Question

这是我的文章结构在我的Post集合中的样子。

     {
        "_id" : ObjectId("51ed845c92964454ee85fa55"),
        "_cls" : "Post",
        "user" : ObjectId("51eb52ba8f24a04aa4761cb8"),
        "tags" : [ 
            {
                "tag" : ObjectId("51eb52be8f24a04aa4761cbe"),
                "name" : "rugby"
            }, 
            {
                "tag" : ObjectId("51eb52ba8f24a04aa4761cb8"),
                "name" : "john smith"
            }
        ],
        "content" : "my fourth post!",
        "upvotes" : 1,
        "downvotes" : 0,
        "rank" : 5343.95577,
        "replies" : [],
        "date_created" : ISODate("2013-07-22T15:13:32.650Z"),
        "date_modified" : ISODate("2013-07-22T15:13:32.650Z")
    },
{
    "_id" : ObjectId("51ec9b188f24a04ff0bd8668"),
    "_cls" : "Post",
    "user" : ObjectId("51eb52ba8f24a04aa4761cb8"),
    "tags" : [ 
        {
            "tag" : ObjectId("51eb52be8f24a04aa4761cba"),
            "name" : "rugby"
        }
    ],
    "content" : "http://www.livememe.com/ebejev5 blah blah",
    "upvotes" : 1,
    "downvotes" : 0,
    "replies" : [],
    "date_created" : ISODate("2013-07-21T22:38:16.000Z"),
    "date_modified" : ISODate("2013-07-21T22:38:16.000Z"),
    "rank" : 5000
}

我要做的是基本上根据每个标签的最高排名来标准化排名。所以我可以以编程方式采取的步骤是：

1. Get posts
2. Group by tags(the first tag in each tags array is fine)
3. Calculate the highest ranked post for each tag(max),(trank), 
4. For each post with said tag, calculate new value in each tag, calling it erank,
   by dividing. basically each post has new value: erank = rank/trank
5.The final output sorted by erank.

我正在尝试在mongodb中计算出一个聚合。这就是我到目前为止只计算每个标签的最高排名。

db.post.aggregate(    {
        "$unwind" : "$tags"
    },{$group: { _id : {tag: "$tag",rank : "$rank"},"tagName" : {
                "$first" : "$tags"
      }}}, 
     {$group: { _id : {tag: "$tagName.tag"}, trank : {$max : "$_id.rank"}}})

编辑：好吧，我相信我弄明白了，仍然需要更多地测试它，但似乎有效。

db.post.aggregate(
    { "$match" : { "tags" : { "$elemMatch" : { "tag" : { "$in" : [ ObjectId("51eb52ba8f24a04aa4761cb8")]}}}}},
    {$unwind : "$tags"},
    {$group: {_id : "$tags.tag",name :{$first: "$tags.name"},"posts" : {$addToSet : {postid : "$_id", content : "$content",
        rank : "$rank",user : "$user", upvotes : "$upvotes", downvotes : "$downvotes", date_created : "$date_created", date_modified : "$date_modified"
        }},
    "trank" : {$max : "$rank"}}},

    {$unwind : "$posts"},
    {$project:{ _id : "$posts.postid", rank : "$posts.rank",content : "$posts.content",user : "$posts.user", tag : {id :"$_id",name : "$name" },
    upvotes : "$posts.upvotes", downvotes : "$posts.downvotes", date_created : "$posts.date_created", date_modified : "$posts.date_modified",
        erank : {$divide : ["$posts.rank","$trank"]}}},


  //   {$project:{content : 1,tags : 1, _id : 1,erank: 1,posts : 1, tag : 1,rank : 1}},
       {$group : {_id : "$_id","tags" : {$addToSet : { tag : "$tag.id",name: "$tag.name"
        }},"erank" : {$max : "$erank"
        },"content" :{$first : "$content"},"rank" :{$first : "$rank"},"user" :{$first : "$user"},
        upvotes : {$first : "$upvotes"}, downvotes : {$first : "$downvotes"}, date_created : {$first : "$date_created"}
        , date_modified :{$first: "$date_modified"}
        }},
        {$sort : {"erank" : -1,"_id" : 1}},
        {$skip : 0},
        {$limit : 25}


    )

我如何聚合我的集合，以便我将每个文档除以mongodb中的“最大”聚合？

0 个答案: