Mongo Aggregation:$ group和$ project数组以对象计数 [英] Mongo Aggregation : $group and $project array to object for counts

问题描述

我有类似的文件:

{
    "platform":"android",
    "install_date":20151029
}

1. 平台-可以具有[android | ios | kindle | facebook]中的一个值.
2. 安装日期-有很多安装日期

也有很多字段.

目标:我正在计算特定日期每个平台的安装量.

Aim : I am calculating installs per platform on particular date.

所以我在聚合框架中使用分组依据，并按平台进行计数.文档应如下所示:

So I am using group by in aggregation framework and make counts by platform. Document should look like like:

{
  "install_date":20151029,
  "platform" : {
   "android":1000,
   "ios": 2000,
   "facebook":1500
  }
}

---

我做得像:

---

I have done like:

db.collection.aggregate([
  { 
      $group: { 
          _id: { platform: "$platform",install_date:"$install_date"},  
            count: { "$sum": 1 } 
        }
  },
  { 
      $group: { 
          _id: { install_date:"$_id.install_date"},
          platform: { $push :  {platform :"$_id.platform", count:"$count" } }  
      }
  },
  { 
     $project : { _id: 0, install_date: "$_id.install_date", platform: 1 } 
  }     
]) 

文件如下:

{
    "platform": [
        {
            "platform": "facebook",
            "count": 1500
        },
        {
            "platform": "ios",
            "count": 2000
        },
        {
            "platform": "android",
            "count": 1000
        }
    ],
    "install_date": 20151027
}

问题:

将数组投影到单个对象作为平台"

Projecting array to single object as "platform"

推荐答案

在MongoDb 3.4和更高版本中，您可以利用

With MongoDb 3.4 and newer, you can leverage the use of $arrayToObject operator to get the desired result. You would need to run the following aggregate pipeline:

db.collection.aggregate([
    { "$group": {
        "_id": {  
            "date": "$install_date",  
            "platform": { "$toLower": "$platform" }
        },
        "count": { "$sum": 1 }
    } },
    { "$group": {
        "_id": "$_id.date",
        "counts": {
            "$push": {
                "k": "$_id.platform",
                "v": "$count"
            }
        }
    } },
    {  "$addFields": {
        "install_date": "$_id", 
        "platform": { "$arrayToObject": "$counts" }
    }  },
    { "$project": { "counts": 0, "_id": 0 } } 
])

---

对于旧版本，请利用

---

For older versions, take advantage of the $cond operator in the $group pipeline step to evaluate the counts based on the platform field value, something like the following:

db.collection.aggregate([    
    { "$group": { 
        "_id": "$install_date",             
        "android_count": {
            "$sum": {
                "$cond": [ { "$eq": [ "$platform", "android" ] }, 1, 0 ]
            }
        },
        "ios_count": {
            "$sum": {
                "$cond": [ { "$eq": [ "$platform", "ios" ] }, 1, 0 ]
            }
        },
        "facebook_count": {
            "$sum": {
                "$cond": [ { "$eq": [ "$platform", "facebook" ] }, 1, 0 ]
            }
        },
        "kindle_count": {
            "$sum": {
                "$cond": [ { "$eq": [ "$platform", "kindle" ] }, 1, 0 ]
            }
        } 
    } },
    { "$project": {
        "_id": 0, "install_date": "$_id",            
        "platform": {
            "android": "$android_count",
            "ios": "$ios_count",
            "facebook": "$facebook_count",
            "kindle": "$kindle_count"
        }
    } }
])

在上面， $cond 将逻辑条件作为第一个参数(如果是)，然后返回第二个参数(如果计算结果为true)(然后)或第三个参数(假)(否则).这会使true/false返回1和0，以分别馈入$sum.

In the above, $cond takes a logical condition as it's first argument (if) and then returns the second argument where the evaluation is true (then) or the third argument where false (else). This makes true/false returns into 1 and 0 to feed to $sum respectively.

例如，如果{ "$eq": [ "$platform", "facebook" ] },为true，则表达式将计算为{ $sum: 1 }，否则它将为{ $sum: 0 }

So for example, if { "$eq": [ "$platform", "facebook" ] }, is true then the expression will evaluate to { $sum: 1 } else it will be { $sum: 0 }

这篇关于Mongo Aggregation:$ group和$ project数组以对象计数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！