I have sample data like below
[
{
brand:"iphone",
category:"mobile"
},
{
brand:"iphone",
category:"laptop"
},
{
brand:"lenova",
category:"laptop"
}
]
and expecting result as
[
{
brand:"iphone",
count:2
},
{
brand:"lenova",
count:1
},
{
category:"laptop",
count:2
},
{
category:"mobile",
count:1
}
]
Here I want group by same object with multiple fields and get there count. Can any one please let me how to do that in the mongoose.
I am not familiarised with Mongoose. Just tried in Mongoshell
db.getCollection('test').aggregate([
{
$group:{
_id:"$brand",
brand:{$first:"$brand"},
category:{$first:"$category"}
}
},
{$project:{_id:0}}
])
Possible only by using two queries.
Group By Brand
db.getCollection('pages').aggregate([
{
$group: {_id: "$brand", category: { $push: "$category" }}
},
{
$project : {
_id : 0, brand : "$_id", count : {$size : "$category"}
}
},
{ $unwind: { path: "$category", preserveNullAndEmptyArrays: true } }
])
Result:-
/* 1 */
{
"brand" : "lenova",
"count" : 1
}
/* 2 */
{
"brand" : "iphone",
"count" : 2
}
Group By Category
db.getCollection('pages').aggregate([
{
$group: {
_id: "$category", brand: { $push: "$brand" },
}
},
{
$project : {
_id : 0, category : "$_id", count : {$size : "$brand"}
}
},
{ $unwind: { path: "$brand", preserveNullAndEmptyArrays: true } },
])
Result:-
/* 1 */
{
"category" : "laptop",
"count" : 2
}
/* 2 */
{
"category" : "mobile",
"count" : 1
}
Merge them for the required output.
We can use $facet to run parallel aggregation on data.
The following query can get us the expected output:
db.collection.aggregate([
{
$facet:{
"brand_group":[
{
$group:{
"_id":"$brand",
"brand":{
$first:"$brand"
},
"count":{
$sum:1
}
}
},
{
$project:{
"_id":0
}
}
],
"category_group":[
{
$group:{
"_id":"$category",
"category":{
$first:"$category"
},
"count":{
$sum:1
}
}
},
{
$project:{
"_id":0
}
}
]
}
},
{
$project:{
"array":{
$concatArrays:["$brand_group","$category_group"]
}
}
},
{
$unwind:"$array"
},
{
$replaceRoot:{
"newRoot":"$array"
}
}
]).pretty()
Data set:
{
"_id" : ObjectId("5da5c0d0795c8651a7f508c2"),
"brand" : "iphone",
"category" : "mobile"
}
{
"_id" : ObjectId("5da5c0d0795c8651a7f508c3"),
"brand" : "iphone",
"category" : "laptop"
}
{
"_id" : ObjectId("5da5c0d0795c8651a7f508c4"),
"brand" : "lenova",
"category" : "laptop"
}
Output:
{ "brand" : "lenova", "count" : 1 }
{ "brand" : "iphone", "count" : 2 }
{ "category" : "laptop", "count" : 2 }
{ "category" : "mobile", "count" : 1 }
Related
I want to group by and count follow_user.tags.tag_id per record, so no matter how many times the same tag_id show up on the same record, it only counts as 1.
My database structure looks like this:
{
"external_userid" : "EXID1",
"follow_user" : [
{
"userid" : "USERID1",
"tags" : [
{
"tag_id" : "TAG1"
}
]
},
{
"userid" : "USERID2",
"tags" : [
{
"tag_id" : "TAG1"
},
{
"tag_id" : "TAG2"
}
]
}
]
},
{
"external_userid" : "EXID2",
"follow_user" : [
{
"userid" : "USERID1",
"tags" : [
{
"tag_id" : "TAG2"
}
]
}
]
}
Here's my query:
[
{ "$unwind": "$follow_user" }, { "$unwind": "$follow_user.tags" },
{ "$group" : { "_id" : { "follow_user᎐tags᎐tag_id" : "$follow_user.tags.tag_id" }, "COUNT(_id)" : { "$sum" : 1 } } },
{ "$project" : { "total" : "$COUNT(_id)", "tagId" : "$_id.follow_user᎐tags᎐tag_id", "_id" : 0 } }
]
What I expected:
{
"total" : 1,
"tagId" : "TAG1"
},
{
"total" : 2,
"tagId" : "TAG2"
}
What I get:
{
"total" : 2,
"tagId" : "TAG1"
},
{
"total" : 2,
"tagId" : "TAG2"
}
$set - Create a new field follow_user_tags.
