Reusing/splitting $match results in a mongo aggregation - mongodb

I have a mongo collection that stores data on page views such as location, user type (e.g. admin, user) and time spent on page. I want to use $match to get a subset of the documents and then use $group to group them by state and also by user type. The $match is rather expensive, so I was wondering if there was a way through the aggregation pipeline to somehow reuse the $match and get two sets of grouped data rather than needing to run two aggregates.
Current js pseudocode:
groupedByState = Views.aggregate([
$match: { ... },
$group: {
_id: '$state',
secondsViewed: { $avg: '$seconds_viewed' },
},
])
groupedByUserType = Views.aggregate([
$match: { ... },
$group: {
_id: '$user_type',
secondsViewed: { $avg: '$seconds_viewed' },
},
])

You can use the $facet aggregation operator to perform multiple $group operations against the output of the $match:
Views.aggregate([
$match: { ... },
$facet: {
byState: [{
$group: {
_id: '$state',
secondsViewed: { $avg: '$seconds_viewed' }
}
}],
byUserType: [
$group: {
_id: '$user_type',
secondsViewed: { $avg: '$seconds_viewed' }
}
}]
}
])

Related

How to filter array (of objects) inside one document in mongo db based on some condition

I have the below docs collection structure.
I'm able to filter the documnents with various approaches, but not able to filter the array inside the documents.
{
"_id": "",
"employee": {
"EmployeeAttributeValues": {
"EmployeeAttributeValue": [
{.....
},
{.....
},
{.....
},
{.....
}
]
}
}
}
Kindly help me on how to filter the MemberAttributeValue array based on some condition.
you can use $where operator for custom filtering
https://docs.mongodb.com/v4.2/reference/operator/query/where/
db.test.aggregate([
{ $match: {_id: <ID>}},
{ $unwind: '$<ARRAY>'},
{ $match: {'<ARRAY>.a': {$gt: 3}}},
{ $group: {_id: '$_id', list: {$push: '$<ARRAY>.a'}}}
])

Count nested wildcard array mongodb query

I have the following data of users and model cars:
[
{
"user_id":"ebebc012-082c-4e7f-889c-755d2679bdab",
"car_1a58db0b-5449-4d2b-a773-ee055a1ab24d":1,
"car_37c04124-cb12-436c-902b-6120f4c51782":0,
"car_b78ddcd0-1136-4f45-8599-3ce8d937911f":1
},
{
"user_id":"f3eb2a61-5416-46ba-bab4-459fbdcc7e29",
"car_1a58db0b-5449-4d2b-a773-ee055a1ab24d":1,
"car_0d15eae9-9585-4f49-a416-46ff56cd3685":1
}
]
I want to see how many users have a car_ with the value 1 using mongodb, something like:
{"car_1a58db0b-5449-4d2b-a773-ee055a1ab24d": 2}
For this example.
The issue is that I will never know how are the fields car_ are going to be, they will have a random structure (wildcard).
Notes:
car_id and user_id are at the same level.
The car_id is not given, I simply want to know for the entire database which are the most commmon cars_ with value 1.
$group by _id and convert root object to array using $objectToArray,
$unwind deconstruct root array
$match filter root.v is 1
$group by root.k and get total count
db.collection.aggregate([
{
$group: {
_id: "$_id",
root: { $first: { $objectToArray: "$$ROOT" } }
}
},
{ $unwind: "$root" },
{ $match: { "root.v": 1 } },
{
$group: {
_id: "$root.k",
count: { $sum: 1 }
}
}
])
Playground

MongoDB aggregation: How to get the index of a document in a collection depending sorted by a document property

