Sort and Group in one MongoDB aggregation query - mongodb

Question

Using $sort and $group in one aggregation query behaving strangely.
Test data:
db.createCollection("test");
db.test.insert({
ts : 100,
category : 1
});
db.test.insert({
ts : 80,
category : 1
});
db.test.insert({
ts : 60,
category : 2
});
db.test.insert({
ts : 40,
category : 3
});
So when sorting it by ts all looks good, but when I use both $sort and $group result goes in a wrong order. Query:
db.test.aggregate([
{
$sort : {ts: 1}
},
{
$group:{"_id":"$category"}
}
]);
And the result in reverse order:
{ "_id" : 1 }
{ "_id" : 2 }
{ "_id" : 3 }
Is it Mongo feature or my misunderstanding? Maby mongo firstly applied grouping and then can't sort by absent field. For this reason probably mongoose prohibits use distinct with sorting.

Answer 1

You need to first $group and $sort the result. Since you only want the _id field you will need the $project stage.
db.test.aggregate(
[
{ "$group": { "_id": "$category" }},
{ "$sort" : { "ts": 1 }},
{ "$project": { "_id": 1 }}
]
);

Answer 2

If you want to sort the other way, do it like this:
db.test.aggregate([
{
$sort : {ts: -1}
},
{
$group:{"_id":"$category"}
}
]);
Notice the - in front of the 1.

Answer 3

When you first $sort by ts, you are basically sorting all the elements from your collection. Thus, if you were to only run the $sort stage in the aggregation pipeline, you would get the following result:
//Query
db.test.aggregate([
{ $sort: { ts: 1} }
]);
//Output
{ "_id" : ObjectId("55141da6e4c260ae9e00832b"), "ts" : 40, "category" : 3 }
{ "_id" : ObjectId("55141d9fe4c260ae9e00832a"), "ts" : 60, "category" : 2 }
{ "_id" : ObjectId("55141d99e4c260ae9e008329"), "ts" : 80, "category" : 1 }
{ "_id" : ObjectId("55141d93e4c260ae9e008328"), "ts" : 100, "category" : 1 }
In your code, when you add the $group stage, you are basically grouping the above results by the category field, producing the output that you get:
{ "_id" : 1 }
{ "_id" : 2 }
{ "_id" : 3 }
In the end it all depends on what you are trying to achieve.
If you want to return the categories filtered by the ts field, you should only use the $sort stage and then manipulate the resulting data set:
var data = db.test.aggregate([
{$sort: { ts: 1}},
{$project: {
_id: 0,
ts: 1,
category: 1
}
}
]).toArray();
for(var i = 0; i < data.length; i++) {
console.log(data[i].category); //Output 3, 2, 1 in that sequence, on different lines
}