How to get item ranking in list sorted by multiple fields in Mongoose

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醉酒成梦
醉酒成梦 2021-02-09 07:14

I have a number of user records (> 10000) in a MongoDB collection which can be sorted by score desc + time asc + bonus desc. How can I get the ranking of one user in the list ac

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  • 2021-02-09 08:14

    Count the number of users that come before this user in your sort order. I'll start with the case of a simple (non-compound sort) because the query in the compound case is more complicated, even though the idea is exactly the same.

    > db.test.drop()
    > for (var i = 0; i < 10; i++) db.test.insert({ "x" : i })
    > db.test.find({ }, { "_id" : 0 }).sort({ "x" : -1 }).limit(5)
    { "x" : 9 }
    { "x" : 8 }
    { "x" : 7 }
    { "x" : 6 }
    { "x" : 5 }
    

    For this order, the ranking of a document { "x" : i } is the number of documents { "x" : j } with i < j

    > var rank = function(id) {
        var i = db.test.findOne({ "_id" : id }).x
        return db.test.count({ "x" : { "$gt" : i } })
    }
    > var id = db.test.findOne({ "x" : 5 }).id
    > rank(id)
    4
    

    The ranking will be based at 0. Similarly, if you want to compute the rank for the document { "x" : i } in the sort { "x" : 1 }, you would count the number of docs { "x" : j } with i > j.

    For a compound sort, the same procedure works, but it is trickier to implement because the order in a compound index is lexicographic, i.e., for the sort { "a" : 1, "b" : 1}, (a, b) < (c, d) if a < c or a = c and b < d, so we need a more complicated query to express this condition. Here's an example for a compound index:

    > db.test.drop()
    > for (var i = 0; i < 3; i++) {
        for (var j = 0; j < 3; j++) {
            db.test.insert({ "x" : i, "y" : j })
        }
    }
    > db.test.find({}, { "_id" : 0 }).sort({ "x" : 1, "y" : -1 })
    { "x" : 0, "y" : 2 }
    { "x" : 0, "y" : 1 }
    { "x" : 0, "y" : 0 }
    { "x" : 1, "y" : 2 }
    { "x" : 1, "y" : 1 }
    { "x" : 1, "y" : 0 }
    { "x" : 2, "y" : 2 }
    { "x" : 2, "y" : 1 }
    { "x" : 2, "y" : 0 }
    

    To find the rank for the document { "x" : i, "y" : j }, you need to find the number of documents { "x" : a, "y" : b } in the order { "x" : 1, "y" : -1 } such that (i, j) < (a, b). Given the sort specification, this is equivalent to the condition i < a or i = a and j > b:

    > var rank = function(id) {
        var doc = db.test.findOne(id)
        var i = doc.x
        var j = doc.y
        return db.test.count({
            "$or" : [
                { "x" : { "$lt" : i } },
                { "x" : i, "y" : { "$gt" : j } }
            ]
        })
    }
    > id = db.test.findOne({ "x" : 1, "y" : 1 })._id
    > rank(id)
    4
    

    Finally, in your case of a three-part compound index

    { "score" : -1, "time" : 1, "bonus" : -1 }
    

    the rank function would be

    > var rank = function(id) {
        var doc = db.test.findOne(id)
        var score = doc.score
        var time = doc.time
        var bonus = doc.bonus
        return db.test.count({
            "$or" : [
                { "score" : { "$gt" : score } },
                { "score" : score, "time" : { "$lt" : time } },
                { "score" : score, "time" : time, "bonus" : { "$gt" : bonus } }
            ]
        })
    }
    
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