Get top 50 records for a certain value w/ mongo and meteor - mongodb

Question

In my meteor project, I have a leaderboard of sorts, where it shows players of every level on a chart, spread across every level in the game. For simplicitys sake, lets say there are levels 1-100. Currently, to avoid overloading meteor, I just tell the server to send me every record newer than two weeks old, but that's not sufficient for an accurate leaderboard.
What I'm trying to do is show 50 records representing each level. So, if there are 100 records at level 1, 85 at level 2, 65 at level 3, and 45 at level 4, I want to show the latest 50 records from each level, making it so I would have [50, 50, 50, 45] records, respectively.
The data looks something like this:
{
snapshotDate: new Date(),
level: 1,
hp: 5000,
str: 100
}
I think this requires some mongodb aggregation, but I couldn't quite figure out how to do this in one query. It would be trivial to do it in two, though - select all records, group by level, sort each level by date, then take the last 50 records from each level. However, I would prefer to do it in one operation, if I could. Is it currently possible to do something like this?

Answer 1

Currently there is no way to pick up n top records of a group, in the aggregation pipeline. There is an unresolved open ticket regarding this: https://jira.mongodb.org/browse/SERVER-9377.
There are two solutions to this:
Keep your document structure as it is now and aggregate, but,
grab the n top records and slice off the remaining records for each group, in the client side.
Code:
var top_records = [];
db.collection.aggregate([
// The sort operation needs to come before the $group,
// because once the records are grouped by level,
// there exists only one document per group.
{$sort:{"snapshotDate":-1}},
// Maintain all the records in an array in sorted order.
{$group:{"_id":"$level","recs":{$push:"$$ROOT"}}},
],{allowDiskUse: true}).forEach(function(level){
level.recs.splice(50); //Keep only the top 50 records.
top_records.push(level);
})
Remember that this loads all the documents for each level and removes the unwanted records in the client side.
Alter your document structure to accomplish what you really need. If
you only need the top n records always, keep them always in sorted
order in the root document.This is accomplished using a sorted capped array.
Your document would look like this:
{
level:1,
records:[{snapshotDate:2,hp:5000,str:100},
{snapshotDate:1,hp:5001,str:101}]
}
where, records is an capped array of size n and always has sub documents sorted in descending order of their snapshotDate.
To make the records array work that way, we always perform an update operation when we need to insert documents to it for any level.
db.collection.update({"level":1},
{$push:{
recs:{
$each:[{snapshotDate:1,hp:5000,str:100},
{snapshotDate:2,hp:5001,str:101}],
$sort:{"snapshotDate":-1},
$slice:50 //Always trim the array size to 50.
}
}},{upsert:true})
What this does is, is always keeps the size of the records array to 50 and always sorts the records whenever new sub documents are inserted at a level.
A simple find, db.collection.find({"level":{$in:[1,2,..]}}), would give you the top 50 records in order, for each selected level.