Question

我尝试根据ArrayList对象的PostingsEntry属性对score PostingsEntry个对象进行排序。该列表位于具有PostingsList方法的sort()对象中。

public class PostingsEntry implements Comparable<PostingsEntry>{

    public int docID;
    public double score = 0;
    private TreeSet<Integer> positions = new TreeSet<Integer>();
    /** 
     *  PostingsEntries are compared by their score (only relevant
     *  in ranked retrieval).
     *
     *  The comparison is defined so that entries will be put in 
     *  descending order.
     */

    public int compareTo( PostingsEntry other ) { 
        return Double.compare( other.score, score );
    } 
}

public class PostingsList{

    private int position = 0;
    /** The postings list */
    private ArrayList<PostingsEntry> list = new ArrayList<PostingsEntry>();

    private class PostingsEntryComparator implements Comparator<PostingsEntry>{
        @Override
        public int compare(PostingsEntry pA, PostingsEntry pB){
            return pA.docID - pB.docID;
        }   
    }   
    /** Number of postings in this list. */
    public int size() {
        return list.size();
    }   

    /** Returns the ith posting. */
    public PostingsEntry get( int i ) { 
    return list.get( i );
    }      

    public void sort(){
        Collections.sort(list, new PostingsEntryComparator());
    }   

}

我试图在此处对列表进行排序：

// sort postingsList
postingsList.sort();

我打印结果：

for(int i=0; i<postingsList.size(); i++){
    System.out.println(index.docNames.get(postingsList.get(i).docID));
    System.out.printf("score: %f\n\n", postingsList.get(i).score);
}

但我明白了：

davisWiki/Zombie_Attack_Response_Guide.f
score: 0,019064

davisWiki/EvanGray.f
score: 0,004368

davisWiki/Mortal_Forever.f
score: 0,002708

davisWiki/JasonRifkind.f
score: 0,767518

davisWiki/Measure_Z.f
score: 0,031980

这表明该列表显然没有排序。我哪里错了？

Answer 1

您对sort的呼唤通过了另一个比较器：

public void sort(){
    Collections.sort(list, new PostingsEntryComparator());
}

出于此类目的，PostingsEntryComparator替换了score的“自然顺序”，由PostingsEntry执行Comparable<PostingsEntry>确定。因此，条目在docID上进行比较。如果您打印docID代替score，您会看到您的列表已根据ID正确排序。

注意：由于整数溢出，减去两个被比较项目的ID可能会产生错误的结果。请改用Integer.compare，方法与在Double.compare中正确使用PostingsEntry.compareTo的方式相同。

Answer 2

致电时

 Collections.sort(list, new PostingsEntryComparator());

您按PostingsEntryComparator排序，比较docID s。

如果您想按分数排序，则需要使用Comparable PostingsEntry的{{1}}实施方式，您可以通过调用Collections.sort()而不传递Comparator来执行此操作：

Collections.sort(list);

Collections.sort（）使我的列表未排序

2 个答案: