generalization of X and support (X?) ? K ? support(X) (K ? 1)
2. ?Y ? X, ? Y? ? X? such that support (Y?) ? K ? support(Y)
Given K (partial completeness level), can determine number of intervals (N)
X
Approximated X
Statistics-based Methods
- Example:
Browser=Mozilla ? Buy=Yes ? Age: ?=23
- Rule consequent consists of a continuous variable, characterized by their statistics
- mean, median, standard deviation, etc.
- Approach:
- Withhold the target variable from the rest of the data
- Apply existing frequent itemset generation on the rest of the data
- For each frequent itemset, compute the descriptive statistics for the corresponding target variable
- Frequent itemset becomes a rule by introducing the target variable as rule consequent
- Apply statistical test to determine interestingness of the rule
Statistics-based Methods
- How to determine whether an association rule interesting?
- Compare the statistics for segment of population covered by the rule vs segment of population not covered by the rule:
A ? B: ? versus A ? B: ??
- Statistical hypothesis testing:
- Null hypothesis: H0: ?? = ? + ?
- Alternative hypothesis: H1: ?? > ? + ?
- Z has zero mean and variance 1 under null hypothesis
Min-Apriori (Han et al)
Example:
W1 and W2 tends to appear together in the same document
Document-term matrix:
Sheet1
| TID | W1 | W2 | W3 | W4 | W5 | TID | W1 | W2 | W3 | W4 | W5 | ||
| D1 | 2 | 2 | 0 | 0 | 1 | D1 | 0.4 | 0.4 | 0.0 | 0.0 | 0.2 | ||
| D2 | 0 | 0 | 1 | 2 | 2 | D2 | 0.0 | 0.0 | 0.2 | 0.4 | 0.4 | ||
| D3 | 2 | 3 | 0 | 0 | 0 | D3 | 0.4 | 0.6 | 0.0 | 0.0 | 0.0 | ||
| D4 | 0 | 0 | 1 | 0 | 1 | D4 | 0.0 | 0.0 | 0.5 | 0.0 | 0.5 | ||
| D5 | 1 | 1 | 1 | 0 | 2 | D5 | 0.2 | 0.2 | 0.2 | 0.0 | 0.4 |
Sheet2
Sheet3
Multi-level Association Rules
- Why should we incorporate concept hierarchy?
- Rules at lower levels may not have enough support to appear in any frequent itemsets
- Rules at lower levels of the hierarchy are overly specific
- e.g., skim milk ? white bread, 2% milk ? wheat bread,
skim milk ? wheat bread, etc.
are indicative of association between milk and bread
Multi-level Association Rules
- How do support and confidence vary as we traverse the concept hierarchy?
- If X is the parent item for both X1 and X2, then
?(X) = ?(X1) + ?(X2)
- If ?(X1 ? Y1) = minsup,
and X is parent of X1, Y is parent of Y1
then ?(X ? Y1) = minsup, ?(X1 ? Y) = minsup
?(X ? Y) = minsup
- If conf(X1 ? Y1) = minconf,
then conf(X1 ? Y) = minconf
Multi-level Association Rules
- Approach 1:
- Extend current association rule formulation by augmenting each transaction with higher level items
Original Transaction: {skim milk, wheat bread}
Augmented Transaction:
{skim milk, wheat bread, milk, bread, food}
- Issues:
- Items that reside at higher levels have much higher support counts
- if support threshold is low, too many frequent patterns involving items from the higher levels
- Increased dimensionality of the data
Multi-level Association Rules
- Approach 2:
- Generate frequent patterns at highest level first
- Then, generate frequent patterns at the next highest level, and so on
- Issues:
- I/O requirements will increase dramatically because we need to perform more passes over the data
- May miss some potentially interesting cross-level association patterns
Formal Definition of a Sequence
- A sequence is an ordered list of elements (transactions)
s = < e1 e2 e3 ? >
- Each element contains a collection of events (items)
ei = {i1, i2, ?, ik}
- Each element is attributed to a specific time or location
- Length of a sequence, |s|, is given by the number of elements of the sequence
- A k-sequence is a sequence that contains k events (items)
Examples of Sequence
- Web sequence:
< {Homepage} {Electronics} {Digital Cameras} {Canon Digital Camera} {Shopping Cart} {Order Confirmation} {Return to Shopping} >
- Sequence of initiating events causing the nuclear accident at 3-mile Island:
(http://stellar-one.com/nuclear/staff_reports/summary_SOE_the_initiating_event.htm)
< {clogged resin} {outlet valve closure} {loss of feedwater}
{condenser polisher outlet valve shut} {booster pumps trip}
{main waterpump trips} {main turbine trips} {reactor pressure increases}>
- Sequence of books checked out at a library:
<{Fellowship of the Ring} {The Two Towers} {Return of the King}>
Sequential Pattern Mining: Definition
- Given:
- a database of sequences
- a user-specified minimum support threshold, minsup
- Task:
- Find all subsequences with support = minsup
Extracting Sequential Patterns
- Given n events: i1, i2, i3, ?, in
- Candidate 1-subsequences:
<{i1}>, <{i2}>, <{i3}>, ?, <{in}>
- Candidate 2-subsequences:
<{i1, i2}>, <{i1, i3}>, ?, <{i1} {i1}>, <{i1} {i2}>, ?, <{in-1} {in}>
- Candidate 3-subsequences:
<{i1, i2 , i3}>, <{i1, i2 , i4}>, ?, <{i1, i2} {i1}>, <{i1, i2} {i2}>, ?,
<{i1} {i1 , i2}>, <{i1} {i1 , i3}>, ?, <{i1} {i1} {i1}>, <{i1} {i1} {i2}>, ?
