CPS 196.03: Information Management and Mining Association Rules and Frequent Itemsets

Apriori Algorithm Method: Let k=1 Generate frequent itemsets of length 1 Repeat until no new frequent itemsets are identified Generate length (k+1) candidate itemsets from length k frequent itemsets Prune candidate itemsets containing subsets of length k that are infrequent Count the support of each candidate by scanning the DB Eliminate candidates that are infrequent, leaving only those that are frequent

Generating Candidate ItemSets Exhaustive enumeration F(k-1) x F(1) Using (lexicographic) ordering F(k-1) x F(k-1) Will hashing help?

Reducing Number of Comparisons Candidate counting: Scan the database of transactions to determine the support of each candidate itemset To 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

Generate Hash Tree Hash function 3,6,9 1,4,7 2,5,8 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) 1,4,7 2,5,8 3,6,9 Hash function

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

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

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

Subset Operation Given a transaction t, what are the possible subsets of size 3?

Subset Operation Using Hash Tree 1,4,7 2,5,8 3,6,9 Hash Function 1 2 3 5 6 transaction 1 + 2 3 5 6 3 5 6 2 + 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 5 6 3 +

Subset Operation Using Hash Tree 1,4,7 2,5,8 3,6,9 Hash Function 1 2 3 5 6 transaction 1 + 2 3 5 6 3 5 6 2 + 3 5 6 1 2 + 5 6 3 + 5 6 1 3 + 2 3 4 6 1 5 + 5 6 7 1 4 5 1 3 6 3 4 5 3 5 6 3 5 7 3 6 7 3 6 8 6 8 9 1 2 4 1 2 5 1 5 9 4 5 7 4 5 8

Subset Operation Using Hash Tree 1,4,7 2,5,8 3,6,9 Hash Function 1 2 3 5 6 transaction 1 + 2 3 5 6 3 5 6 2 + 3 5 6 1 2 + 5 6 3 + 5 6 1 3 + 2 3 4 6 1 5 + 5 6 7 1 4 5 1 3 6 3 4 5 3 5 6 3 5 7 3 6 7 3 6 8 6 8 9 1 2 4 1 2 5 1 5 9 4 5 7 4 5 8 Match transaction against 11 out of 15 candidates

Factors Affecting Performance Choice of minimum support threshold lowering support threshold results in more frequent itemsets this may increase number of candidates and max length of frequent itemsets Dimensionality (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 increase Size of database since Apriori makes multiple passes, run time of algorithm may increase with number of transactions Average transaction width number of subsets in a transaction increases with its width this may increase max length of frequent itemsets and traversals of hash tree

Compact Representation of Frequent Itemsets Some itemsets are redundant because they have identical support as their supersets Number of frequent itemsets Need a compact representation

Compression of Itemset Information Transaction Ids Not supported by any transactions

Maximal Frequent Itemset An itemset is maximal frequent if none of its immediate supersets is frequent Maximal Itemsets Infrequent Itemsets Border

Closed Itemset An itemset is closed if none of its immediate supersets has the same support as the itemset

Maximal vs Closed Itemsets Transaction Ids Not supported by any transactions

Maximal vs Closed Frequent Itemsets Closed but not maximal Minimum support = 2 Closed and maximal # Closed = 9 # Maximal = 4

Maximal vs Closed Itemsets

Why Do We Care About Closed Itemsets? Compact representation of frequent itemsets Helps if there are many frequent itemsets There are efficient algorithms that can efficiently find the closed itemsets (and only them) HW Exercise: Come up with an algorithm to generate all info about frequent itemsets given info about closed frequent itemsets? Closed itemsets help identify redundant association rules E.g., if {b} and {b,c} have the same support: then would you care about {b} -> {d} or {b,c} -> {d}?

Toivonen’s Algorithm --- (1) Start as in the simple algorithm, but lower the threshold slightly for the sample. Example: if the sample is 1% of the baskets, use s/125 as the support threshold rather than s /100. Goal is to avoid missing any itemset that is frequent in the full set of baskets.

