Mongodb aggregation $group followed by $limit for pagination - mongodb

In MongoDB aggregation pipeline, record flow from stage to stage happens one/batch at a time (or) will wait for the current stage to complete for whole collection before passing it to next stage?
For e.g., I have a collection classtest with following sample records
{name: "Person1", marks: 20}
{name: "Person2", marks: 20}
{name: "Person1", marks: 20}
I have total 1000 records for about 100 students and I have following aggregate query
db.classtest.aggregate(
[
{$sort: {name: 1}},
{$group: {_id: '$name',
total: {$sum: '$marks'}}},
{$limit: 5}
])
I have following questions.
The sort order is lost in final results. If I place another sort after $group, then results are sorted properly. Does that mean $group not maintains the previous sort order?
I would like to limit the results to 5. Does group operation has to be completely done (for all 1000 records) before passing to the limit. (or) The group operation passes the records to limit stage as and when it has record and stops processing when the requirement for limit stage is met?
My actual idea is to do pagination on results of aggregate. In above scenario, if $group maintains sort order and processes only required number of records, I want to apply $match condition {$ge: 'lastPersonName'} in subsequent page queries.
I do not want to apply $limit before $group as I want results for 5 students not first 5 records.
I may not want to use $skip as that means effectively traversing those many records.

I have solved the problem without need of maintaining another collection or even without $group traversing whole collection, hence posting my own answer.
As others have pointed:
$group doesn't retain order, hence early sorting is not of much help.
$group doesn't do any optimization, even if there is a following $limit, i.e., runs $group on entire collection.
My usecase has following unique features, which helped me to solve it:
There will be maximum of 10 records per each student (minimum of 1).
I am not very particular on page size. The front-end capable of handling varying page sizes.
The following is the aggregation command I have used.
db.classtest.aggregate(
[
{$sort: {name: 1}},
{$limit: 5 * 10},
{$group: {_id: '$name',
total: {$sum: '$marks'}}},
{$sort: {_id: 1}}
])
Explaining the above.
if $sort immediately precedes $limit, the framework optimizes the amount of data to be sent to next stage. Refer here
To get a minimum of 5 records (page size), I need to pass at least 5 (page size) * 10 (max records per student) = 50 records to the $group stage. With this, the size of final result may be anywhere between 0 and 50.
If the result is less than 5, then there is no further pagination required.
If the result size is greater than 5, there may be chance that last student record is not completely processed (i.e., not grouped all the records of student), hence I discard the last record from the result.
Then name in last record (among retained results) is used as $match criteria in subsequent page request as shown below.
db.classtest.aggregate(
[
{$match: {name: {$gt: lastRecordName}}}
{$sort: {name: 1}},
{$limit: 5 * 10},
{$group: {_id: '$name',
total: {$sum: '$marks'}}},
{$sort: {_id: 1}}
])
In above, the framework will still optimize $match, $sort and $limit together as single operation, which I have confirmed through explain plan.

The first few things to consider here is that the aggregation framework works with a "pipeline" of stages to be applied in order to get a result. If you are familiar with processing things on the "command line" or "shell" of your operating system, then you might have some experience with the "pipe" or | operator.
Here is a common unix idiom:
ps -ef | grep mongod | tee "out.txt"
In this case the output of the first command here ps -ef is being "piped" to the next command grep mongod which in turn has it's output "piped" to the tee out.txt which both outputs to terminal as well as the specified file name. This is a "pipeline" wher each stage "feeds" the next, and in "order" of the sequence they are written in.
The same is true of the aggregation pipeline. A "pipeline" here is in fact an "array", which is an ordered set of instructions to be passed in processing the data to a result.
db.classtest.aggregate([
{ "$group": {
"_id": "$name",
"total": { "$sum": "$marks"}
}},
{ "$sort": { "name": 1 } },
{ "$limit": 5 }
])
So what happens here is that all of the items in the collection are first processed by $group to get their totals. There is no specified "order" to grouping so there is not much sense in pre-ordering the data. Neither is there any point in doing so because you are yet to get to your later stages.
Then you would $sort the results and also $limit as required.
For your next "page" of data you will want ideally $match on the last unique name found, like so:
db.classtest.aggregate([
{ "$match": { "name": { "$gt": lastNameFound } }},
{ "$group": {
"_id": "$name",
"total": { "$sum": "$marks"}
}},
{ "$sort": { "name": 1 } },
{ "$limit": 5 }
])
It's not the best solution, but there really are not alternatives for this type of grouping. It will however notably get "faster" with each iteration towards the end. Alternately, storing all the unqiue names ( or reading that out of another collection ) and "paging" through that list with a "range query" on each aggregation statement may be a viable option, if your data permits it.
Something like:
db.classtest.aggregate([
{ "$match": { "name": { "$gte": "Allan", "$lte": "David" } }},
{ "$group": {
"_id": "$name",
"total": { "$sum": "$marks"}
}},
{ "$sort": { "name": 1 } },
])
Unfortunately there is not a "limit grouping up until x results" option, so unless you can work with another list, then you are basically grouping up everything ( and possibly a a gradually smaller set each time ) with each aggregation query you send.

