Mongodb : 如何使用聚合在嵌套字段上建立排名

Question

这是我的 collection :

[
  {userId: "u1", data: { score1: 1, score2: 2, score3: 3 }, day: 1},
  {userId: "u1", data: { score1: 1, score2: 0, score3: 0 }, day: 2}, 
  {userId: "u1", data: { score1: 5, score2: 3, score3: 2 }, day: 3},
  {userId: "u2", data: { score1: 2, score2: 5, score3: 1 }, day: 1}, 
  {userId: "u2", data: { score1: 1, score2: 1, score3: 6 }, day: 2},
  {userId: "u2", data: { score1: 3, score2: 5, score3: 3 }, day: 3},
  {userId: "u3", data: { score1: 4, score2: 1, score3: 1 }, day: 1},
  {userId: "u3", data: { score1: 0, score2: 1, score3: 1 }, day: 2},
  {userId: "u3", data: { score1: 0, score2: 1, score3: 10 }, day: 3}
]

我想建立以下排行榜表：

{
  score1: [
    {"u1": 7}, // sum of all score1 for u1
    {"u2": 6}, // sum of all score1 for u2
    {"u3": 4}, // sum of all score1 for u3
  ],
  score2: [
    {"u2": 11}, // sum of all score2 for u2
    {"u1": 5}, // sum of all score2 for u1
    {"u3": 3}, // sum of all score2 for u3
  ],
  score3: [
    {"u3": 12}, // sum of all score3 for u3
    {"u2": 10}, // sum of all score3 for u2
    {"u1": 5}, // sum of all score3 for u1
  ],
}

到目前为止，我可以按 userId 分组并计算其中 3 个的每个分数的总和：

db.myCollection.aggregate([
  {
    $group: {
      _id: "$userId",
      score1: { $sum: "$score1" },
      score2: { $sum: "$score2" },
      score3: { $sum: "$score3" }
    }
  }
])

这给了我：

[
  { 
    _id: "u1",
    score1: 7,
    score2: 5,
    score3: 5
  },
  { 
    _id: "u2",
    score1: 6,
    score2: 11,
    score3: 10
  },
  { 
    _id: "u3",
    score1: 4,
    score2: 3,
    score3: 12
  },
]

如何提取每种类型的分数并建立相应的排行榜？

提前致谢。

Answer 1

我会首先在 data 字段上使用 $objectToArray，然后 $unwind 它因此每个文档都有 1 个用户和 1 个分数。然后按 userId 和 data.k 分组（其中将包含 "score1"、"score2" 等）并计算总和。然后按分数名称重新分组并将带有 k:userId, v:<score> 的对象推送到数组。然后在 null 上再次分组并将 k:scoreName, v:<object with user scores> 推入数组。最后 $arrayToObject 将该数组转换为您想要的对象：

db.collection.aggregate([
    {$addFields: {data: {$objectToArray: "$data"}}},
    {$unwind: "$data"},
    {$group: {
         _id: {userId: "$userId", scoreName: "$data.k"},
         score: {$sum:"$data.v"}
    }},
    {$group: {
         _id:"$_id.scoreName",
         data:{$push:{k:"$_id.userId", v:"$score"}}
    }},
    {$group: {
         _id: null,
         scores:{$push:{k:"$_id", v:{$arrayToObject:"$data"}}}
    }},
    {$replaceRoot:{newRoot:{$arrayToObject:"$scores"}}}
])

Playground

Mongodb : 如何使用聚合在嵌套字段上建立排名

Mongodb : how to use aggregate to build standings on nested fields

leaderboard

mongodb

aggregation-framework