How to use aggregate for group by in mongodb - mongodb

I have collection that contains documents with below schema.
Schema
{
"categoryId": "1234",
"sellerId": "2323",
"productId": "121",
"rating": 1
},
{
"categoryId": "1235",
"sellerId": "2323",
"productId": "122",
"rating": -1
},
{
"categoryId": "1234",
"sellerId": "2323",
"productId": "123",
"rating": -1
},
{
"categoryId": "1235",
"sellerId": "2323",
"productId": "124",
"rating": 1
},
{
"categoryId": "1234",
"sellerId": "2323",
"productId": "125",
"rating": 1
},
{
"categoryId": "1234",
"sellerId": "2325",
"productId": "125",
"rating": 1
}
The rating can have values 1 or -1. I want to find all documents grouped by categoryId and sum of the ratings.
Example Result:
{categoryId: 1234, positiveRatingCount: 2, negativeRatingCount: 1}
This is what I have done so far:
ratingsCollection.aggregate(
{
$match: {sellerId: "2323" }
},
{
$group: {
_id: "$categoryId",
count: { $sum: "rating" }
}
}
);
I get the following result. I am able to group by category but not able to figure out to get count of positive and negative ratings.
[
{
"_id": "1234",
"count": 3
},
{
"_id": "1235",
"count": 2
}
]

You need to use $sum with the conditions($cond) where the rating is $gt or $lt then 0
db.collection.aggregate([
{ "$match": { "sellerId": "2323" } },
{ "$group": {
"_id": "$categoryId",
"positiveRatingCount": {
"$sum": { "$cond": [{ "$gt": [ "$rating", 0 ] }, "$rating", 0 ] }
},
"negativeRatingCount": {
"$sum": { "$cond": [{ "$lt": [ "$rating", 0 ] }, "$rating", 0 ] }
}
}}
])
Output
[
{
"_id": "1235",
"negativeRatingCount": -1,
"positiveRatingCount": 1
},
{
"_id": "1234",
"negativeRatingCount": -2,
"positiveRatingCount": 3
}
]

Related

Aggregate occurrences of events in nested array

Given the following input:
[
{
"statuses": [
{
"status": "allowed",
"count": 3,
"events_count": [
"2001",
"1001",
"1001"
]
}
],
"date": "2022-09-10 15:00",
"_id": "2022-09-10 15:00"
}
]
I need count the number of occurrences of stauses.events_count, so the output would be:
[
{
"statuses": [
{
"status": "allowed",
"count": 3,
"events_count": [
{"type": "2001", "count": 1},
{"type": "1001", "count": 2},
]
}
],
"date": "2022-09-10 15:00",
"_id": "2022-09-10 15:00"
}
]
What I've tried
This is what I got so far:
db.collection.aggregate([
{
"$unwind": "$statuses"
},
{
"$unwind": "$statuses.events_count"
},
{
"$group": {
"_id": {
"event_count": "$statuses.events_count",
"status": "$statuses.status",
"date": "$date",
"count": "$statuses.count"
},
"occurences": {
"$sum": 1
}
}
}
])
Which produces:
[
{
"_id": {
"count": 3,
"date": "2022-09-10 15:00",
"event_count": "2001",
"status": "allowed"
},
"occurences": 1
},
{
"_id": {
"count": 3,
"date": "2022-09-10 15:00",
"event_count": "1001",
"status": "allowed"
},
"occurences": 2
}
]
I'm having difficulties grouping everything back together. I tried grouping by date and pushing back to a 'statuses' array, but it produces two items in the array (with status==allowed), rather than 1 item with status==allowed
You did 2 $unwinds, so it should be 2 $groups in reverse order:
{
"$group": {
"_id": {
"status": "$_id.status",
"count": "$_id.count",
"date": "$_id.date"
},
"event_count": {
"$push": {
"type": "$_id.event_count",
"count": "$occurences"
}
}
}
},
{
"$group": {
"_id": "$_id.date",
"date": {"$last": "$_id.date"},
"statuses": {
"$push": {
"status": "$_id.status",
"count": "$_id.count",
"event_count": "$event_count"
}
}
}
}

MongoDB Select By Group along with that Count Unique match exclude array and object fields Get data sort by latest objects

