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CSE 634/590 Data mining Extra Credit: Classification by Association rules: Example Problem Muhammad Asiful Islam, SBID: 106506983

Original Data Outlook Humidity Wind PlayTenis Sunny High Weak No Sunny High Strong No Overcast Normal Weak Yes Rain High Weak Yes Rain Normal Weak Yes Rain High Strong No Overcast Normal Strong Yes Sunny High Weak No Rain Normal Weak Yes Sunny Normal Strong Yes Overcast High Strong Yes Overcast Normal Weak Yes Rain High Strong No

Converted Data Outlook =Sunny (I1) Outlook =Overcast (I2) Outlook = Rain (I3) Humidity =High (I4) Humidity =Normal (I5) Wind =Weak (I6) Wind =Strong (I7) PlayTenis =Yes (I8) PlayTenis =No (I9) + - - + - + - - + + - - + - - + - + - + - - + + - + - - - + + - + - + - - - + - + + - + - - - + + - - + - + - + - - + - + + - + - - + - + - - + - - + - + + - + - + - - - + - + + - - + - + - - + + - - + - - + + - + - - - + + - - + - +

Generating 1-itemset Frequent Pattern Scan D for count of each candidate Item Set Support Count {I1} 4 {I2} 4 {I3} 5 Compare candidate support count with minimum support count Item Set Support Count {I1} 4 {I2} 4 {I3} 5 {I4} 7 {I4} 7 {I5} 6 {I5} 6 {I6} 7 {I6,} 7 {I7} 6 {I7} 6 {I8} 8 {I8} 8 {I9} 5 {I9} 5 C1 L1 Let, the minimum support count be 4. So, min_sup = 4/13 = 30.76% Let, minimum confidence required is 70%.

Generating 2-itemset Frequent Pattern Generate C2 candidates from L1 Item Set {I1,I2} {I1,I3} {I1,I4} {I1,I5} {I1,I6} {I1,I7} {I1,I8} {I1,I9} {I2,I3} {I2,I4} {I2,I5} {I2,I6} {I2,I7} {I2,I8} {I2,I9} Scan D for count of each candidate Item Set Support Count {I1,I2} 0 {I1,I3} 0 {I1,I4} 3 {I1,I5} 1 {I1,I6} 2 {I1,I7} 2 {I1,I8} 1 {I1,I9} 3 {I2,I3} 0 {I2,I4} 1 {I2,I5} 3 {I2,I6} 2 {I2,I7} 2 {I2,I8} 4 Compare candidate support count with minimum support count Item Set Support Count {I2,I8} 4 {I4,I7} 4 {I3,I4} {I2,I9} 0 {I4,I9} 5 {I3,I5} {I3,I4} 3 {I5,I6} 4 {I3,I6} {I3,I7} {I3,I8} {I3,I5} 2 {I3,I6} 3 {I3,I7} 2 {I5,I8} 6 {I6,I8} 5 {I3,I9} {I3,I8} 3 {I4,I5} {I3,I9} 2 {I4,I6} {I4,I7} {I4,I5} 0 {I4,I6} 3 L2 {I4,I8} {I4,I7} 4 {I4,I9} {I4,I8} 2 {I5,I6} {I4,I9} 5 {I5,I7} {I5,I6} 4 {I5,I8} {I5,I7} 2 {I5,I9} {I5,I8} 6 {I6,I7} {I5,I9} 0 {I6,I8} {I6,I9} {I7,I8} {I7,I9} {I8,I9} C2 {I6,I7} 0 {I6,I8} 5 {I6,I9} 2 {I7,I8} 3 {I7,I9} 3 {I8,I9} 0 C2

Join Operation L2 = {{I2, I8}, {I4, I7}, {I4, I9}, {I5, I6}, {I5, I8}, {I6, I8}}. -To find C3, we compute L2 join L2 -Similar to natural join operation in Data Base Join Operation illustrated: -Consider an itemset {I2,I8} from L2, -Now search for other itemsets containing I2 or I8 in L2; -this gives us itemsets{i5,i8} and {I6,I8} join = {I2,I8} {I5,I8} {I2,I5,I8} So, it s like joining two database tuples with common column value I8 So, {I2,I8} joining with {I5,I8} results {I2,I5,I8} and, {I2,I8} joining with {I6,I8} results {I2,I6,I8} Similarly, using the same technique for every itemset in L2 we get;- C3= L2 Join L2 = {{I2, I5, I8}, {I2, I6, I8}, {I4, I7, I9}, {I5, I6, I8}}.