Assume I have a collection with millions of documents. Below is a sample of how the documents look like
[
{ _id:"1a1", points:[2,3,5,6] },
{ _id:"1a2", points:[2,6] },
{ _id:"1a3", points:[3,5,6] },
{ _id:"1b1", points:[1,5,6] },
{ _id:"1c1", points:[5,6] },
// ... more documents
]
I want to query a document by _id and return a document that looks like below:
{
_id:"1a1",
totalPoints: 16,
rank: 29
}
I know I can query the whole document, sort by descending order then get the index of the document I want by _id and add one to get its rank. But I have worries about this method.
If the documents are in millions won't this be 'overdoing' it. Querying a whole collection just to get one document? Is there a way to achieve what I want to achieve without querying the whole collection? Or the whole collection has to be involved because of the ranking?
I cannot save them ranked because the points keep on changing. The actual code is more complex but the take away is that I cannot save them ranked.
Total points is the sum of the points in the points array. The rank is calculated by sorting all documents in descending order. The first document becomes rank 1 and so on.
an aggregation pipeline like the following can get the result you want. but how it operates on a collection of millions of documents remains to be seen.
db.collection.aggregate(
[
{
$group: {
_id: null,
docs: {
$push: { _id: '$_id', totalPoints: { $sum: '$points' } }
}
}
},
{
$unwind: '$docs'
},
{
$replaceWith: '$docs'
},
{
$sort: { totalPoints: -1 }
},
{
$group: {
_id: null,
docs: { $push: '$$ROOT' }
}
},
{
$set: {
docs: {
$map: {
input: {
$filter: {
input: '$docs',
as: 'x',
cond: { $eq: ['$$x._id', '1a3'] }
}
},
as: 'xx',
in: {
_id: '$$xx._id',
totalPoints: '$$xx.totalPoints',
rank: {
$add: [{ $indexOfArray: ['$docs._id', '1a3'] }, 1]
}
}
}
}
}
},
{
$unwind: '$docs'
},
{
$replaceWith: '$docs'
}
])

MongoDB use $last with $cond

Is there any way to get the last non-zero value in an aggregation. Is that possible?
Scenario:
I have an events collection, in which I store all the events from users. I want to fetch the list of users with last purchased items cost is $1.99 and logged in at least once last week.
My events collection will have records like
{_id:ObjectId("58af54d5ab7df73d71822708"),uid:1,event:"login"}
{_id:ObjectId("58db7189296fdedde1c04bc1"),uid:2,event:"login"}
{_id:ObjectId("5888419bfa4b69dc4af7c76c"),uid:2,event:"purchase",amount:3}
{_id:ObjectId("5888419bfa4b69dc4af7d45c"),uid:1,event:"purchase",amount:1.9}
{_id:ObjectId("5888819bfa4b69dc4af7c76c"),uid:1,event:"custom",type:3,value:2}
What am trying to do:
db.events.aggregate([{
{
$group: {
_id: uid,
last_login: {
$max: {
$cond: [{
$eq: ['$event', 'login']
}, '$_id', 0]
}
},
last_amount: {
$last: {
$cond: [{
$eq: ['$event', 'login']
}, '$_id', 0]
}
}
}
}
}, {
$match: {
last_purchase: {
$gte: ObjectId("58af54d50000000000000000")
},
last_amount: 1.9
}
}])
which obviously will fail because last will have 0 as the last item.
The output am expecting is
{_id:1,last_login:_id:ObjectId("58af54d5ab7df73d71822708"),last_amount:1.9}
The query is system generated. Please help.
To answer your question, you cannot modify the behaviour of $last using $cond. i.e. 'not the last entry because of a criteria'.
There are few alternatives that you could try:
Alter your document schema for the access pattern, especially if this is a frequently used query in your application. You should utilise the schema as a leverage to optimise your application queries.
Use one of the MongoDB supported drivers to process the expected outcome.
Depending on the use case, you could execute 2 separate aggregation queries; first query all the users uid that logged in at least once last week, and second to query only those users that have the last purchased item value of $1.99.
I found a workaround.Instead of $last, I used $push in the $group and added $slice with $setDifference in $project to remove null values. The query will now look something like
db.events.aggregate([{
{
$group: {
_id: uid,
last_login: {
$max: {
$cond: [{
$eq: ['$event', 'login']
}, '$_id', 0]
}
},
last_amount: {
$push: {
$cond: [{
$eq: ['$event', 'login']
}, '$_id', null]
}
}
}
}
},
{$project:{last_login:{$slice:[{$setDifference,'$last_login',[null]},-1,1]}}}
, {
$match: {
last_purchase: {
$gte: ObjectId("58af54d50000000000000000")
},
last_amount: 1.9
}
}])