I have a collection where from the backend user can input multiple same name bikes but with different registration number but in front-End I want them to be grouped by matching the same name but as user updates separately display image changes but I want only one display image as it is 1 vehicle
provided there is a node created I will implement it we can sort it by the latest and take the price and image of it
Activa -2 Count
KTM -1 Count
but there is a catch.
Activa 2 bikes but I want only count 2 and the price as it is the same in an array I want only 1 and the same applies to displayimage here display image file path is different but I want the latest one only Sharing data below
Data:
[
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62e69ee3edfe4d0f3cb4994a",
"bikename": "KTM",
"bikenumber": "KA05HM2034",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-2020-honda-city-exterior-8-1659281111883.jpg",
"file_path": "https://www.example.com/images/upload/bike-2020-honda-city-exterior-8-1659281111883.jpg",
"idx": 1
}
},
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62dba8418ef8f51f454ed757",
"bikename": "Activa",
"bikenumber": "KA05HM2033",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658562557459.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658562557459.jpg",
"idx": 0
}
},
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62d7ff7e70b9ab38c6ab0cb1",
"bikename": "Activa",
"bikenumber": "KA05HM2223",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"afterfreekmprice": 22,
"descreption": "Activa",
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658322798414.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658322798414.jpg",
"idx": 0
}
}
]
Expected:
[
{
"_id":{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62dba8418ef8f51f454ed757",
"bikename": "Activa",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658562557459.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658562557459.jpg",
"idx": 0
}
},
"count": 2
},
{
"_id":{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62e69ee3edfe4d0f3cb4994a",
"bikename": "KTM",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-2020-honda-city-exterior-8-1659281111883.jpg",
"file_path": "https://www.example.com/images/upload/bike-2020-honda-city-exterior-8-1659281111883.jpg",
"idx": 1
}
}
"count": 1
}
]
You can use the aggregation pipeline,
$sort by _id in descending order
$group by bikename and get the first root document that is latest one in root and count total documents in count
$project to show required documents
db.collection.aggregate([
{ $sort: { _id: -1 } },
{
$group: {
_id: "$bikename",
root: { $first: "$$ROOT" },
count: { $sum: 1 }
}
},
{
$project: {
_id: "$root",
count: 1
}
}
])
Playground
You can use $group for this:
db.collection.aggregate([
{$group: {
_id: "$bikename",
count: {$sum: 1},
data: {$first: "$$ROOT"}
}
},
{$set: {"data.count": "$count"}},
{$replaceRoot: {newRoot: "$data"}}
])
See how it works on the playground example

Group by an optional field in mongodb

I would like to independently group the results of an or clause, including overlap. The data set is rather large so running 2 queries sequentially will result in an undesirable wait time. I am hoping I can somehow project which clause returned the corresponding data. Given this data set:
[
{
"_id": 1,
"item": "abc",
"name": "Michael",
"price": NumberDecimal("10"),
"quantity": NumberInt("2"),
"date": ISODate("2014-03-01T08:00:00Z")
},
{
"_id": 2,
"item": "jkl",
"name": "Toby",
"price": NumberDecimal("20"),
"quantity": NumberInt("1"),
"date": ISODate("2014-03-01T09:00:00Z")
},
{
"_id": 3,
"item": "xyz",
"name": "Keith",
"price": NumberDecimal("5"),
"quantity": NumberInt("10"),
"date": ISODate("2014-03-15T09:00:00Z")
},
{
"_id": 4,
"item": "abc",
"name": "Dwight",
"price": NumberDecimal("5"),
"quantity": NumberInt("20"),
"date": ISODate("2014-04-04T11:21:39.736Z")
},
{
"_id": 5,
"item": "abc",
"name": "Ryan",
"price": NumberDecimal("10"),
"quantity": NumberInt("10"),
"date": ISODate("2014-04-04T21:23:13.331Z")
},
{
"_id": 6,
"item": "def",
"name": "Jim",
"price": NumberDecimal("7.5"),
"quantity": NumberInt("5"),
"date": ISODate("2015-06-04T05:08:13Z")
},
{
"_id": 7,
"item": "abc",
"name": "Keith",
"price": NumberDecimal("7.5"),
"quantity": NumberInt("10"),
"date": ISODate("2015-09-10T08:43:00Z")
},
{
"_id": 8,
"item": "abc",
"name": "Michael",
"price": NumberDecimal("10"),
"quantity": NumberInt("5"),
"date": ISODate("2016-02-06T20:20:13Z")
},
]
I would like to receive this result:
[{
"_id": {
"name": "Keith"
},
"count": 2
},
{
"_id": {
"item": "abc",
},
"count": 5
}]
Here is what I have tried so far:
db.collection.aggregate([
{
$match: {
$or: [
{
item: "abc"
},
{
name: "Keith"
}
]
}
},
{
$group: {
_id: {
item: "$item",
name: "$name"
},
count: {
$sum: 1
}
}
}
])
You can use $facet to get multiple aggregation pipelines into the same stage in this way:
Using $facet there are two "outputs" one group by name and other by item.
In each one there are multiple stages:
First $match to process only documents you want.
Then $group with _id name or item, and $count to get the total.
db.collection.aggregate([
{
"$facet": {
"groupByName": [
{
"$match": {"name": "Keith"}
},
{
"$group": {"_id": "$name","count": {"$sum": 1}}
}
],
"groupByItem": [
{
"$match": {"item": "abc"}
},
{
"$group": {"_id": "$item","count": {"$sum": 1}}
}
]
}
}
])
Example here
The output is:
{
"groupByItem": [
{
"_id": "abc",
"count": 5
}
],
"groupByName": [
{
"_id": "Keith",
"count": 2
}
]
}
Here it is:
mongos> db.n.aggregate([ { $facet:{ names:[ {$match:{name:"Keith"}} , {$group:{_id:{name:"$name"}, count:{$sum:1}}} ] , items:[ {$match:{item:"abc"}},{ $group:{_id:{item:"$item"}, count:{$sum:1}} } ] } } , {$project:{ "namesANDitems":{$concatArrays:[ "$names","$items" ]} }} ,{$unwind:"$namesANDitems"} ,{$replaceRoot:{newRoot:"$namesANDitems"} } ]).pretty()
{ "_id" : { "name" : "Keith" }, "count" : 2 }
{ "_id" : { "item" : "abc" }, "count" : 5 }
mongos>
explained:
You create two pipes via $facet
Match in every facet pipe what you need to group pipe1=names , pipe2=items
Join the arrays from the two pipes in single array named "namesANDitems"
Convert the array to object with $unwind
Remove the temporary object name namesANDitems so you have only the two objects as requested

