MongoDB 在事先不知道所有字段的情况下聚合字段 [英] MongoDB aggregate fields without knowing all the fields before hand

问题描述

How can I compute the aggregate of the following metrics without knowing all the metrics before hand? Can I accomplish this using the aggregate framework or MapReduce?

[
  {
   player_id: '123',
   timestamp: <sometime>,
   metrics: {
     points_per_game: 1,
     rebounds_per_game: 2,
     assist_per_game: 3,
   }
  },
  {
    player_id: '123',
    timestamp: <sometime>,
    metrics: {
      points_per_game: 1,
      rebounds_per_game: 2,
    }
  },
  {
    player_id: '345',
    timestamp: <sometime>,
    metrics: {
      points_per_game: 1,
      rebounds_per_game: 2,
      point_in_the_paint_per_game: 2
    }
  }
]

I would like to have the following result

[
 {
   player_id: '123',
   metrics: {
     points_per_game: 2,
     rebounds_per_game: 4,
     assist_per_game: 3,
   }
 },
 {
   player_id: '345',
   metrics: {
     points_per_game: 1,
     rebounds_per_game: 2,
     point_in_the_paint_per_game: 2
   }
 }
]

I cannot do something like the following since it would require me to know every metrics:

db.stats.aggregate([
   {$group: {
     _id: {player: "$player_id"},
     points_per_game: { $sum: "$metrics.points_per_game"}
     ...
])

解决方案

You can try below aggregation.

Convert the object into array of key value pairs followed by $unwind+$group to group by each key and accumulate the count. Final step to go back to named key value object.

db.colname.aggregate([
  {"$addFields":{"metrics":{"$objectToArray":"$metrics"}}},
  {"$unwind":"$metrics"},
  {"$group":{
    "_id":{"id":"$player_id","key":"$metrics.k"},
    "count":{"$sum":"$metrics.v"}
  }},
  {"$group":{
    "_id":"$_id.id",
    "metrics":{"$mergeObjects":{"$arrayToObject":[[["$_id.key","$count"]]]}}
  }}
])

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