我正在构建倒排索引,但在检查数据库时,我似乎无法获得正确的频率。我到处读到你应该使用HashMap,但我不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));
}
}
答案 0 :(得分:2)
您应首先获取令牌的值,递增并再次放置。
在你的循环中这样:
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);
答案 1 :(得分:1)
这个想法是正确的,但据我所知,你没有正确使用HashMap
。您必须获得与密钥关联的值,即
Integer i = map.get(token);
i += 1;
map.put(token, i);
修改强>
另一种选择是使用AtomicInteger
代替Integer
,因为AtomicInteger
是可变的。
Map<String, AtomicInteger> map = new HashMap<String, AtomicInteger>();
map.get(token).getAndIncrement();