Group items inside of another group

I want to group the following collection by category and sum its total value, then create a subcategory attribute, on same structure, summing subcategories if more than one equal.
[
{
"_id": 1,
"date": ISODate("2019-10-10T00:00:00.000Z"),
"description": "INTERNET BILL",
"credit": "",
"debit": "-100.00",
"category": "home",
"subcategory": "internet",
"__v": 0
},
{
"_id": 2,
"date": ISODate("2019-10-10T00:00:00.000Z"),
"description": "WATER BILL",
"credit": "",
"debit": "-150.00",
"category": "home",
"subcategory": "water",
"__v": 0
},
{
"_id": 3,
"date": ISODate("2019-10-10T00:00:00.000Z"),
"description": "MC DONALDS",
"credit": "",
"debit": "-30.00",
"category": "food",
"subcategory": "restaurants",
"__v": 0
},
{
"_id": 4,
"date": ISODate("2019-10-10T00:00:00.000Z"),
"description": "BURGER KING",
"credit": "",
"debit": "-50.00",
"category": "food",
"subcategory": "restaurants",
"__v": 0
},
{
"_id": 5,
"date": ISODate("2019-10-10T00:00:00.000Z"),
"description": "WALMART",
"credit": "",
"debit": "-20.00",
"category": "food",
"subcategory": "groceries",
"__v": 0
},
]
Desireble output:
[
{
"_id": "home",
"total": "-250.00",
"subcategory" : [
{"id": "internet", "total": "-100"},
{"id": "water", "total": "-150"}
]
},
{
"_id": "food",
"total": "-100.00",
"subcategory" : [
{"id": "restaurants", "total": "-80"},
{"id": "groceries", "total": "-20"}
]
}
]
With the following query, I've almost achieved it, but I haven't find a way to sum values on subcategories.
db.getCollection('expenses').aggregate([
{$match:
{"date" : { "$gte" : new Date("11-10-2019"), "$lte": new Date("2019-10-11") }}
},
{$group: {
_id: "$category",
total: { $sum: { $toDouble: { $cond: { if: { $ne: [ "$debit", ""] }, then: "$debit", else: "$credit" } } } },
subcategories: { $addToSet: {id: "$subcategory" }},
}}
])
You can to $group twice (by subcategory first):
db.collection.aggregate([
{
$group: {
_id: { category: "$category", subcategory: "$subcategory" },
total: { $sum: { $toDouble: "$debit" } }
}
},
{
$group: {
_id: "$_id.category",
total: { $sum: "$total" },
subcategories: { $push: { id: "$_id.subcategory", total: "$total" } }
}
}
])
Mongo Playground

MongoDB, How to associate array fields for statistics

Example JSON:
{
"groups": [
{
"_id": 1,
"name": "g1"
},
{
"_id": 2,
"name": "g2"
}
],
"items": [
{
"_id": 1,
"name": "item1",
"gid": 1
},
{
"_id": 2,
"name": "item2",
"gid": 2
}
]
}
How to associate two arrays and count ?I tried to use aggregate, I didn't get the results I wanted.
Required Result:
Or can directly find all the items associated with it, perfect....
{"groups": [
{
"_id": 1,
"name": "g1",
"count": 1,
"items": [
{
"_id": 1,
"name": "item1"
}
]
},
{
"_id": 2,
"name": "g2",
"count": 1,
"items": [
{
"_id": 2,
"name": "item2"
}
]
}
]}
db.getCollection('collection').aggregate([
{$unwind:{
path:"$groups",
preserveNullAndEmptyArrays:true
}},
{$unwind:{
path:"$items",
preserveNullAndEmptyArrays:true
}},
{$redact: {$cond: [{
$eq: [
"$groups._id",
"$items.gid"
]
},
"$$KEEP",
"$$PRUNE"
]
}
},
{$project:{
_id:1,
groups_id:"$groups._id",
group_name:"$groups.name",
item_data:{
_id:"$items._id",
name:"$items.name",
}
}},
{
$group:{
_id:"$groups_id",
name:{$first:"$group_name"},
count:{$sum:1},
items:{$push:"$item_data"}
}
}
])