如何在 MongoDB 中根据重复 ID 获取评分平均值?
mongodbdatabasebig data analytics更新于 2026/2/11 5:22:17
在 MongoDB 中，使用 $avg 函数获取平均值。我们创建一个包含文档的集合。这里，每个文档都有重复的 ID 和评分。−
> db.demo606.insertOne({id:1,rating:5});{
"acknowledged" : true, "insertedId" : ObjectId("5e972dfbf57d0dc0b182d623")
}
> db.demo606.insertOne({id:1,rating:4});{
"acknowledged" : true, "insertedId" : ObjectId("5e972dfef57d0dc0b182d624")
}
> db.demo606.insertOne({id:2,rating:3});{
"acknowledged" : true, "insertedId" : ObjectId("5e972e09f57d0dc0b182d625")
}
> db.demo606.insertOne({id:1,rating:null});{
"acknowledged" : true, "insertedId" : ObjectId("5e972e0ef57d0dc0b182d626")
}
> db.demo606.insertOne({id:2,rating:null});{
"acknowledged" : true, "insertedId" : ObjectId("5e972e15f57d0dc0b182d627")
}
> db.demo606.insertOne({id:2,rating:3});{
"acknowledged" : true, "insertedId" : ObjectId("5e972e1bf57d0dc0b182d628")
}
借助 find() 方法显示集合中的所有文档 −
> db.demo606.find();
这将产生以下输出 −
{ "_id" : ObjectId("5e972dfbf57d0dc0b182d623"), "id" : 1, "rating" : 5 }
{ "_id" : ObjectId("5e972dfef57d0dc0b182d624"), "id" : 1, "rating" : 4 }
{ "_id" : ObjectId("5e972e09f57d0dc0b182d625"), "id" : 2, "rating" : 3 }
{ "_id" : ObjectId("5e972e0ef57d0dc0b182d626"), "id" : 1, "rating" : null }
{ "_id" : ObjectId("5e972e15f57d0dc0b182d627"), "id" : 2, "rating" : null }
{ "_id" : ObjectId("5e972e1bf57d0dc0b182d628"), "id" : 2, "rating" : 3 }
以下是根据重复 ID 获取平均评分的查询 −
> db.demo606.aggregate(
... [
... { "$group": {
... "_id": "$id",
... "AverageRating": { "$avg": { "$ifNull": ["$rating",0 ] } }
... }}
... ]
... );
这将产生以下输出 −
{ "_id" : 2, "AverageRating" : 2 }
{ "_id" : 1, "AverageRating" : 3 }