1.1. $setUnion - To distinct the value from the Result 1.1.1.
1.1.1. $reduce - Add the value of follow_user.tags.tag_id into array.
$unwind - Deconstruct follow_user_tags array field to multiple documents.
$group - Group by follow_user_tags and perform total count via $sum.
$project - Decorate output document.
db.collection.aggregate([
{
$set: {
follow_user_tags: {
$setUnion: {
"$reduce": {
"input": "$follow_user.tags",
"initialValue": [],
"in": {
"$concatArrays": [
"$$value",
"$$this.tag_id"
]
}
}
}
}
}
},
{
$unwind: "$follow_user_tags"
},
{
$group: {
_id: "$follow_user_tags",
total: {
$sum: 1
}
}
},
{
$project: {
_id: 0,
tagId: "$_id",
total: 1
}
}
])
Sample Mongo Playground
How to get percentage total of data with group by date in MongoDB ?
Link example : https://mongoplayground.net/p/aNND4EPQhcb
I have some collection structure like this
{
"_id" : ObjectId("5ccbb96706d1d47a4b2ced4b"),
"date" : "2019-05-03T10:39:53.108Z",
"id" : 166,
"update_at" : "2019-05-03T10:45:36.208Z",
"type" : "image"
}
{
"_id" : ObjectId("5ccbb96706d1d47a4b2ced4c"),
"date" : "2019-05-03T10:39:53.133Z",
"id" : 166,
"update_at" : "2019-05-03T10:45:36.208Z",
"type" : "image"
}
{
"_id" : ObjectId("5ccbb96706d1d47a4b2ced4d"),
"date" : "2019-05-03T10:39:53.180Z",
"id" : 166,
"update_at" : "2019-05-03T10:45:36.208Z",
"type" : "image"
}
{
"_id" : ObjectId("5ccbb96706d1d47a4b2ced4e"),
"date" : "2019-05-03T10:39:53.218Z",
"id" : 166,
"update_at" : "2019-05-03T10:45:36.208Z",
"type" : "image"
}
And I have query in mongodb to get data of collection, how to get percentage of total data. in bellow example query to get data :
db.name_collection.aggregate(
[
{ "$match": {
"update_at": { "$gte": "2019-11-04T00:00:00.0Z", "$lt": "2019-11-06T00:00:00.0Z"},
"id": { "$in": [166] }
} },
{
"$group" : {
"_id": {
$substr: [ '$update_at', 0, 10 ]
},
"count" : {
"$sum" : 1
}
}
},
{
"$project" : {
"_id" : 0,
"date" : "$_id",
"count" : "$count"
}
},
{
"$sort" : {
"date" : 1
}
}
]
)
and this response :
{
"date" : "2019-11-04",
"count" : 39
},
{
"date" : "2019-11-05",
"count" : 135
}
how to get percentage data total from key count ? example response to this :
{
"date" : "2019-11-04",
"count" : 39,
"percentage" : "22%"
},
{
"date" : "2019-11-05",
"count" : 135,
"percentage" : "78%"
}
You have to group by null to get total count and then use $map to calculate the percentage. $round will be a useful operator in such case. Finally you can $unwind and $replaceRoot to get back the same number of documents:
db.collection.aggregate([
// previous aggregation steps
{
$group: {
_id: null,
total: { $sum: "$count" },
docs: { $push: "$$ROOT" }
}
},
{
$project: {
docs: {
$map: {
input: "$docs",
in: {
date: "$$this.date",
count: "$$this.count",
percentage: { $concat: [ { $toString: { $round: { $multiply: [ { $divide: [ "$$this.count", "$total" ] }, 100 ] } } }, '%' ] }
}
}
}
}
},
{
$unwind: "$docs"
},
{
$replaceRoot: { newRoot: "$docs" }
}
])
Mongo Playground
Look my problen hehe. I don't know how I can do it.
