所以我正在尝试使用MapDB做某事,但遇到了麻烦。我会尽力描述一下:
我有四个数据,我们会说它像这样:
1) String action; //the name of the action itself
2) String categoryOfAction; //the category of the action
3) Integer personWhoPerformedAction; //the person that did the action
4) Long timeOfOccurrence; //the time the action was done
同一操作可以在此数据库中由不同的人员,不同的时间以及不同的类别多次执行。我想要三个独立的地图,每个地图将数据组织成这样的形式:
String[] actionOccurances = map1.get(action); //returns every occurrence of that action (possibly in an array), including who did that occurrence, time the occurrence occurred, and the category of that occurrence
Long latestOccurance = map2.get(action); //returns the latest occurrence of that action
String[] actionsPerformedByPerson = map3.get(personWhoPerformedAction); //returns every action that this person has done, including the category of that action, the time they performed that action, and the name of the action itself
因此,我想尽可能有效地做到这一点。我知道我可以做这样的事情:
DB thedb = DBMaker.newTempFileDB().make();
NavigableSet<Object[]> map1 = thedb.createTreeSet("actionOccurences").comparator(Fun.COMPARABLE_ARRAY_COMPARATOR).make();
HTreeMap<String, Long> map2 = thedb.getHashMap("lastOccurrence");
NavigableSet<Object[]> map3 = thedb.createTreeSet("actionsPerformedByPerson").comparator(Fun.COMPARABLE_ARRAY_COMPARATOR).make();
但是我觉得那是错误的。必须有一种更有效的方法,使我不必多次存储相同的数据,是吗?
我在Bind类及其功能(secondaryValues,mapInverse等)上已经玩了很多,但是我似乎找不到找到将这组数据映射成我想要的方式的方法。
有什么帮助吗?谢谢。
最佳答案
啊!经过一段时间的研究,我找到了解决方案。我基本上为每个记录分配一个唯一的ID,然后使用MapDB的secondaryKey绑定。它是这样的:
static class Record implements Serializable
{
final String action;
final String categoryOfAction;
final String personWhoPerformedAction;
final Long timeOfOccurrence;
public record(String actn, String cat, String person, Long time)
{
action = actn;
categoryOfAction = cat;
personWhoPerformedAction = person;
timeOfOccurence = time;
}
}
static void main(String[] args)
{
DB thedb = DBMaker.newTempFileDB().make();
//primaryMap maps each record to a unique ID
BTreeMap<Integer,Record> primaryMap = thedb.createTreeMap("pri")
.keySerializer(BTreeKeySerializer.INTEGER)
.makeOrGet();;
//this map holds the unique ID of every record in primaryMap with a common action
NavigableSet<Object[]> map_commonAction = thedb.createTreeSet("com_a")
.comparator(Fun.COMPARABLE_ARRAY_COMPARATOR)
.makeOrGet();
//this map holds the unique ID of every record in primaryMap with a common person
NavigableSet<Object[]> map_commonPerson = thedb.createTreeSet("com_p")
.comparator(Fun.COMPARABLE_ARRAY_COMPARATOR)
.makeOrGet();
//binding map_commonAction to primaryMap so it is updated with primary
Bind.secondaryKey(primaryMap, map_commonAction, new Fun.Function2<String, Integer, Record>() {
@Override
public String run(Integer recordID, Record r) {
return r.action;
}
});
//binding map_commonPerson to primaryMap so it is updated with primary
Bind.secondaryKey(primaryMap, map_commonPerson, new Fun.Function2<String, Integer, Record>() {
@Override
public String run(Integer recordID, Record r) {
return r.personWhoPerformedAction;
}
});
primaryMap.put(1, new Record("a", "abc", "person1", 123434L));
primaryMap.put(2, new Record("a", "abc", "person2", 322443L));
primaryMap.put(3, new Record("b", "def", "person2", 124243L));
primaryMap.put(4, new Record("b", "abc", "person1", 983243L));
primaryMap.put(5, new Record("c", "def", "person2", 999993L));
//this is how we attain all records with some action
for (Object[] k : Fun.filter(map_commonAction, "someAction"))
{
Record obtainedRecord = primary.get(k[1]);
}
//this is how we attain all records with some person
for (Object[] k : Fun.filter(map_commonPerson, "somePerson"))
{
Record obtainedRecord = primary.get(k[1]);
}
}
但是,如果您认为可以改进此解决方案,请继续关注。谢谢!