java - 如何确定每个文档中术语的术语频率？

Question

我正在构建倒排索引，但在检查数据库时似乎无法获得正确的频率。我到处都读到您应该使用 HashMap，但我不太确定这是否是正确的方法。有什么想法吗？

public class Tokenize {

    public static void createIndex() throws Exception{
        
        ArrayList<Dokument> dok = new QueryHandler().getDokuments();
        ArrayList<String> queries = new ArrayList<String>();
        ArrayList<String> queries2 = new ArrayList<String>();
        HashMap<String, Integer> frek = new HashMap<String, Integer>();
        
        for(int d = 0; d < dok.size(); d++){
            String token = "";
            int frekvens = 0;
            
            
            try{
                
                Dokument document = dok.get(d);
                StringTokenizer st = new StringTokenizer(document.dokument());
                while (st.hasMoreTokens()) {
                    
                    
                    token = st.nextToken();
                    token.replaceAll("[']", "");
                    token.replaceAll("[,]", "");
                    token.replaceAll("[)]", "");
                    token.replaceAll("[(]", "");
                    token.replaceAll("[.]", "");
                    frekvens ++;
                    frek.put(token, frekvens);
                    
                
                        queries.add("INSERT IGNORE INTO termindeks (docID, term) values ("+document.docID()+", '"+token+"')");
                        queries2.add("INSERT IGNORE INTO invertedindeks (term, docID, termfrekvens) values ('"+token+"', "+document.docID()+", "+ frekvens+")");
                                        
                            
                }
            }
            

            catch (Exception e) {
            e.printStackTrace();
            System.out.println(token);
            }
        }
        
        String[] ffs = new String[queries.size()];
        ffs = queries.toArray(ffs);
        getDB().runQueriesIgnoreException(queries.toArray(ffs));
        
        String[] ffs2 = new String[queries2.size()];
        ffs2 = queries2.toArray(ffs2);
        getDB().runQueriesIgnoreException(queries2.toArray(ffs2));

    }
}

Answer 1

您应该首先获取 token 的值，递增它并再次放置它。

在循环中像这样:

Integer frekvens = frek.get(token); //remove the other frekvens as it's not needed - or find a better name for this one ;)
if( frekvens == null ) { frekvens = 0 };
frekvens++;
frek.put(token, frekvens);