mongodb aggregation framework group + project

I have the following issue:
this query return 1 result which is what I want:
> db.items.aggregate([ {$group: { "_id": "$id", version: { $max: "$version" } } }])
{
"result" : [
{
"_id" : "b91e51e9-6317-4030-a9a6-e7f71d0f2161",
"version" : 1.2000000000000002
}
],
"ok" : 1
}
this query ( I just added projection so I can later query for the entire document) return multiple results. What am I doing wrong?
> db.items.aggregate([ {$group: { "_id": "$id", version: { $max: "$version" } }, $project: { _id : 1 } }])
{
"result" : [
{
"_id" : ObjectId("5139310a3899d457ee000003")
},
{
"_id" : ObjectId("513931053899d457ee000002")
},
{
"_id" : ObjectId("513930fd3899d457ee000001")
}
],
"ok" : 1
}
found the answer
1. first I need to get all the _ids
db.items.aggregate( [
{ '$match': { 'owner.id': '9e748c81-0f71-4eda-a710-576314ef3fa' } },
{ '$group': { _id: '$item.id', dbid: { $max: "$_id" } } }
]);
2. then i need to query the documents
db.items.find({ _id: { '$in': "IDs returned from aggregate" } });
which will look like this:
db.items.find({ _id: { '$in': [ '1', '2', '3' ] } });
( I know its late but still answering it so that other people don't have to go search for the right answer somewhere else )
See to the answer of Deka, this will do your job.
Not all accumulators are available in $project stage. We need to consider what we can do in project with respect to accumulators and what we can do in group. Let's take a look at this:
db.companies.aggregate([{
$match: {
funding_rounds: {
$ne: []
}
}
}, {
$unwind: "$funding_rounds"
}, {
$sort: {
"funding_rounds.funded_year": 1,
"funding_rounds.funded_month": 1,
"funding_rounds.funded_day": 1
}
}, {
$group: {
_id: {
company: "$name"
},
funding: {
$push: {
amount: "$funding_rounds.raised_amount",
year: "$funding_rounds.funded_year"
}
}
}
}, ]).pretty()
Where we're checking if any of the funding_rounds is not empty. Then it's unwind-ed to $sort and to later stages. We'll see one document for each element of the funding_rounds array for every company. So, the first thing we're going to do here is to $sort based on:
funding_rounds.funded_year
funding_rounds.funded_month
funding_rounds.funded_day
In the group stage by company name, the array is getting built using $push. $push is supposed to be part of a document specified as the value for a field we name in a group stage. We can push on any valid expression. In this case, we're pushing on documents to this array and for every document that we push it's being added to the end of the array that we're accumulating. In this case, we're pushing on documents that are built from the raised_amount and funded_year. So, the $group stage is a stream of documents that have an _id where we're specifying the company name.
Notice that $push is available in $group stages but not in $project stage. This is because $group stages are designed to take a sequence of documents and accumulate values based on that stream of documents.
$project on the other hand, works with one document at a time. So, we can calculate an average on an array within an individual document inside a project stage. But doing something like this where one at a time, we're seeing documents and for every document, it passes through the group stage pushing on a new value, well that's something that the $project stage is just not designed to do. For that type of operation we want to use $group.
Let's take a look at another example:
db.companies.aggregate([{
$match: {
funding_rounds: {
$exists: true,
$ne: []
}
}
}, {
$unwind: "$funding_rounds"
}, {
$sort: {
"funding_rounds.funded_year": 1,
"funding_rounds.funded_month": 1,
"funding_rounds.funded_day": 1
}
}, {
$group: {
_id: {
company: "$name"
},
first_round: {
$first: "$funding_rounds"
},
last_round: {
$last: "$funding_rounds"
},
num_rounds: {
$sum: 1
},
total_raised: {
$sum: "$funding_rounds.raised_amount"
}
}
}, {
$project: {
_id: 0,
company: "$_id.company",
first_round: {
amount: "$first_round.raised_amount",
article: "$first_round.source_url",
year: "$first_round.funded_year"
},
last_round: {
amount: "$last_round.raised_amount",
article: "$last_round.source_url",
year: "$last_round.funded_year"
},
num_rounds: 1,
total_raised: 1,
}
}, {
$sort: {
total_raised: -1
}
}]).pretty()
In the $group stage, we're using $first and $last accumulators. Right, again we can see that as with $push - we can't use $first and $last in project stages. Because again, project stages are not designed to accumulate values based on multiple documents. Rather they're designed to reshape documents one at a time. Total number of rounds is calculated using the $sum operator. The value 1 simply counts the number of documents passed through that group together with each document that matches or is grouped under a given _id value. The project may seem complex, but it's just making the output pretty. It's just that it's including num_rounds and total_raised from the previous document.