Removing duplicates in mongodb with aggregate query

Question

db.games.aggregate([ 
{ $unwind : "$rounds"},
{ $match: {
 "rounds.round_values.gameStage": "River",
 "rounds.round_values.decision": "BetPlus" }
 },
 { $project: {"FinalFundsChange":1, "GameID":1}
    }])

The resulting output is:

{ "_id" : ObjectId("57cbce66e281af12e4d0731f"), "GameID" : "229327202", "FinalFundsChange" : 0.8199999999999998 }
{ "_id" : ObjectId("57cbe2fce281af0f34020901"), "FinalFundsChange" : -0.1599999999999997, "GameID" : "755030199" }
{ "_id" : ObjectId("57cbea3ae281af0f340209bc"), "FinalFundsChange" : 0.10000000000000009, "GameID" : "231534683" }
{ "_id" : ObjectId("57cbee43e281af0f34020a25"), "FinalFundsChange" : 1.7000000000000002, "GameID" : "509975754" }
{ "_id" : ObjectId("57cbee43e281af0f34020a25"), "FinalFundsChange" : 1.7000000000000002, "GameID" : "509975754" }

As you can see the last element is a duplicate, that's because the unwind creates two elements of it, which it should. How can I (while keeping the aggregate structure of the query) keep the first element of the duplicate or keep the last element of the duplicate only?

I have seen that the ways to do it seem to be related to either $addToSet or $setUnion (any details how this works exactly are appreciated as well), but I don't understand how I can choose the 'subset' by which I want to identify the duplicates (in my case that's the 'GameID', other values are allowed to be different) and how I can select whether I want the first or the last element.

Answer 1

You could group by _id via $group and then use the $last and $first operator respectively to keep the last or first values.

db.games.aggregate([ 
{ $unwind : "$rounds"},
{ $match: {
 "rounds.round_values.gameStage": "River",
 "rounds.round_values.decision": "BetPlus" }
 },
 { $group: { 
     _id: "$_id", 
     "FinalFundsChange": { $first: "$FinalFundsChange" }, 
     "GameID": { $last: "$GameID" }
   }
 }
])