Generating 3-itemset Frequent Pattern Generate C3 candidates from L2 Item Set {I5,I6,I8} Scan D for count of each candidate Item Set Support Count {I5,I6,I8} 4 Compare candidate support count with minimum support count Item Set Support Count {I5,I6,I8} 4 C3 C3 L3 -Use Apriori Property -- all subsets of a frequent itemset must also be frequent C3= L2 Join L2 = {{I2, I5, I8}, {I2, I6, I8}, {I4, I7, I9}, {I5, I6, I8}}. -For example, lets take {I5, I6, I8}.The 2-item subsets of it are {I5, I6}, {I5, I8} & {I6, I8}. Since all 2- item subsets of {I5, I6, I8} are members of L2, We will keep {I5, I6, I8} in C3. Lets take another example of {I2, I5, I8}which shows how the pruning is performed. The 2-item subsets are {I2, I5}, {I2, I8} & {I5,I8}. -BUT, {I2, I5} is not a member of L2and hence it is not frequent, violating Apriori Property. Thus We will have to remove {I2, I5, I8} from C3. Therefore, C3= {{I5, I6, I8}} after checking for all members of result of Join operation for Pruning.

Generating 4-itemset Frequent Pattern The algorithm uses L3 Join L3 to generate a candidate set of 4-itemsets, C4. Thus, C4= φ, and algorithm terminates, having found all of the frequent items. This completes our Apriori Algorithm. Next? These frequent itemsets will be used to generate strong association rules (where strong association rules satisfy both minimum support & minimum confidence). We had L = {{I1}, {I2}, {I3}, {I4}, {I5}, {I6}, {I7}, {I8}, {I9} {I2,I8}, {I4,I7}, {I4,I9}, {I5,I6}, {I5,I8}, {I6,I8}, {I5,I6,I8}}. Now, to generate classification rules, we need to consider only the frequent item sets which contains I8 or I9.

Generating Association Rules from Frequent Itemsets Consider only the frequent item sets which contains I8 or I9: {I2,I8}, {I4,I9}, {I5,I8}, {I6,I8}, {I5,I6,I8} Consider {I2,I8} - R1: I2 I8 Confidence = sc{i2,i8}/sc{i2} = 4/4 = 100% selected. Consider {I4,I9} - R2: I4 I9 Confidence = sc{i4,i9}/sc{i4} = 5/7 = 71.42% selected. Consider {I5,I8} - R3: I5 I8 Confidence = sc{i5,i8}/sc{i5} = 6/6 = 100% selected. Consider {I6,I8} - R4: I6 I8 Confidence = sc{i6,i8}/sc{i6} = 5/7 = 71.42% selected. Consider{I5,I6,I8} - R5: I5 ^ I6 I8 Confidence = sc{i5,i6,i8}/sc{i5,i6} = 4/4 = 100% selected.

Classification Rules Rule (A B) [support, confidence] R1: I2 I8 [30.76%, 100%] R2: I4 I9 [38.46%, 71.42%] R3: I5 I8 [46.15%, 100%] R4: I6 I8 [38.46%, 71.42%] R6: I5 ^ I6 I8 [30.76%, 100%] Alternatively:- 1. Outlook = overcast PlayTenis = yes [30.76%, 100%] 2. Humidity = high PlayTenis = no [38.46%, 71.42%] 3. Humidity = normal PlayTenis = yes [46.15%, 100%] 4. Wind = weak PlayTenis = yes [38.46%, 71.42%] 5. Humidity = normal AND Wind = weak PlayTenis = yes [30.76%, 100%]

Test Data Set Outlook Humidity Wind PlayTenis Overcast Normal Strong Yes Rain Normal Weak Yes Rain Normal Strong No Sunny Normal Weak Yes Sunny High Strong No Given this test data, lets classify each of them and determine the accuracy:- 1: Outlook = overcast, And using rule 1, we get PlayTenis = Yes, So, this example is correctly classified. Note that rule 3 also applies. 2: Humidity = normal ^ Wind = weak, And using rule 5, we get PlayTenis = Yes, So, this example is correctly classified. 3: Humidity = normal, And using rule 3, we get PlayTenis = Yes. But actual class is No. So, this example is incorrectly classified. 4: Humidity = normal ^ Wind = weak, And using rule 5, we get PlayTenis = Yes, So, this example is correctly classified. 5: Humidity = high, And using rule 2, we get PlayTenis = No, So, this example is correctly classified. So, the accuracy of our association rule based classifier is = 4/5 = 80%

Thank You

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