java - not getting exact result in closed itemsets -
in variable total(instance of array list class) have list of frequent item sets. while finding closed frequent item sets displaying wrong result.
in intersection method going find whether 2 array lists same or not.
arraylist close=new arraylist(); system.out.println("total: "+total); for(int i1=0;i1<total.size();i1++) { list list1=(list)total.get(i1); if(close.size()==0) close.add(list1); else { list tp=intersection(list1,close); if(tp.size()!=0) close.add(list1); } } system.out.println("close: "+close); for(int i2=0;i2<total.size();i2++) { list list1=(list)total.get(i2); if(close.size()!=0) { for(int i1=0;i1<close.size();i1++) { list list2=(list)close.get(i1); list tp=intersect(list1,list2); if(tp.size()!=0) { if(!close.contains(list1)) { //here hm containing list of frequent item sets support count. if(((integer)hm.get(list1)>(integer)hm.get(list2))&&((integer)hm.get(list1)!=(integer)hm.get(list2))) { close.add(list1); //system.out.println(list1+": "+hm.get(list1)+" "+list2+": "+hm.get(list2)); } } } } } } public <t> list<t> intersection(list<t> list1, list<t> list2) { list<t> list = new arraylist<t>(); (t t : list1) { list<t> list3=(list<t>)list2.get(0); if(!list3.contains(t)) { return list1; } if(list.size()!=0) return list1; } return list; } in total object values total: [[1, 3, 7, 9, 12], [1, 3, 7, 9], [1, 3, 9, 12], [1, 7, 9, 12], [1, 3, 7, 12], [3, 7, 9, 12], [3, 9, 12], [3, 7, 9], [3, 7, 12], [1, 7, 12], [1, 9, 12], [ 7, 9, 12], [1, 7, 9], [1, 3, 12], [1, 3, 9], [1, 3, 7], [1, 9], [1, 12], [1, 3], [12, 15], [1, 7], [3, 12], [7, 9], [9, 12], [7, 12], [3, 15], [3, 9], [3, 7], [ 3], [1], [7], [12], [9], [15]] in close close: [[1, 3, 7, 9, 12], [1, 3, 9, 12], [3, 9, 12], [1, 9, 12], [1, 3, 12], [1, 3, 9], [12, 15], [3, 15], [1, 9], [1, 12], [1, 3], [3, 12], [9, 12], [7, 12], [ 3, 9], [15], [3], [1], [7], [12], [9]]
your intersection code badly badly broken.
use unit tests.
try code on computing intersection of [1,2,3], [3,4,5].
if i'm not mistaken, returns first list.
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