编写聚合 MongoDB 查询来计算字段计数 ID
writing aggregate MongoDB query to calculate field count ids
我正在为以下记录和输出编写聚合查询。
数据:
[
{
"_id" : ObjectId("5f3b2626927b18001db86884"),
"collections" : [
Art, Craft
]
},{
"_id" : ObjectId("5f3b2626927b18001db86885"),
"collections" : [
Craft
]
},{
"_id" : ObjectId("5f3b2626927b18001db86886"),
"collections" : [
Apex, Art
]
},
...
]
预期输出:
集合 ID 的数量
{
Art : 2,
Craft : 2,
Apex : 1
}
现在,我们正在循环遍历集合以计算每个集合的计数作为所需的输出,但它的性能很低,因为这个集合由 10,000 条记录组成。
所以,我正在考虑构建一个聚合查询,如果有人可以帮助我开始或指出正确的方向,我将不胜感激。谢谢。
查了一段时间想出了解决办法
db.getCollection("collectionName").aggregate(
[
// get all the records with at least one collection name
{
$match: {
"collections.0": { $exists: true }
}
},
// populate the collection record
{
$lookup: {
from: "from_collection",
localField: "localField",
foreignField: "foreignField",
as: "collections"
}
},
// unwind
{ $unwind: "$collections" },
// group by the collections._id
{ $group: { _id: "$collections._id", collections: { $push: "$$ROOT.ID" } } },
// project with collection contains _id, and count
{
$project : {
collections: "$collections",
count: { $size: "$collections" }
}
}
]
).toArray();
输出:
[
{
"_id" : ObjectId("61c4c42d68579f00311dd3e1"),
"collections" : [
"015151",
"015152",
"015153"
],
"count" : 3.0
},
{
"_id" : ObjectId("615f38016f40710033699939"),
"collections" : [
"014871"
],
"count" : 1.0
},
{
"_id" : ObjectId("611fed5ee0d12c00337cb009"),
"collections" : [
"014788",
"014786",
"014789",
"014787",
"014884",
"014893",
"014967",
"014968",
"015016",
"015017"
],
"count" : 10.0
}
...
]
$unwind
$group
$group
$replaceRoot
db.collection.aggregate([
{
$unwind: "$collections"
},
{
"$group": {
"_id": "$collections",
"v": {
"$sum": 1
}
}
},
{
"$group": {
"_id": null,
"collections": {
"$push": {
$arrayToObject: [
[ { "k": "$$ROOT._id", "v": "$$ROOT.v" } ]
]
}
}
}
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: "$collections"
}
}
}
])
我正在为以下记录和输出编写聚合查询。
数据:
[
{
"_id" : ObjectId("5f3b2626927b18001db86884"),
"collections" : [
Art, Craft
]
},{
"_id" : ObjectId("5f3b2626927b18001db86885"),
"collections" : [
Craft
]
},{
"_id" : ObjectId("5f3b2626927b18001db86886"),
"collections" : [
Apex, Art
]
},
...
]
预期输出:
集合 ID 的数量
{
Art : 2,
Craft : 2,
Apex : 1
}
现在,我们正在循环遍历集合以计算每个集合的计数作为所需的输出,但它的性能很低,因为这个集合由 10,000 条记录组成。
所以,我正在考虑构建一个聚合查询,如果有人可以帮助我开始或指出正确的方向,我将不胜感激。谢谢。
查了一段时间想出了解决办法
db.getCollection("collectionName").aggregate(
[
// get all the records with at least one collection name
{
$match: {
"collections.0": { $exists: true }
}
},
// populate the collection record
{
$lookup: {
from: "from_collection",
localField: "localField",
foreignField: "foreignField",
as: "collections"
}
},
// unwind
{ $unwind: "$collections" },
// group by the collections._id
{ $group: { _id: "$collections._id", collections: { $push: "$$ROOT.ID" } } },
// project with collection contains _id, and count
{
$project : {
collections: "$collections",
count: { $size: "$collections" }
}
}
]
).toArray();
输出:
[
{
"_id" : ObjectId("61c4c42d68579f00311dd3e1"),
"collections" : [
"015151",
"015152",
"015153"
],
"count" : 3.0
},
{
"_id" : ObjectId("615f38016f40710033699939"),
"collections" : [
"014871"
],
"count" : 1.0
},
{
"_id" : ObjectId("611fed5ee0d12c00337cb009"),
"collections" : [
"014788",
"014786",
"014789",
"014787",
"014884",
"014893",
"014967",
"014968",
"015016",
"015017"
],
"count" : 10.0
}
...
]
$unwind
$group
$group
$replaceRoot
db.collection.aggregate([
{
$unwind: "$collections"
},
{
"$group": {
"_id": "$collections",
"v": {
"$sum": 1
}
}
},
{
"$group": {
"_id": null,
"collections": {
"$push": {
$arrayToObject: [
[ { "k": "$$ROOT._id", "v": "$$ROOT.v" } ]
]
}
}
}
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: "$collections"
}
}
}
])