There are a lot of assetState 1..n I would like do aggregation for get last asset state group by asset.
Mongo collection : assetState
[
{
"lsd" : {
"$id" : ObjectId("lucas")
},
"stateDate" : ISODate("2018-09-10T16:26:44.501Z"),
"assetId" : ObjectId("5b96b7645f2b3c0101520s60")
},
{
"lsd" : {
"$id" : ObjectId("denner")
},
"stateDate" : ISODate("2018-09-10T17:26:44.501Z"),
"assetId" : ObjectId("5b96b7645f2b3c0101520s60")
},
{
"lsd" : {
"$id" : ObjectId("denner")
},
"stateDate" : ISODate("2018-09-10T18:26:44.501Z"),
"assetId" : ObjectId("5b96b7645f2a8c0001530f61")
},
{
"lsd" : {
"$id" : ObjectId("lermen")
}
},
"stateDate" : ISODate("2018-09-10T20:26:44.501Z"),
"assetId" : ObjectId("5b96b7645f2a8c0001530f61")
},
{
"lsd" : {
"$id" : ObjectId("floripa")
},
"stateDate" : ISODate("2018-09-10T19:26:44.501Z"),
"assetId" : ObjectId("5b96b7645f2a8c0001530f61")
}
]
I would like get max "stateDate", so I need get LSD from same row(document).
Expected result:
{
"lsd" : {
"$id" : ObjectId("lermen")
},
"stateDate" : ISODate("2018-09-10T20:26:44.501Z")
}
I tried to do:
db.getCollection('assetState').aggregate([
{
$group: {
"_id": {"assetId": "$assetId"},
"stateDate": {
"$max": "$stateDate"
},
"lsd": {$last: "$lsd"} // I tried change $max to $min and $last it din't work :(
}
]);
Result:
{
"lsd" : {
"$id" : ObjectId("floripa")
},
"stateDate" : ISODate("2018-09-10T20:26:44.501Z")
}
Many Thanks
Try this query
db.getCollection('assetState').aggregate([
{$sort:{"stateDate":-1}},
]).limit(1)
you can $sort descending before $group and gets the first item of each group with $arrayElemAt
db.getCollection('assetState').aggregate([
{ $sort: { stateDate: -1 } },
{ $group: { _id: { "assetId" : "$assetId" },
states: { $push: "$$ROOT" }
}
},
{ $project: { "last_asset": { $arrayElemAt: [ "$states", 0 ] }, _id:0 } },
])
Result:
/* 1 */
{
"last_asset" : {
"_id" : ObjectId("5db2b34fa1b70230bba9c4d9"),
"lsd" : "denner",
"stateDate" : ISODate("2018-09-10T17:26:44.501Z"),
"assetId" : "5b96b7645f2b3c0101520s60"
}
}
/* 2 */
{
"last_asset" : {
"_id" : ObjectId("5db2b34fa1b70230bba9c4db"),
"lsd" : "lermen",
"stateDate" : ISODate("2018-09-10T20:26:44.501Z"),
"assetId" : "5b96b7645f2a8c0001530f61"
}
}
You could use $unwind (aggregation)
https://docs.mongodb.com/manual/reference/operator/aggregation/unwind/
I have this for the group by phase:
{
_id: {"sourceOfEvent":"$sourceOfEvent","vehicleId":"$vehicleId"},
maxTimeOfEvent: { $max: "$timeOfEvent" }
}
So group by on 2 fields and I use the max to get the max of the timeOfEvent (which is a date)
UserDetails
{
"_id" : "5c23536f807caa1bec00e79b",
"UID" : "1",
"name" : "A",
},
{
"_id" : "5c23536f807caa1bec00e78b",
"UID" : "2",
"name" : "B",
},
{
"_id" : "5c23536f807caa1bec00e90",
"UID" : "3",
"name" : "C"
}
UserProducts
{
"_id" : "5c23536f807caa1bec00e79c",
"UPID" : "100",
"UID" : "1",
"status" : "A"
},
{
"_id" : "5c23536f807caa1bec00e79c",
"UPID" : "200",
"UID" : "2",
"status" : "A"
},
{
"_id" : "5c23536f807caa1bec00e52c",
"UPID" : "300",
"UID" : "3",
"status" : "A"
}
Groups
{
"_id" : "5bb20d7556db6915846da55f",
"members" : {
"regularStudent" : [
"200" // UPID
],
}
},
{
"_id" : "5bb20d7556db69158468878",
"members" : {
"regularStudent" : {
"0" : "100" // UPID
}
}
}
Step 1
I have to take UID from UserDetails check with UserProducts then take UPID from UserProducts
Step 2
we have to check this UPID mapped to Groups collection or not ?.