"$group does not order its output documents." See http://docs.mongodb.org/manual/reference/operator/aggregation/group/
$limit limits the number of processed elements of an immediately preceding $sort operation, not only the number of elements passed to the next stage. See the note at http://docs.mongodb.org/manual/reference/operator/aggregation/limit/
For the very first question you asked, I am not sure, but it appears (see 1.) that a stage n+1 can influence the behaviour of stage n : the limit will limit the sort operation to its first n elements, and the sort operation will not complete just as if the following limit stage did not exist.

pagination on group data mongodb -
in $group items you can't directly apply pagination, but below trick will be used ,
if you want pagination on group data -
for example- i want group products categoryWise and then i want only 5 product per category then
step 1 - write aggregation on product table, and write groupBY
{ $group: { _id: '$prdCategoryId', products: { $push: '$$ROOT' } } },
step 2 - prdSkip for skipping , and limit for limiting data , pass it
dynamically
{
$project: {
// pagination for products
products: {
$slice: ['$products', prdSkip, prdLimit],
}
}
},
finally query looks like -
params - limit , skip - for category pagination
and prdSkip and PrdLimit for products pagination
db.products.aggregate([
{ $group: { _id: '$prdCategoryId', products: { $push: '$$ROOT' } } },
{
$lookup: {
from: 'categories',
localField: '_id',
foreignField: '_id',
as: 'categoryProducts',
},
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: [{ $arrayElemAt: ['$categoryProducts', 0] }, '$$ROOT'],
},
},
},
{
$project: {
// pagination for products
products: {
$slice: ['$products', prdSkip, prdLimit],
},
_id: 1,
catName: 1,
catDescription: 1,
},
},
])
.limit(limit) // pagination for category
.skip(skip);
I used replaceRoot here to pullOut category.

Related

How to get best 5 results in $group method in mongodb?

On production server I use mongodb 4.4
I have a query that works well
db.step_tournaments_results.aggregate([
{ "$match": { "tournament_id": "6377f2f96174982ef89c48d2" } },
{ "$sort": { "total_points": -1, "time_spent": 1 } },
{
$group: {
_id: "$club_name",
'total_points': { $sum: "$total_points"},
'time_spent': { $sum: "$time_spent"}
},
},
])
But the problem is in $group operator, because it sums all the points of every group for total_points, but I need only best 5 of every group. How to achieve that?
Query
like your query, match and sort
on group instead of sum, gather all members inside one array
(i collected the $ROOT but you can collect only the 2 fields you need inside a {}, if the documents have many fields)
take the first 5 of them
take the 2 sums you need from the first 5
remove the temp fields
*with mongodb 6, you can do this in the group, without need to collect th members in an array, in mongodb 5 you can also do those with window-fields without group, but for mongodb 4.4 i think this is a way to do it
aggregate(
[{"$match": {"tournament_id": {"$eq": "6377f2f96174982ef89c48d2"}}},
{"$sort": {"total_points": -1, "time_spent": 1}},
{"$group": {"_id": "$club_name", "group-members": {"$push": "$$ROOT"}}},
{"$set":
{"first-five": {"$slice": ["$group-members", 5]},
"group-members": "$$REMOVE"}},
{"$set":
{"total_points": {"$sum": "$first-five.total_points"},
"time_spent": {"$sum": "$first-five.time_spent"},
"first-five": "$$REMOVE"}}])

