mongo query/aggregation with equality inside array - mongodb

Question

I am trying to formulate a query over the sample bios collection http://docs.mongodb.org/manual/reference/bios-example-collection/:
Retrieve all the persons who received two awards on the same year.
The expected answers are "Ole-Johan Dahl" and "Kristen Nygaard" as for instance the doc for Ole-Johan Dahl is
{
"_id" : 5,
"name" : {
"first" : "Ole-Johan",
"last" : "Dahl"
},
"birth" : ISODate("1931-10-12T04:00:00Z"),
"death" : ISODate("2002-06-29T04:00:00Z"),
"contribs" : [
"OOP",
"Simula"
],
"awards" : [
{
"award" : "Rosing Prize",
"year" : 1999,
"by" : "Norwegian Data Association"
},
{
"award" : "Turing Award",
"year" : 2001,
"by" : "ACM"
},
{
"award" : "IEEE John von Neumann Medal",
"year" : 2001,
"by" : "IEEE"
}
]
}
So far, the best query that I could come up with is the following query using aggregation framework:
db.bios.aggregate([
{$project : { "first_name": "$name.first", "last_name": "$name.last" , "award1" :"$awards", "award2" :"$awards" } },
{$unwind : "$award1"},
{$unwind : "$award2"},
{$project : { "first_name": 1, "last_name": 1, "award1" : 1, "award2" : 1,
"super" : { $and : [ {$eq : ["$award1.year", "$award2.year"]},
{$lt: ["$award1.award", "$award2.award"]}
]
}}
},
{$match : {"super": true}}
])
However I am not happy with this solution because
the query projects awards twice and unwind them in the following step. This will generate quadratic many intermediate documents;
the query computes an auxiliary field "super" which is only used for filtering afterwards.
Is there a better way to formulate this query?

Answer 1

Try the following aggregation pipeline:
db.bios.aggregate([
{
"$unwind": "$awards"
},
{
"$group": {
"_id": {
"year": "$awards.year",
"firstName": "$name.first",
"lastName": "$name.last"
},
"count": { "$sum": 1 },
"award_recepients": { "$push": "$name" }
}
},
{
"$match": { "count": 2 }
},
{
"$project": {
"_id": 0,
"year": "$_id.year",
"award_recepients": 1,
"count": 1
}
}
])