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
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 } }
]
})
}