Question

我遵循循环遍历数组列表（mainItems）的代码，找到最相似的两个数组并将它们放在sortedTransactions中。它适用于小数据（10000个事务），但它永远运行88000个事务。可以采取哪些措施使其适用于大数据。

import java.util.*;

    public class Sort {

    static private  List<Transactions> trans = ReadFile.transactions;
    static public   List<int[]> mainItems;
    static public   ArrayList<int[]> sortedTransactions = new ArrayList<int[]>();

    static {
        mainItems = new ArrayList<int[]>();

        for (Transactions t : trans) {
            mainItems.add(t.getItems());
        }
    }

    static private  double jaccardSimilarity(int[] a, int[] b) {

        Set<Integer> s1 = new LinkedHashSet<Integer>();
        for(int i =0; i< a.length; i++){
            s1.add(a[i]);
        }
        Set<Integer> s2 = new LinkedHashSet<Integer>();
        for(int i =0; i< b.length; i++){
            s2.add(b[i]);
        }

        Set<Integer> intersection = new LinkedHashSet<>(s1);
        intersection.retainAll(s2);

        Set<Integer> union = new LinkedHashSet<Integer>(s1); 
        union.addAll(s2); 

        double jaccardSimilarity = (double)intersection.size()/ (double)union.size();
        //System.out.println(intersection);
        return jaccardSimilarity;
    }

    static private boolean isAllEqual(List<Double> a){

        for(int i=1; i<a.size(); i++){
            if(a.get(0) != a.get(i)){
                return false;
            }
        }

        return true;
    }


    static public void generatePairs() {

        for (int i = 0; i < mainItems.size() - 1; i++) {

            if (!sortedTransactions.contains(mainItems.get(i))) {

                List<Double> myd = new ArrayList<Double>();
                List<int[]> mys = new ArrayList<int[]>();

                for (int j = i + 1; j < mainItems.size(); j++) {

                    if (!sortedTransactions.contains(mainItems.get(j))) {

                        myd.add(jaccardSimilarity(mainItems.get(i),mainItems.get(j)));
                        mys.add(mainItems.get(j));
                    }
                }

                if (isAllEqual(myd) == false) {

                    sortedTransactions.add(mainItems.get(i));
                    sortedTransactions.add(mys.get(maxValue(myd)));
                }
            }
        }
    }

    static private int maxValue(List<Double> d) {

        double max = d.get(0);
        int f = 0;

        for(int i =1; i< d.size(); i++){

            if(d.get(i) > max){

                max= d.get(i); 
                f= i; 
            }
        }
        return f;
    }
}

Answer 1

您不必创建联合集（union（s1，s2）.size（）是s1.size（）+ s2.size（） - intersection（s1，s2）.size（））。< / p>

static private double jaccardSimilarity(int[] a, int[] b) {

    Set<Integer> s1 = new HashSet<Integer>();
    for (int i = 0; i < a.length; i++) {
        s1.add(a[i]);
    }
    Set<Integer> s2 = new HashSet<Integer>();
    for (int i = 0; i < b.length; i++) {
        s2.add(b[i]);
    }

    final int sa = s1.size();
    final int sb = s2.size();
    s1.retainAll(s2);
    final int intersection = s1.size();
    return 1d / (sa + sb - intersection) * intersection;
}

在Java中计算Jaccard的相似性

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