How to access overall document count during arithmetic aggregation expression

I have a collection of documents in this format:
{
_id: ObjectId,
items: [
{
defindex: number,
...
},
...
]
}
Certain parts of the schema not relevant are omitted, and each item defindex within the items array is guaranteed to be unique for that array. The same defindex can occur in different documents' items fields, but will only occur once in each respective array if present.
I currently call $unwind upon the items field, followed by $sortByCount upon items.defindex to get a sorted list of items with the highest count.
I now want to add a new field to this final sorted list using $set called usage, that shows the item's usage as a percentage of the initial number of total documents in the collection.
(i.e. if the item's count is 1300, and the overall document count pre-$unwind was 2600, the usage value will be 0.5)
My initial plan was to use $facet upon the initial collection, creating a document as so:
{
total: number (achieved using $count),
documents: [{...}] (achieved using an empty $set)
}
And then calling $unwind on the documents field to add the total document count to each document. Calculating the usage value is then trivial using $set, since the total count is a field in the document itself.
This approach ran into memory issues though, since my collection is far larger than the 16MB limit.
How would I solve this?
One way to do it is use $setWindowFields:
db.collection.aggregate([
{
$setWindowFields: {
output: {
totalCount: {$count: {}}
}
}
},
{
$unwind: "$items"
},
{
$group: {
_id: "$items.defindex",
count: {$sum: 1},
totalCount: {$first: "$totalCount"}
}
},
{
$project: {
count: 1,
usage: {$divide: ["$count", "$totalCount"]
}
}
},
{$sort: {count: -1}}
])
As you can see here

Find documents based on a property values progressively in MongoDB

Assume I have a collection with many documents which they have a property called "status". Status accepts any Int value. I want to find all documents that have status with value "1". If there are not any, find all documents that have status with value "2" and so on... Is there any solution to do such action in a single query?
That is possible.
If you create an index on {status:1}, you can run a query with limit:1 and sort:{status:1}, and project to return only the status field (excluding _id). This would be a covered query that is quite efficient and should only examine a single index key.
Then use that value to query for matching status. This query would also use the index to minimize the number of document examined.
The difference between doing this in 2 queries vs 1 is likely small.
You could peform both in an aggregation:
db.collection.aggregate([
{$sort: {status: 1}},
{$limit: 1},
{$project: {
_id: 0,
status: 1
}},
{$lookup: {
as: "matched",
from: "collection",
let: {target: "$status"},
pipeline: [
{$match: {
$expr: {
$eq: [
"$status",
"$$target"
]
}
}}
]
}},
{$unwind: "$matched"},
{$replaceRoot: {newRoot: "$matched"}}
])
It is not completely clear whether the $lookup part will be able to use the index, so you should test to see if that actually performs better than running 2 queries from the client.
Playground

How can I limit the result in a category using moongose

I have a collection in mongodb, the collection has a field displayInCategories. The collection contains 1000's of data wrt to different displayInCategories.
Is it possible to limit the records to <=5 for all the available displayInCategories.
I didn't want to limit the record on whole result, I need to limit the record as per the displayInCategories
This might get you going:
db.collection.aggregate([{
$group: {
_id: "$displayInCategories", // group by displayInCategories
"docs": { $push: "$$ROOT" } // remember all documents for this category
}
}, {
$project: {
"docs": { $slice: [ "$docs", 5 ] } // limit the items in each "docs" array to 5
}
}])
You might want to apply a $sort stage at the start to make sure you don't get random documents but rather the "top 5" based on some criteria.

Mongodb how to aggregate the number of occurencies(count) of distinct values?

I have a set with 2m hashtags. However only around 200k are distinct values. I want to know wich hashtags are more repeated on my data.
I used this to find how many times each hashtag is repeated on my dataset:
db.hashtags.aggregate([{ "$group": {"_id": "$hashtag","count": { "$sum": 1 }}}]);
However, I would like to save the values in a distinct collection only with the unique values and its correspondency number of occurency.
How should I do that?
Please, if possible provide me some information in order that I can UNDERSTAND how to do it not only the code.
Thank you.
You can use the $out pipeline operator to write the output of the aggregation to another collection.
db.hashtags.aggregate([
{ "$group": {"_id": "$hashtag", "count": { "$sum": 1 }}},
{ "$out": "newcoll" }
]);
Note that this feature was added in MongoDB 2.6
Using the aggregation framework the following will, for hashtag with multiple records, return the duplicate hashtag and the corresponding record count:
db.hashtags.aggregate([
{
$group: {
_id: "$hashtag",
count: { $sum: 1 }
}
},
{ $match: { count: { $gt: 1 } } },
{ $sort : { count : -1} },
{ $limit : 200 },
{ $out: "duphashtags" }
])
The $sum operator adds up the values of the fields passed to it, in this case the constant 1 - thereby counting the number of grouped records into the count field. The $match filters documents with a count greater than 1, i.e. duplicates. $sort sorts the most frequent duplicates first, and limit the results to the top 200. The $out operator writes the documents returned by the aggregation pipeline to a specified collection, say "duphashtags".