Question

我在mongodb中有一个如下集合：

{
"_id" : ObjectId("54901212f315dce7077204af"),
"Date" : ISODate("2014-10-20T04:00:00.000Z"),
"Type" : "Twitter",
"Entities" : [ 
    {
        "ID" : 2,
        "Name" : "test1",
        "Sentiment" : {
            "Value" : 20,
            "Neutral" : 1
        },
        {
        "ID" : 1,
        "Name" : "test1",
        "Sentiment" : {
            "Value" : 1,
            "Neutral" : 1
        }
    },
     {
        "ID" : 3,
        "Name" : "test1",
        "Sentiment" : {
            "Value" : 2,
            "Neutral" : 1
        }
]

}

我有几个，例如在2014-10-20日期你可能会发现5条推文，每条推文都有不同的情绪价值，现在我想做的是按日期分组，然后得到总和每个日期的情绪值乘以每个日期的集合数量，例如，如果我们在2014-10-20中有2个集合，情绪值为20,1,2，如上面显示的集合，而另一个集合仅为5 2014-10-20的值是（20 + 1 + 2 + 5） 3（因为此推文重复3个推荐） 2（因为我们在此日期有2条推文文件）= 168 ，如果我不考虑集合的频率，我的代码可以很好地工作如下：

DBObject unwind = new BasicDBObject("$unwind", "$Entities"); // "$unwind" converts object with array into many duplicate objects, each with one from array
    collectionG = db.getCollection("GraphDataCollection");
    DBObject groupFields = new BasicDBObject( "_id", "$Date");
   groupFields.put("value", new BasicDBObject( "$sum", "$Entities.Sentiment.Value"));
    DBObject groupBy = new BasicDBObject("$group", groupFields );
    DBObject sort = new BasicDBObject("$sort", new BasicDBObject("Date", 1));
    stages.add(unwind);
    stages.add(groupBy);
    DBObject project = new BasicDBObject("_id",0);
    project.put("Date","$_id");
     project.put("value",1);
     stages.add(new BasicDBObject("$project",project));
    stages.add(sort);
    AggregationOutput output = collectionG.aggregate(stages);

现在结果例如2014-10-20返回28，但我想要168 谁能帮我？

更新：我使用的代码的最后一个版本如下：

DBCollection collectionG;
    collectionG = db.getCollection("GraphDataCollection");
    List<DBObject> stages = new ArrayList<DBObject>();
    ArrayList<DBObject> andArray = null;
         DBObject groupFields = new BasicDBObject( "_id", "$_id");
            groupFields.put("value", new BasicDBObject( "$sum", "$Entities.Sentiment.Value"));
            groupFields.put("date", new BasicDBObject( "$first", "$Date"));
            DBObject groupBy = new BasicDBObject("$group", groupFields );
            stages.add(groupBy);
            DBObject groupByDate = new BasicDBObject( "_id", "$date");
            groupByDate.put("value",new BasicDBObject("$sum","$value"));
            groupByDate.put("count",new BasicDBObject("$sum",1));
            DBObject dtGrp = new BasicDBObject("$group", groupByDate );
            stages.add(dtGrp);

            DBObject project = new BasicDBObject("_id",1);
            project.put("value",new BasicDBObject("$multiply",
                                            new Object[]{"$value","$count"}));
            stages.add(new BasicDBObject("$project",project));
            AggregationOutput output = collectionG.aggregate(stages);
            System.out.println(output.results());

Answer 1

Unwind实体：

    DBObject unwind = new BasicDBObject("$unwind", "$Entities");
    stages.add(unwind);

<Group _id

    DBObject groupFields = new BasicDBObject( "_id", "$_id");
    groupFields.put("value", new BasicDBObject( "$sum", "$Entities.Sentiment.Value"));
    groupFields.put("date", new BasicDBObject( "$first", "$Date"));
    DBObject groupBy = new BasicDBObject("$group", groupFields );
    stages.add(groupBy);

查找所有实体情绪值每个文档的总和。

Group

Date现在count，获取实体总价值的总和，以及每组的文档DBObject groupByDate = new BasicDBObject( "_id", "$date"); groupByDate.put("value",new BasicDBObject("$sum","$value")); groupByDate.put("count",new BasicDBObject("$sum",1)); DBObject dtGrp = new BasicDBObject("$group", groupByDate ); stages.add(dtGrp);。

Project

count值作为value和DBObject project = new BasicDBObject("_id",1); project.put("value",new BasicDBObject("$multiply", new Object[]{"$value","$count"})); stages.add(new BasicDBObject("$project",project));的乘法结果，适用于每个组。

{{1}}

如果您的日期相差几毫秒，您需要按日期，年份和月份分组，在第二组阶段，并在必要时添加排序阶段。

mongodb中的乘法和分组

1 个答案: