Question

给定集合中的以下项目，如何获得所有记录的id == 1的assessment.answer字段的平均值？此外，我需要返回subs []的第一个索引，每个记录像subs [0]

  {
    "_id" : ObjectId("57b5e55e8426706357a90728"),
    "updatedAt" : ISODate("2016-08-20T00:21:16.497Z"),
    "createdAt" : ISODate("2016-08-18T16:42:06.173Z"),
    "password" : "password",
    "email" : "test@user.com",
    "lastName" : "User",
    "firstName" : "Test",
    "assessment" : [ 
        {
            "subs" : [ 
                100, 
                100, 
                100, 
                100, 
                100
            ],
            "answer" : 2,
            "id" : 1
        }, 
        {
            "subs" : [ 
                80, 
                40, 
                100, 
                60, 
                20
            ],
            "answer" : 2,
            "id" : 2
        }, 
        {
            "subs" : [ 
                80, 
                20, 
                20, 
                20, 
                20
            ],
            "answer" : 4,
            "id" : 3
        }, 
        {
            "subs" : [ 
                80, 
                100, 
                60, 
                100, 
                100
            ],
            "answer" : 5,
            "id" : 4
        }, 
        {
            "subs" : [ 
                60, 
                100, 
                100, 
                100, 
                100
            ],
            "answer" : 1,
            "id" : 5
        }, 
        {
            "subs" : [ 
                100, 
                100, 
                100, 
                100, 
                100
            ],
            "answer" : 3,
            "id" : 6
        }, 
        {
            "subs" : [ 
                80, 
                20, 
                100, 
                80, 
                60
            ],
            "answer" : 3,
            "id" : 7
        }, 
        {
            "subs" : [ 
                20, 
                20, 
                20, 
                20, 
                20
            ],
            "answer" : 4,
            "id" : 8
        }, 
        {
            "subs" : [ 
                60, 
                60, 
                60, 
                60, 
                40
            ],
            "answer" : 5,
            "id" : 9
        }, 
        {
            "subs" : [ 
                60, 
                60, 
                60, 
                60, 
                40
            ],
            "answer" : 3,
            "id" : 10
        }, 
        {
            "subs" : [ 
                100, 
                100, 
                100, 
                100, 
                100
            ],
            "answer" : 4,
            "id" : 11
        }, 
        {
            "subs" : [ 
                80, 
                80, 
                80, 
                80, 
                80
            ],
            "answer" : 1,
            "id" : 12
        }, 
        {
            "subs" : [ 
                60, 
                60, 
                60, 
                60, 
                60
            ],
            "answer" : 2,
            "id" : 13
        }, 
        {
            "subs" : [ 
                40, 
                40, 
                40, 
                40, 
                40
            ],
            "answer" : 3,
            "id" : 14
        }, 
        {
            "subs" : [ 
                80, 
                20, 
                20, 
                20, 
                100
            ],
            "answer" : 3,
            "id" : 15
        }
    ],
    "__v" : 51
}

Answer 1

你快到了。您只需选择符合条件的文档$match和$in

var ids = [ 1, 2, 3, 4, 5 ];
var matchStage = { "$match": { "assessment.id": { "$in": ids } } };

db.accounts.aggregate(
    [
        matchStage,
        { "$unwind": "$assessment" }, 
        matchStage,
        { "$group": { 
            "_id": null, 
            "avg": { "$avg": "$assessment.answer" }
        }} 
    ]
)

从MongoDB 3.2开始，您可以这样做：

db.accounts.aggregate(
    [
        matchStage,
        { "$project": { 
            "preAvg": { 
                "$avg": { 
                    "$map": { 
                        "input": { 
                            "$filter": { 
                                "input": "$assessment", 
                                "as": "el", 
                                "cond": { "$setIsSubset": [ [ "$$el.id" ], ids ] }
                            }  
                        }, 
                        "as": "val", 
                        "in": "$$val.answer" 
                    } 
                } 
            } 
        }}, 
        { "$group": {  
            "_id": null, 
            "avg": { "$sum": "$preAvg" } 
        }}
    ]
)

Mongo DB在集合的子字段上获得平均值

1 个答案: