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Author: Admin | 2025-04-28
Eliminate bucket number 4 since its count (1) is smaller than the minimum support 2. Accordingly, we can eliminate all itemsets associated with bucket number 4, namely {Butter, Jam} and {Cheese, Jam}.Confidence is an indication of how often the rule has been found to be true. The confidence value of a rule, X -> Y, with respect to a set of transactions D, is the proportion of the transactions that contains X which also contains Y. It is defined as seen in the left. Thus, we’ll have these value calculated:support(Milk) = 6/9, support (Bread) = 7/9, support(Milk U Bread) = 4/9confidence(Milk -> Bread) = (4/9) / (6/9) = 6/9lift(Milk -> Bread) = (4/9) / ( (6/9) * (7/9) ) = 6/7Rule LatticeFor each frequent itemset with k > 1, we consider all possible association rules. For example, if we have a set of values including {A, B, C, D}. Here are all the association rules:If the rule {Milk, Butter} -> {Bread} does not meet the minimum support requirement, then the 2 association rules at the level below connected to it ({Milk} -> {Bread, Butter} and {Butter} -> {Milk, Bread}) can’t also be eliminated. Maximal Frequent ItemsetsThe last concept that I’ll cover in this post is maximal frequent item sets. Looking at the tables below, let’s say we have a 3-itemsets set {Milk, Bread, Butter} with the support of 2. That means we do not need to consider any 2-itemsets and 1-itemset sets that have these items because that would be redundant. Thus, I can remove {Milk}, {Bread}, {Butter}, {Milk, Bread}, {Milk, Butter}, and {Bread, Butter} from my calculation.
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