Toivonen’s Algorithm --- (2) Add to the itemsets that are frequent in the sample the negative border of these itemsets. An itemset is in the negative border if it is not deemed frequent in the sample, but all its immediate subsets are.

Example ABCD is in the negative border if and only if it is not frequent, but all of ABC, BCD, ACD, and ABD are.

Toivonen’s Algorithm --- (3) In a second pass, count all candidate frequent itemsets from the first pass, and also count the negative border. If no itemset from the negative border turns out to be frequent, then the candidates found to be frequent in the whole data are exactly the frequent itemsets.

Toivonen’s Algorithm --- (4) What if we find something in the negative border is actually frequent? We must start over again! Try to choose the support threshold so the probability of failure is low, while the number of itemsets checked on the second pass fits in main- memory.

Alternative Methods for Frequent Itemset Generation Traversal of Itemset Lattice General-to-specific vs Specific-to-general

Alternative Methods for Frequent Itemset Generation Traversal of Itemset Lattice Equivalent Classes

Alternative Methods for Frequent Itemset Generation Traversal of Itemset Lattice Breadth-first vs Depth-first

Alternative Methods for Frequent Itemset Generation Representation of Database horizontal vs vertical data layout

ECLAT For each item, store a list of transaction ids (tids) TID-list

ECLAT Determine support of any k-itemset by intersecting tid-lists of two of its (k-1) subsets. 3 traversal approaches: top-down, bottom-up and hybrid Advantage: very fast support counting Disadvantage: intermediate tid-lists may become too large for memory  

Bottleneck of Frequent-pattern Mining Multiple database scans are costly Mining long patterns needs many passes of scanning and generates lots of candidates To find frequent itemset i1i2…i100 # of scans: 100 # of Candidates: (1001) + (1002) + … + (110000) = 2100-1 = 1.27*1030 ! Bottleneck: candidate-generation-and-test Can we avoid candidate generation?

FP-growth Algorithm Use 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 After reading TID=1: A:1 B:1

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

Construct FP-tree from Transaction Database TID Items bought (ordered) frequent items 100 {f, a, c, d, g, i, m, p} {f, c, a, m, p} 200 {a, b, c, f, l, m, o} {f, c, a, b, m} 300 {b, f, h, j, o, w} {f, b} 400 {b, c, k, s, p} {c, b, p} 500 {a, f, c, e, l, p, m, n} {f, c, a, m, p} min_support = 3 {} f:4 c:1 b:1 p:1 c:3 a:3 m:2 p:2 m:1 Header Table Item frequency head f 4 c 4 a 3 b 3 m 3 p 3 Scan DB once, find frequent 1-itemset (single item pattern) Sort frequent items in frequency descending order, f-list Scan DB again, construct FP-tree F-list=f-c-a-b-m-p

Benefits of the FP-tree Structure Completeness Preserve complete information for frequent pattern mining Never break a long pattern of any transaction Compactness Reduce irrelevant info—infrequent items are gone Items in frequency descending order: the more frequently occurring, the more likely to be shared Never be larger than the original database (not count node-links and the count field)

Partition Patterns and Databases Frequent patterns can be partitioned into subsets according to f-list F-list=f-c-a-b-m-p Patterns containing p Patterns having m but no p … Patterns having c but no a nor b, m, p Pattern f Completeness and non-redundancy

Find Patterns From P-conditional Database Starting at the frequent item header table in the FP-tree Traverse the FP-tree by following the link of each frequent item p Accumulate all of transformed prefix paths of item p to form p’s conditional pattern base {} f:4 c:1 b:1 p:1 c:3 a:3 m:2 p:2 m:1 Header Table Item frequency head f 4 c 4 a 3 b 3 m 3 p 3 Conditional pattern bases item cond. pattern base c f:3 a fc:3 b fca:1, f:1, c:1 m fca:2, fcab:1 p fcam:2, cb:1

