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Mining Associations

Mining Associations

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Mining Associations

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  1. Mining Associations Apriori Algorithm

  2. Item Item Basket 1 Item Item Item Item Basket 2 Item Item Item Item Basket 3 Item Item Etc. Computation Model • Typically, data is kept in a flat file rather than a database system. • Stored on disk. • Stored basket-by-basket. • The true cost of mining disk-resident data is usually the number of disk I/O’s. • In practice, association-rule algorithms read the data in passes – all baskets read in turn. • Thus, we measure the cost by the number of passes an algorithm takes.

  3. Main-Memory Bottleneck • For many frequent-itemset algorithms, main memory is the critical resource. • As we read baskets, we need to count something, e.g., occurrences of pairs. • The number of different things we can count is limited by main memory. • Swapping counts in/out is a disaster.

  4. Finding Frequent Pairs • The hardest problem often turns out to be finding the frequent pairs. • We’ll concentrate on how to do that, then discuss extensions to finding frequent triples, etc.

  5. Naïve Algorithm • Read file once, counting in main memory the occurrences of each pair. • From each basket of n items, generate its n (n -1)/2 pairs by two nested loops. • Fails if (#items)2 exceeds main memory. • Remember: #items can be 100K (Wal-Mart) or 10B (Web pages).

  6. Details of Main-Memory Counting • Two approaches: • Count all pairs, using a triangular matrix. • Keep a table of triples [i, j, c] = the count of the pair of items {i,j } is c. • (1) requires only 4 bytes/pair. • Note: assume integers are 4 bytes. • (2) requires 12 bytes, but only for those pairs with count > 0.

  7. 4 per pair 12 per occurring pair Method (1) Method (2)

  8. Triangular-Matrix Approach – (1) • Number items 1, 2, …,n • Count {i, j } only if i < j. • Keep pairs in the order • {1,2}, {1,3},…, {1,n}, • {2,3}, {2,4},…,{2,n}, • {3,4},…, {3, n}, • … • {n -1,n}.

  9. Triangular-Matrix Approach – (2) • Let n be the number of items. Count for pair {i, j } is at position T(i,j) = (i-1)n - i(i+1)/2 + j {1,2}, {1,3}, {1,4}, {2,3}, {2,4} {3,4} • Total number of pairs n (n –1)/2; total bytes about 2n 2.

  10. Details of Approach #2 • Total bytes used is about 12p, where p is the number of pairs that actually occur. • Beats triangular matrix if at most 1/3 of possible pairs actually occur. • May require extra space for retrieval structure, e.g., a hash table.

  11. Apriori Algorithm for pairs– (1) • A two-pass approach called a-priori limits the need for main memory. • Key idea: monotonicity : if a set of items appears at least s times, so does every subset. • Contrapositive for pairs: if item i does not appear in s baskets, then no pair including i can appear in s baskets.

  12. Item counts Frequent items Counts of candidate pairs Pass 1 Pass 2 Apriori Algorithm for pairs– (2) • Pass 1: Read baskets and count in main memory the occurrences of each item. • Requires only memory proportional to #items. • Pass 2: Read baskets again and count in main memory only those pairs whose both elements were found in Pass 1 to be frequent. • Requires memory proportional to square of frequent items only.

  13. Detail for A-Priori • You can use the triangular matrix method with n = number of frequent items. • Trick: number frequent items 1,2,… and keep a table relating new numbers to original item numbers.

  14. All pairs of items from F1 Count the items Count the pairs To be explained All items Filter Filter Construct Construct C1 F1 C2 F2 C3 First pass Second pass Frequent Triples, Etc. • For each k, we construct two sets of k –itemsets: • Ck= candidate k - itemsets = those that might be frequent (support >s ) based on information from the pass for k –1. • Fk = the set of truly frequent k - itemsets.

  15. Full Apriori Algorithm • Let k=1 • Generate frequent itemsets of length 1 • Repeat until no new frequent itemsets are found k=k+1 • Generatelength kcandidate itemsets from length k-1frequent itemsets • Prune candidate itemsets containing subsets of length k-1 that are infrequent • Count the support of each candidate by scanning the DB and eliminate candidates that are infrequent, leaving only those that are frequent

  16. Illustrating Apriori Suppose AB is not in F2. All these will either not be generated by Step 1 as candidates in C3, or will be pruned in Step 2.

  17. Candidate generation • Must ensure that the candidate set is complete. • Should not generate the same candidate itemset more than once.

  18. Data Set Example s=3

  19. Fk-1F1 Method • Extend each frequent (k - 1)­itemset with a frequent 1-itemset. • Is it complete? • Yes, because every frequent k­itemset is composed of • a frequent (k-1)­itemset and • a frequent 1­itemset. • However, it doesn’t prevent the same candidate itemset from being generated more than once. • E.g., {Bread, Diapers, Milk} can be generated by merging • {Bread, Diapers} with {Milk}, • {Bread, Milk} with {Diapers}, or • {Diapers, Milk} with {Bread}.

  20. Lexicographic Order • Keep frequent itemset sorted in lexicographic order. • Each frequent (k-1)­itemsetX is extended with frequent items that are lexicographically larger than the items in X. Example • {Bread, Diapers} can be extended with {Milk} • {Bread, Milk} can’t be extended with {Diapers} • {Diapers, Milk} can’t be extended with {Bread} • Why is it complete?

  21. Prunning • Merging {Beer, Diapers} with {Milk} is unnecessary. Why? • Because one of its subsets, {Beer, Milk}, is infrequent. • Solution: Prune! • How?

  22. Fk-1F1 Example {Beer,Diapers,Bread} and {Bread,Milk,Beer} aren't in fact generated if lexicographical ord. is considered.

  23. Fk-1Fk-1 Method • Merge a pair of frequent (k-1) itemsets only if their first k-2 items are identical. • E.g. frequent itemsets {Bread, Diapers} and {Bread, Milk} are merged to form a candidate 3­itemset {Bread, Diapers, Milk}.

  24. Fk-1Fk-1 Method • We don’t merge {Beer, Diapers} with {Diapers, Milk} because the first item in both itemsets is different. But, is this "don't merge" decision Ok? • Indeed, if {Beer, Diapers, Milk} is a viable candidate, it would have been obtained by merging {Beer, Diapers} with {Beer, Milk} instead. Pruning • Because each candidate is obtained by merging a pair of frequent (k-1)­itemsets, an additional candidate pruning step is needed to ensure that the remaining k-2 subsets of k-1 elements are frequent.

  25. Fk-1Fk-1 Example

  26. C1 F1 Itemset Itemset Sup. count {I1} {I1} 6 {I2} {I2} 7 {I3} {I3} 6 {I4} {I4} 2 {I5} {I5} 2 Another Example Min_sup_count = 2

  27. C2 Itemset {I1,I2} {I1,I3} {I1,I4} {I1,I5} {I2,I3} {I2,I4} {I2,I5} {I3,I4} {I3,I5} {I4,I5} Generate C2 from F1F1 Min_sup_count = 2 F1

  28. Generate C3 from F2F2 Min_sup_count = 2 F2 Prune C3 F3

  29. Generate C4 from F3F3 Min_sup_count = 2 C4 {I1,I2,I3,I5} is pruned because {I2,I3,I5} is infrequent F3