MongoDB - sort by subdocument match - mongodb

Question

Say I have a users collection in MongoDB. A typical user document contains a name field, and an array of subdocuments, representing the user's characteristics. Say something like this:
{
"name": "Joey",
"characteristics": [
{
"name": "shy",
"score": 0.8
},
{
"name": "funny",
"score": 0.6
},
{
"name": "loving",
"score": 0.01
}
]
}
How can I find the top X funniest users, sorted by how funny they are?
The only way I've found so far, was to use the aggregation framework, in a query similar to this:
db.users.aggregate([
{$project: {"_id": 1, "name": 1, "characteristics": 1, "_characteristics": '$characteristics'}},
{$unwind: "$_characteristics"},
{$match: {"_characteristics.name": "funny"}},
{$sort: {"_characteristics.score": -1}},
{$limit: 10}
]);
Which seems to be exactly what I want, except for the fact that according to MongoDB's documentation on using indexes in pipelines, once I call $project or $unwind in an aggregation pipeline, I can no longer utilize indexes to match or sort the collection, which renders this solution somewhat unfeasible for a very large collection.

Answer 1

I think you are half way there. I would do
db.users.aggregate([
{$match: { 'characteristics.name': 'funny' }},
{$unwind: '$characteristics'},
{$match: {'characteristics.name': 'funny'}},
{$project: {_id: 0, name: 1, 'characteristics.score': 1}},
{$sort: { 'characteristics.score': 1 }},
{$limit: 10}
])
I add a match stage to get rid of users without the funny attribute (which can be easily indexed).
unwind and match again to keep only the certain part of the data
keep only the necessary data with project
sort by the correct score
and limit the results.
that way you can use an index for the first match.
The way I see it, if the characteristics you are interested about are not too many, IMO it would be better to have your structure as
{
"name": "Joey",
"shy": 0.8
"funny": 0.6
"loving": 0.01
}
That way you can use an index (sparse or not) to make your life easier!