Mongodb聚合：从键值对象返回不同值的计数答案

【问题标题】：Mongo aggregation: Return count of distinct values from Key Value objectMongodb聚合：从键值对象返回不同值的计数
【发布时间】：2021-07-21 21:13:09
【问题描述】：

我真的希望有人可以在这里帮助我，我正在为这个问题发疯:-d

所以我有多个文档 (+100,000)，如下所示：

 "_id" : ObjectId("60e5ae42fcc92f14c3a41208"),
    "userId" : "xxxx",
    "projectCreator" : {
        "userId" : "xxx|xxxx"
    },
    "hashTags" : [
        "Spring",
        "Java"
    ],
    "projectCategories" : {
        "60d76ef0597444095b8ab4b2" : "Backend",
        "60d76ef0597444095b8ab232" : "Infrastructure" 
    },
    "createdDate" : ISODate("2021-07-07T13:38:10.655Z"),
    "updatedAt" : ISODate("2021-07-08T11:48:36.200Z"),

    "_class" : "xxxx.model.project.Project"
}

我想要一个执行以下操作的查询：

从集合中的所有文档中提取所有唯一的 projectCategories 值（字符串值而不是 id）
计算每个值的出现次数

所以结果应该是这样的：

Backend : NUMBER OF OCCURRENCES
FrontEnd : NUMBER OF OCCURRENCES
Infrastructure: NUMBER OF OCCURRENCES

我“认为”我需要进行聚合并将值分组，然后进行计数，但老实说我无法理解这一点。

我试过这个查询：

db.projects.aggregate([ { $match: { isDeleted : {$ne: true} }},{ $match: { projectCategories: { $exists:true, $ne: null }} },{ $project: { result: { $objectToArray: "$projectCategories" } } },{ $unwind : "$result"}])

这将返回：

{ "_id" : ObjectId("60c313e2905d344c7dd117f1"), "result" : { "k" : "60d76f295974444b818ab4bc", "v" : "Apps" } }
{ "_id" : ObjectId("60c313e2905d344c7dd117f1"), "result" : { "k" : "60d76f1759744461468ab4b8", "v" : "Development Tools" } }
{ "_id" : ObjectId("60c313e2905d344c7dd117f1"), "result" : { "k" : "60d76eeb597444b9da8ab4b1", "v" : "Frontend" } }
{ "_id" : ObjectId("60cfb59f30b2647610a6c931"), "result" : { "k" : "60d76eeb597444b9da8ab4b1", "v" : "Frontend" } }
{ "_id" : ObjectId("60cfb59f30b2647610a6c931"), "result" : { "k" : "60d76ef659744422d68ab4b3", "v" : "Fullstack" } }
{ "_id" : ObjectId("60cfb69730b2647610a6c932"), "result" : { "k" : "60d76f295974444b818ab4bc", "v" : "Apps" } }
{ "_id" : ObjectId("60df83e84d8b6341d49cff4e"), "result" : { "k" : "60d76ef0597444095b8ab4b2", "v" : "Backend" } }
{ "_id" : ObjectId("60df83e84d8b6341d49cff4e"), "result" : { "k" : "60d76eeb597444b9da8ab4b1", "v" : "Frontend" } }
{ "_id" : ObjectId("60df83e84d8b6341d49cff4e"), "result" : { "k" : "60d76ef659744422d68ab4b3", "v" : "Fullstack" } }
{ "_id" : ObjectId("60e5ae42fcc92f14c3a41208"), "result" : { "k" : "60d76ef0597444095b8ab4b2", "v" : "Backend" } }
{ "_id" : ObjectId("60f0abf9f5c82b27af712ad7"), "result" : { "k" : "60d76f2559744477168ab4bb", "v" : "Games" } }
{ "_id" : ObjectId("60f0abf9f5c82b27af712ad7"), "result" : { "k" : "60d76ef659744422d68ab4b3", "v" : "Fullstack" } }
{ "_id" : ObjectId("60f68d2df9710f58c1e9c872"), "result" : { "k" : "60d76f295974444b818ab4bc", "v" : "Apps" } }
{ "_id" : ObjectId("60f68d2df9710f58c1e9c872"), "result" : { "k" : "60d76f0e5974448f038ab4b7", "v" : "Open Source" } }
{ "_id" : ObjectId("60f68d2df9710f58c1e9c872"), "result" : { "k" : "60d76eeb597444b9da8ab4b1", "v" : "Frontend" } }

所以我现在卡住的地方是我如何放松并获得如下输出：

Backend : NUMBER OF OCCURRENCES
FrontEnd : NUMBER OF OCCURRENCES
Infrastructure: NUMBER OF OCCURRENCES

有人可以帮我吗？

谢谢！

更新：我已经设法通过这个查询关闭：

db.projects.aggregate([ { $match: { isDeleted : {$ne: true} }},{ $match: { projectCategories: { $exists:true, $ne: null }} },{ $project: { result: { $objectToArray: "$projectCategories" } } },{ $unwind : "$result"}, { $group: { _id: "$result.v", count: { $sum: 1 } } }  ] )

但是现在的输出是这样的：

{ "_id" : "Development Tools", "count" : 1 }
{ "_id" : "Games", "count" : 1 }
{ "_id" : "Fullstack", "count" : 3 }
{ "_id" : "Open Source", "count" : 1 }
{ "_id" : "Frontend", "count" : 4 }
{ "_id" : "Apps", "count" : 3 }
{ "_id" : "Backend", "count" : 2 }

是否可以删除_id？

【问题讨论】：

标签： mongodb aggregation-framework

【解决方案1】：

您可以再次分组以将键和值推送到数组中，然后将 $replaceRoot 与 $arrayToObject 一起使用：

{$group:{
   _id:null,
   results:{$push: {k:"$_id", v:"$count"}}
}},
{$replaceRoot:{ newRoot: {$arrayToObject:"$results"}}}

【讨论】：

其实乔，如果你不介意的话。还有一个问题。如果我想同时获得 k 和 V 怎么办，它看起来像这样： { "_id" : "60d76f295974444b818ab4bc", "name: "backend", "count" : 2 } 我怎样才能更改我的查询以取回它？