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private int compareItems(String item1, String item2) { //int compare = (int)( mapItemToTWU.get(item1) - mapItemToTWU.get(item2)); // if the same, use the lexical order otherwise use the TWU //return (compare == 0)? item1 - item2 : compare; if (item1 == null) { return -1; } if (item2 == null) { return 1; } if (item1.equals( item2 )) { return 0; } return item1.compareTo(item2); }
private int compareItems(String item1, String item2) { // ... }
private int compareItems(String item1, String item2) { int compare = (int)( mapItemToTWU.get(item1) - mapItemToTWU.get(item2)); // if the same, use the lexical order otherwise use the TWU return (compare == 0)? item1.compareTo(item2); }
please, correct me if I am wrong.Quote
webmasterphilfv
Yes, it is possible that the itemset {5} is not a high utility itemsets but that the itemset {1,2,3,4,5} is a high utility itemset.Quote
Mohammad S. Alodadi
is it possible to have an itemset that one of its items' utility is not >=minutil.
the reason I asked is that I have these high utility frequent itemsets:
1 2 3 4 #UTIL: 228.80042749643326
1 2 3 4 5 #UTIL: 228.80735443532467
Property 3 (pruning). Let X be an itemset. If TW U(X) < minutil, then the itemset X is a low-utility itemset as well as all its supersets. Proof. This directly follows from Property 1 and Property 2.