From Conditional Pattern-bases to Conditional FP-trees For each pattern-base Accumulate the count for each item in the base Construct the FP-tree for the frequent items of the pattern base m-conditional pattern base: fca:2, fcab:1 {} Header Table Item frequency head f 4 c 4 a 3 b 3 m 3 p 3 f:4 c:1 All frequent patterns relate to m m, fm, cm, am, fcm, fam, cam, fcam {} f:3 c:3 a:3 m-conditional FP-tree c:3 b:1 b:1   a:3 p:1 m:2 b:1 p:2 m:1

Recursion: Mining Each Conditional FP-tree {} f:3 c:3 am-conditional FP-tree Cond. pattern base of “am”: (fc:3) {} f:3 c:3 a:3 m-conditional FP-tree {} Cond. pattern base of “cm”: (f:3) f:3 cm-conditional FP-tree {} Cond. pattern base of “cam”: (f:3) f:3 cam-conditional FP-tree

Mining Frequent Patterns With FP-trees Idea: Frequent pattern growth Recursively grow frequent patterns by pattern and database partition Method For each frequent item, construct its conditional pattern-base, and then its conditional FP-tree Repeat the process on each newly created conditional FP-tree Until the resulting FP-tree is empty, or it contains only one path—single path will generate all the combinations of its sub-paths, each of which is a frequent pattern

Rule Generation Given a frequent itemset L, find all non-empty subsets f  L such that f  L – f satisfies the minimum confidence requirement If {A,B,C,D} is a frequent itemset, candidate rules: ABC D, ABD C, ACD B, BCD A, A BCD, B ACD, C ABD, D ABC AB CD, AC  BD, AD  BC, BC AD, BD AC, CD AB, If |L| = k, then there are 2k – 2 candidate association rules (ignoring L   and   L)

Rule Generation How to efficiently generate rules from frequent itemsets? In general, confidence does not have an anti- monotone property c(ABC D) can be larger or smaller than c(AB D) But confidence of rules generated from the same itemset has an anti-monotone property e.g., L = {A,B,C,D}: c(ABC  D)  c(AB  CD)  c(A  BCD) Confidence is anti-monotone w.r.t. number of items on the RHS of the rule

Rule Generation for Apriori Algorithm Lattice of rules Pruned Rules Low Confidence Rule

Pattern Evaluation Association rule algorithms tend to produce too many rules many of them are uninteresting or redundant Redundant if {A,B,C}  {D} and {A,B}  {D} 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 Interestingness Measure Given a rule X  Y, information needed to compute rule interestingness can be obtained from a contingency table Contingency table for X  Y Y X f11 f10 f1+ f01 f00 fo+ f+1 f+0 |T| f11: support of X and Y f10: support of X and Y f01: support of X and Y f00: support of X and Y Used to define various measures support, confidence, lift, Gini, J-measure, etc.

Drawback of Confidence Coffee Tea 15 5 20 75 80 90 10 100 Association Rule: Tea  Coffee Confidence= P(Coffee|Tea) = 0.75 but P(Coffee) = 0.9 Although confidence is high, rule is misleading P(Coffee|Tea) = 0.9375

Statistical Independence Population of 1000 students 600 students know how to swim (S) 700 students know how to bike (B) 420 students know how to swim and bike (S,B) P(SB) = 420/1000 = 0.42 P(S)  P(B) = 0.6  0.7 = 0.42 P(SB) = P(S)  P(B) => Statistical independence P(SB) > P(S)  P(B) => Positively correlated P(SB) < P(S)  P(B) => Negatively correlated

Statistical-based Measures Measures that take into account statistical dependence

Example: Lift/Interest Coffee Tea 15 5 20 75 80 90 10 100 Association Rule: Tea  Coffee Confidence= P(Coffee|Tea) = 0.75 but P(Coffee) = 0.9 Lift = 0.75/0.9= 0.8333 (< 1, therefore is negatively associated)

Drawback of Lift & Interest Y X 10 90 100 Y X 90 10 100 Statistical independence: If P(X,Y)=P(X)P(Y) => Lift = 1