Organize outcomes based on the size of the array

Question

Organize outcomes based on the size of the array

I have a collection called 'article' and I need to sort the objects based on the size of the array it contains. What is the most efficient way to achieve this? One option is to retrieve the entire list of objects and manually sort them in JavaScript. However, since I am returning articles 10 at a time, sorting the array every time the API is called seems like unnecessary work.

Article.find({})
     .limit(10)
     .skip(req.params.page*10)
     //Possibly using $project to create a new variable 'votecount' that counts objects in a given array.
     .sort({votecount:-1})
     .exec(function(err,arts){
      articleObj.articles = arts;
        if (arts.length<10){
          articleObj.reachedEnd = true;
        }
        res.json(articleObj);
     });

I also need to calculate the number of votes received up. Here is an example object:

{
"_id" : ObjectId("55f50cfddcf1ad6931fb8dd4"),
"timestamp" : "2015-09-13T00:58:57-5:00",
"url" : "http://www.nytimes.com/2015/09/13/sports/floyd-mayweather-finishes-bout-and-maybe-his-career-with-lopsided-win-over-andre-berto.html",
"abstract" : "Mayweather’s victory by unanimous decision gave him a record of 49-0, the same as the legendary heavyweight Rocky Marciano.",
"title" : "Mayweather Wins Easily in What He Calls Last Bout",
"section" : "Sports",
"comments" : [ ],
"votes" : {
    "up" : [
        ObjectId("55e5e16934d355d61c471e48")
    ],
    "down" : [ ]
},
"image" : {
    "caption": "Floyd Mayweather Jr. after learning he defeated Andre Berto in a unanimous decision.",
    "url": "http://static01.nyt.com/images/2015/09/14/sports/13fight/13fight-mediumThreeByTwo210.jpg"
},
"__v" : 0
}

javascript mongodb mongoose mongodb-query aggregation-framework

Answer 1

Answer №1

To properly sort the documents based on the size of an array field, you should use the .aggregate() method and project the $size of the array into the document for sorting:

Article.aggregate(
    [
        { "$project": {
            "timestamp": 1,
            "url": 1,
            "abstract": 1,
            "title": 1,
            "section": 1,
            "comments": 1,
            "votes": 1,
            "image": 1,
            "voteCount": { 
                "$subtract": [
                    { "$size": "$votes.up" },
                    { "$size": "$votes.down" }
                ]
            }
        }},
        { "$sort": { "voteCount": -1 } },
        { "$skip": req.params.page*10 },
        { "$limit": 10 },
    ],
    function(err,results) {
        // results here
    }
);

It's important to note that this method may have a performance cost as it requires calculating the size on every iteration. A better approach would be to maintain the vote count within the document using Bulk Operations for efficiency across all scenarios:

var bulk = Article.collection.intializeOrderedBulkOp();

// Update downvote if present
bulk.find({ 
    "_id": id, 
    "votes.up": { "$ne": userId },
    "votes.down": userId
}).updateOne({
    "$push": { "votes.up": userId },
    "$pull": { "votes.down": userId }
    "$inc": { "voteCount": 2 }
});

// Add upvote if not present
bulk.find({ 
    "_id": id, 
    "votes.up": { "$ne": userId },
    "votes.down": { "$ne": userId }
}).updateOne({
    "$push": { "votes.up": userId },
    "$inc": { "voteCount": 1 }
});

bulk.execute(function(err,response) {
    // Handle response here
});

Consider implementing the opposite process for downvoting.

The key benefit is that the "count" or "score" stays updated with each vote processed, eliminating the need to recalculate when accessing the data again. This approach ensures optimal performance in handling vote counts.

Answer 2

To properly sort the documents based on the size of an array field, you should use the .aggregate() method and project the $size of the array into the document for sorting:

Article.aggregate(
    [
        { "$project": {
            "timestamp": 1,
            "url": 1,
            "abstract": 1,
            "title": 1,
            "section": 1,
            "comments": 1,
            "votes": 1,
            "image": 1,
            "voteCount": { 
                "$subtract": [
                    { "$size": "$votes.up" },
                    { "$size": "$votes.down" }
                ]
            }
        }},
        { "$sort": { "voteCount": -1 } },
        { "$skip": req.params.page*10 },
        { "$limit": 10 },
    ],
    function(err,results) {
        // results here
    }
);

It's important to note that this method may have a performance cost as it requires calculating the size on every iteration. A better approach would be to maintain the vote count within the document using Bulk Operations for efficiency across all scenarios:

var bulk = Article.collection.intializeOrderedBulkOp();

// Update downvote if present
bulk.find({ 
    "_id": id, 
    "votes.up": { "$ne": userId },
    "votes.down": userId
}).updateOne({
    "$push": { "votes.up": userId },
    "$pull": { "votes.down": userId }
    "$inc": { "voteCount": 2 }
});

// Add upvote if not present
bulk.find({ 
    "_id": id, 
    "votes.up": { "$ne": userId },
    "votes.down": { "$ne": userId }
}).updateOne({
    "$push": { "votes.up": userId },
    "$inc": { "voteCount": 1 }
});

bulk.execute(function(err,response) {
    // Handle response here
});

Consider implementing the opposite process for downvoting.

The key benefit is that the "count" or "score" stays updated with each vote processed, eliminating the need to recalculate when accessing the data again. This approach ensures optimal performance in handling vote counts.

Organize outcomes based on the size of the array

Answer №1

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