mongodb 中的乘法和分组

Question

我在 mongodb 中有一个集合如下：

{
"_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，例如在日期 2014-10-20 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，然后得到每个日期的情绪值总和乘以每个日期的集合数量，例如，如果我们在 2014 年 10 月 20 日有 2 个集合，情绪值为 20、1、2，如上面显示的集合，而另一个只有 5 collection 那么 2014-10-20 的值为 (20+1+2+5)3（因为这条推文重复了 3 个实体） 2（因为我们有 2 个推文文档这个日期）=168，如果我不考虑收集频率，我的代码运行良好，如下所示：

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);

现在的结果例如 2014-10-20 returns 28 但我想要 168 谁能帮我？

更新：我使用的最新版本代码如下：

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 个实体：

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

Group 通过 _id 找到所有实体情绪值的总和 每个文档.

    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);

Group Date 现在，得到总实体值的总和，以及每组 文档的 count 。

    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 值作为每个组的 count 和 value 的乘法结果。

    DBObject project = new BasicDBObject("_id",1);
    project.put("value",new BasicDBObject("$multiply",
                                    new Object[]{"$value","$count"}));
    stages.add(new BasicDBObject("$project",project));

如果您的日期相差毫秒，您需要在第二个分组阶段同时按日期、年份和月份进行分组，并在必要时添加排序阶段。

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