Generalized Sequential Pattern (GSP)
- Step 1:
- Make the first pass over the sequence database D to yield all the 1-element frequent sequences
- Step 2:
Repeat until no new frequent sequences are found
- Candidate Generation:
- Merge pairs of frequent subsequences found in the (k-1)th pass to generate candidate sequences that contain k items
- Candidate Pruning:
- Prune candidate k-sequences that contain infrequent (k-1)-subsequences
- Support Counting:
- Make a new pass over the sequence database D to find the support for these candidate sequences
- Candidate Elimination:
- Eliminate candidate k-sequences whose actual support is less than minsup
Candidate Generation
- Base case (k=2):
- Merging two frequent 1-sequences <{i1}> and <{i2}> will produce two candidate 2-sequences: <{i1} {i2}> and <{i1 i2}>
- General case (k>2):
- A frequent (k-1)-sequence w1 is merged with another frequent
(k-1)-sequence w2 to produce a candidate k-sequence if the subsequence obtained by removing the first event in w1 is the same as the subsequence obtained by removing the last event in w2 - The resulting candidate after merging is given by the sequence w1 extended with the last event of w2.
If the last two events in w2 belong to the same element, then the last event in w2 becomes part of the last element in w1
Otherwise, the last event in w2 becomes a separate element appended to the end of w1
Candidate Generation Examples
- Merging the sequences
w1=<{1} {2 3} {4}> and w2 =<{2 3} {4 5}>
will produce the candidate sequence < {1} {2 3} {4 5}> because the last two events in w2 (4 and 5) belong to the same element - Merging the sequences
w1=<{1} {2 3} {4}> and w2 =<{2 3} {4} {5}>
will produce the candidate sequence < {1} {2 3} {4} {5}> because the last two events in w2 (4 and 5) do not belong to the same element - We do not have to merge the sequences
w1 =<{1} {2 6} {4}> and w2 =<{1} {2} {4 5}>
to produce the candidate < {1} {2 6} {4 5}> because if the latter is a viable candidate, then it can be obtained by merging w1 with
< {1} {2 6} {5}>
Mining Sequential Patterns with Timing Constraints
- Approach 1:
- Mine sequential patterns without timing constraints
- Postprocess the discovered patterns
- Approach 2:
- Modify GSP to directly prune candidates that violate timing constraints
- Question:
- Does Apriori principle still hold?
Contiguous Subsequences
- s is a contiguous subsequence of
w = <e1>< e2>?< ek>
if any of the following conditions hold:
s is obtained from w by deleting an item from either e1 or ek
s is obtained from w by deleting an item from any element ei that contains more than 2 items
s is a contiguous subsequence of s? and s? is a contiguous subsequence of w (recursive definition)
- Examples: s = < {1} {2} >
- is a contiguous subsequence of
< {1} {2 3}>, < {1 2} {2} {3}>, and < {3 4} {1 2} {2 3} {4} > - is not a contiguous subsequence of
< {1} {3} {2}> and < {2} {1} {3} {2}>
Modified Candidate Pruning Step
- Without maxgap constraint:
- A candidate k-sequence is pruned if at least one of its (k-1)-subsequences is infrequent
- With maxgap constraint:
- A candidate k-sequence is pruned if at least one of its contiguous (k-1)-subsequences is infrequent
Modified Support Counting Step
- Given a candidate pattern: <{a, c}>
- Any data sequences that contain
<? {a c} ? >,
<? {a} ? {c}?> ( where time({c}) ? time({a}) = ws)
<?{c} ? {a} ?> (where time({a}) ? time({c}) = ws)
will contribute to the support count of candidate pattern
Challenges
- Node may contain duplicate labels
- Support and confidence
- How to define them?
- Additional constraints imposed by pattern structure
- Support and confidence are not the only constraints
- Assumption: frequent subgraphs must be connected
- Apriori-like approach:
- Use frequent k-subgraphs to generate frequent (k+1) subgraphs
- What is k?
Challenges?
- Support:
- number of graphs that contain a particular subgraph
- Apriori principle still holds
- Level-wise (Apriori-like) approach:
- Vertex growing:
- k is the number of vertices
- Edge growing:
- k is the number of edges
Apriori-like Algorithm
- Find frequent 1-subgraphs
- Repeat
- Candidate generation
- Use frequent (k-1)-subgraphs to generate candidate k-subgraph
- Candidate pruning
- Prune candidate subgraphs that contain infrequent
(k-1)-subgraphs - Support counting
- Count the support of each remaining candidate
- Eliminate candidate k-subgraphs that are infrequent
In practice, it is not as easy. There are many other issues
Candidate Generation
- In Apriori:
- Merging two frequent k-itemsets will produce a candidate (k+1)-itemset
- In frequent subgraph mining (vertex/edge growing)
- Merging two frequent k-subgraphs may produce more than one candidate (k+1)-subgraph
Graph Isomorphism
- Test for graph isomorphism is needed:
- During candidate generation step, to determine whether a candidate has been generated
- During candidate pruning step, to check whether its
(k-1)-subgraphs are frequent - During candidate counting, to check whether a candidate is contained within another graph
Session
Id
Country Session
Length
(sec)
Number of
Web Pages
viewed
Gender
Browser
Type
Buy
1 USA 982 8 Male IE No
2 China 811 10 Female Netscape No
3 USA 2125 45 Female Mozilla Yes
4 Germany 596 4 Male IE Yes
5 Australia 123 9 Male Mozilla No
? ? ? ? ? ? ?
10
2
2
2
1
2
1
‘
n
s
n
s
Z
+
D
–
–
=
m
m
TIDW1W2W3W4W5
D122001
D200122
D323000
D400101
D511102
