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חוקי Association ד " ר אבי רוזנפלד

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המוטיבציה מה הם הדברים שהולכים ביחד ? –איזה מוצרים בסופר שווה לשים ביחד –מערכות המלצה – Recommendation Systems שבוע הבא –חיפוש באינטרנט בעוד שבועיים

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אבל איך מגדירים ביחד ? האם זה Supervised או Unsupervised – Unsupervised מה כמה פעמים דברים חייבים להיות ביחד לפני שאני מחשב אותם ? – Threshold של Confidence ו Support –תכף נגדיר אותם...

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December 2008©GKGupta4 אבל קודם כל, דוגמא

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December 2008©GKGupta5 Frequency of Items

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6 מה אני מחפש... If-then rules about the contents of baskets. {i 1, i 2,…,i k } → j means: “if a basket contains all of i 1,…,i k then it is likely to contain j.” אני רוצה ללמוד אם יש תלות למשהו – j, בהינתן {i 1, i 2,…,i k }

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7 Support Simplest question: find sets of items that appear “frequently” in the baskets. Support for itemset I = the number of baskets containing all items in I. –בדרך כלל, ההסתברות למצוא את הדברים ביחד Given a support threshold s, sets of items that appear in > s baskets are called frequent itemsets.

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Confidence כמה פעמים היה גם I וגם j Confidence of this association rule is the probability of j given i 1,…,i k. – I = i 1,…,i k באופן פורמאלי : Confidence = באופן פורמאלי : Lift =

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December 2008©GKGupta9 Frequent Items Assume 25% support. In 25 transactions, a frequent item must occur in at least 7 transactions (25*1/4=6.25). The frequent 1-itemset or L 1 is now given below.

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December 2008©GKGupta10 L2L2 The following pairs are frequent.

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11 Example: Confidence B 1 = {m, c, b}B 2 = {m, p, j} B 3 = {m, b}B 4 = {c, j} B 5 = {m, p, b}B 6 = {m, c, b, j} B 7 = {c, b, j}B 8 = {b, c} An association rule: {m, b} → c. – Confidence = 2/4 = 50%. + _ +

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December 2008©GKGupta12 Rules The full set of rules are given below. Could some rules be removed? Comment: Study the above rules carefully.

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13 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 (why?).

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14 Finding Frequent Pairs The hardest problem often turns out to be finding the frequent pairs. – Why? Often frequent pairs are common, frequent triples are rare. We’ll concentrate on how to do that, then discuss extensions to finding frequent triples, etc.

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html 111111 2111 3111 4111 511 611 7111 811 911 1011 1111 121 1311 1411 1511 161 1711 1811111 1911111 201 ID a, b, c, d, e, f, g, h, i E.g. 3-itemset {a,b,h} has support 15% 2-itemset {a, i} has support 0% 4-itemset {b, c, d, h} has support 5% If minimum support is 10%, then {b} is a large itemset, but {b, c, d, h} Is a small itemset!

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16 Finding Frequent Pairs The hardest problem often turns out to be finding the frequent pairs. – Why? Often frequent pairs are common, frequent triples are rare. We’ll concentrate on how to do that, then discuss extensions to finding frequent triples, etc.

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17 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).

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18 Example: Counting Pairs Suppose 10 5 items. Suppose counts are 4-byte integers. Number of pairs of items: 10 5 (10 5 -1)/2 = 5*10 9 (approximately). Therefore, 2*10 10 (20 gigabytes) of main memory needed.

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html The Apriori algorithm for finding large itemsets efficiently in big DBs 1: Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) 3{C k = apriori-gen (L k-1 ) 4 For each c in C k, initialise c.count to zero 5 For all records r in the DB 6 {C r = subset (C k, r); For each c in C r, c.count++ } 7 Set L k := all c in C k whose count >= minsup 8 } /* end -- return all of the L k sets.

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1: Find all large 1-itemsets To start off, we simply find all of the large 1- itemsets. This is done by a basic scan of the DB. We take each item in turn, and count the number of times that item appears in a basket. In our running example, suppose minimum support was 60%, then the only large 1-itemsets would be: {a}, {b}, {c}, {d} and {f}. So we get L 1 = { {a}, {b}, {c}, {d}, {f}}

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1 : Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) We already have L 1. This next bit just means that the remainder of the algorithm generates L 2, L 3, and so on until we get to an L k that’s empty. How these are generated is like this:

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1 : Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) 3 {C k = apriori-gen (L k-1 ) Given the large k-1-itemsets, this step generates some candidate k-itemsets that might be large. Because of how apriori-gen works, the set C k is guaranteed to contain all the large k-itemsets, but also contains some that will turn out not to be `large’.

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1 : Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) 3 {C k = apriori-gen (L k-1 ) 4 For each c in C k, initialise c.count to zero We are going to work out the support for each of the candidate k-itemsets in C k, by working out how many times each of these itemsets appears in a record in the DB.– this step starts us off by initialising these counts to zero.

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1 : Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) 3 {C k = apriori-gen (L k-1 ) 4 For each c in C k, initialise c.count to zero 5 For all records r in the DB 6 {C r = subset (C k, r); For each c in C r, c.count++ } We now take each record r in the DB and do this: get all the candidate k-itemsets from C k that are contained in r. For each of these, update its count.

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1 : Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) 3 {C k = apriori-gen (L k-1 ) 4 For each c in C k, initialise c.count to zero 5 For all records r in the DB 6 {C r = subset (C k, r); For each c in C r, c.count++ } 7 Set L k := all c in C k whose count >= minsup Now we have the count for every candidate. Those whose count is big enough are valid large itemsets of the right size. We therefore now have L k, We now go back into the for loop of line 2 and start working towards finding L k+1

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David Corne, and Nick Taylor, Heriot-Watt University - dwcorne@gmail.com These slides and related resources: http://www.macs.hw.ac.uk/~dwcorne/Teac hing/dmml.html Explaining the Apriori Algorithm … 1 : Find all large 1-itemsets 2: For (k = 2 ; while L k-1 is non-empty; k++) 3 {C k = apriori-gen (L k-1 ) 4 For each c in C k, initialise c.count to zero 5 For all records r in the DB 6 {C r = subset (C k, r); For each c in C r, c.count++ } 7 Set L k := all c in C k whose count >= minsup 8 } /* end -- return all of the L k sets. We finish at the point where we get an empty L k. The algorithm returns all of the (non-empty) L k sets, which gives us an excellent start in finding interesting rules (although the large itemsets themselves will usually be very interesting and useful.

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פתרון נוסף : FP-Growth בונה עץ לפי התדירות של הביטויים –מהגדול לקטן –יש לי אלגוריתם כזה... ( שיש עליו פטנט - אבל הוא מיועד לחיפוש ולא Association)

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