members.regularStudent we are mapped UPID
Step 3
Suppose UPID not mapped means i want to print the UPID from from UserProducts
I have tried but couldn't complete this, kindly help me out on this.
Expected Output:
["300"]
Note: Expected Output is ["300"] , because UserProducts having UPID 100 & 200 but Groups collection mapped only 100& 200.
My Code
var queryResult = db.UserDetails.aggregate(
{
$lookup: {
from: "UserProducts",
localField: "UID",
foreignField: "UID",
as: "userProduct"
}
},
{ $unwind: "$userProduct" },
{ "$match": { "userProduct.status": "A" } },
{
"$project": { "_id" : 0, "userProduct.UPID" : 1 }
},
{
$group: {
_id: null,
userProductUPIDs: { $addToSet: "$userProduct.UPID" }
}
});
let userProductUPIDs = queryResult.toArray()[0].userProductUPIDs;
db.Groups.aggregate([
{
$unwind: "$members.regularStudent"
},
{
$group: {
_id: null,
UPIDs: { $addToSet: "$members.regularStudent" }
}
},
{
$project: {
members: {
$setDifference: [ userProductUPIDs , "$UPIDs" ]
},
_id : 0
}
}
])
My Output
{
"members" : [
"300",
"100"
]
}
You need to fix that second aggregation and get all UPIDs as an array. To achieve that you can use $cond and based on $type either return an array or use $objectToArray to run the conversion, try:
db.Groups.aggregate([
{
$project: {
students: {
$cond: [
{ $eq: [ { $type: "$members.regularStudent" }, "array" ] },
"$members.regularStudent",
{ $map: { input: { "$objectToArray": "$members.regularStudent" }, as: "x", in: "$$x.v" } }
]
}
}
},
{
$unwind: "$students"
},
{
$group: {
_id: null,
UPIDs: { $addToSet: "$students" }
}
},
{
$project: {
members: {
$setDifference: [ userProductUPIDs , "$UPIDs" ]
},
_id : 0
}
}
])
I am trying to push the data who is having minSalary using $push at group aggregate.
Query:
db.users.aggregate([
{ $match: { experience: { $gte:3, $lte:10} } },
{ $group: { _id: {totalExperience:"$experience"}, "count": {$sum:1},"minSalary": {$min:"$expected_salary"}, "minUsers": {$push:"$_id"}, "maxSalary": {$max:"$expected_salary"} } },
{ $sort: { '_id.totalExperience': -1 } }
])
Result
{
"_id" : {
"totalExperience" : 9
},
"count" : 549.0,
"minSalaryCount" : 120000,
"maxSalary" : 180000
}
Also i am expecting following result
{
"_id" : {
"totalExperience" : 9
},
"count" : 549.0,
"minSalaryCount" : 120000,
"maxSalary" : 180000,
"minSalaryUsers":[
ObjectId('5355345345sdrrw234234'),
ObjectId('5355345345sdeee234234'),
ObjectId('5355345345sdertw234234')
]
}
Thank you.