Sorting based on the number of elements in a filtered subdocument array

Question

Sorting based on the number of elements in a filtered subdocument array

I have a MongoDB collection structured like this:

{
    {
        "_id": ObjectId,
        "user_id": Number,
        "updates": [
            {
                "_id": ObjectId,
                "mode": Number,
                "score": Number
            },
            {
                "_id": ObjectId,
                "mode": Number,
                "score": Number
            },
            {
                "_id": ObjectId,
                "mode": Number,
                "score": Number
            }
        ]
    }
}

My goal is to identify users with the highest number of updates per mode. For example, if I specify mode 0, I want to retrieve users based on the maximum number of updates with mode: 0.

Can MongoDB handle this type of query? While speed isn't a top priority, since the data will be cached and processed asynchronously for an extended period.

javascript mongodb mongoose mongodb-query aggregation-framework

Answer 1

Answer №1

To increase efficiency, consider storing a count for each "mode" in the document as an additional field. This way, you can easily sort based on that count:

var update = { 
   "$push": { "updates": updateDoc },
};

var countDoc = {};
countDoc["counts." + updateDoc.mode] = 1;

update["$inc"] = countDoc;

Model.update(
    { "_id": id },
    update,
    function(err,numAffected) {

    }
);

This approach utilizes $inc to increment a "counts" field corresponding to each "mode" value when a new entry is added to the "updates" array. Sorting becomes efficient with this set-up:

Model.find({ "updates.mode": 0 }).sort({ "counts.0": -1 }).exec(function(err,users) {

});

If storing a count field is not feasible, another option is to calculate at query time using .aggregate():

Model.aggregate(
    [
        { "$match": { "updates.mode": 0 } },
        { "$project": {
            "user_id": 1,
            "updates": 1,
            "count": {
                "$size": {
                    "$setDifference": [
                        { "$map": {
                            "input": "$updates",
                            "as": "el",
                            "in": {
                                "$cond": [
                                    { "$eq": [ "$$el.mode", 0 ] },
                                    "$$el",
                                    false
                                ]
                            }
                        }},
                        [false]
                    ]
                }
            }
        }},
        { "$sort": { "count": -1 } }
    ],
    function(err,results) {

    }
);

While querying and calculating in real-time through .aggregate() is effective, it may not be as fast as having a pre-calculated count field.

The $map operator allows inline processing of array elements, combined with $cond for validation. Utilizing $setDifference ensures efficient filtering without unnecessary computation, unlike $unwind.

In conclusion, storing a count field is recommended for faster performance and potential index utilization.

Answer 2

To increase efficiency, consider storing a count for each "mode" in the document as an additional field. This way, you can easily sort based on that count:

var update = { 
   "$push": { "updates": updateDoc },
};

var countDoc = {};
countDoc["counts." + updateDoc.mode] = 1;

update["$inc"] = countDoc;

Model.update(
    { "_id": id },
    update,
    function(err,numAffected) {

    }
);

This approach utilizes $inc to increment a "counts" field corresponding to each "mode" value when a new entry is added to the "updates" array. Sorting becomes efficient with this set-up:

Model.find({ "updates.mode": 0 }).sort({ "counts.0": -1 }).exec(function(err,users) {

});

If storing a count field is not feasible, another option is to calculate at query time using .aggregate():

Model.aggregate(
    [
        { "$match": { "updates.mode": 0 } },
        { "$project": {
            "user_id": 1,
            "updates": 1,
            "count": {
                "$size": {
                    "$setDifference": [
                        { "$map": {
                            "input": "$updates",
                            "as": "el",
                            "in": {
                                "$cond": [
                                    { "$eq": [ "$$el.mode", 0 ] },
                                    "$$el",
                                    false
                                ]
                            }
                        }},
                        [false]
                    ]
                }
            }
        }},
        { "$sort": { "count": -1 } }
    ],
    function(err,results) {

    }
);

While querying and calculating in real-time through .aggregate() is effective, it may not be as fast as having a pre-calculated count field.

The $map operator allows inline processing of array elements, combined with $cond for validation. Utilizing $setDifference ensures efficient filtering without unnecessary computation, unlike $unwind.

In conclusion, storing a count field is recommended for faster performance and potential index utilization.

Answer 3

Answer №2

It appears that this question may be a duplicate of the following:

Querying MongoDB for longest arrays inside an object

The solution provided in the accepted answer seems to address your query.

db.collection.aggregate( [
  { $unwind : "$values" },
  { $group : { _id : "$_id", count : { $sum : 1 } } },
  { $sort : { count : -1 } }
] )

Simply update "$values" in the code with "$entries".

[update:] Additionally, if you do not wish to limit the results to 25, you can remove the line { $limit : 25 } from the code snippet.

Answer 4

It appears that this question may be a duplicate of the following:

Querying MongoDB for longest arrays inside an object

The solution provided in the accepted answer seems to address your query.

db.collection.aggregate( [
  { $unwind : "$values" },
  { $group : { _id : "$_id", count : { $sum : 1 } } },
  { $sort : { count : -1 } }
] )

Simply update "$values" in the code with "$entries".

[update:] Additionally, if you do not wish to limit the results to 25, you can remove the line { $limit : 25 } from the code snippet.

Sorting based on the number of elements in a filtered subdocument array

Answer №1

Answer №2

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