Scala / MongoDB - removing duplicate - mongodb

Question

I have seen very similar questions with solutions to this problem, but I am unsure how I would incorporate it in to my own query. I'm programming in Scala and using a MongoDB Aggregates "framework".
val getItems = Seq (
Aggregates.lookup(Store...)...
Aggregates.lookup(Store.STORE_NAME, "relationship.itemID", "uniqueID", "item"),
Aggregates.unwind("$item"),
// filter duplicates here ?
Aggregates.lookup(Store.STORE_NAME, "item.content", "ID", "content"),
Aggregates.unwind("$content"),
Aggregates.project(Projections.fields(Projections.include("store", "item", "content")))
)
The query returns duplicate objects which is undesirable. I would like to remove these. How could I go about incorporating Aggregates.group and "$addToSet" to do this? Or any other reasonable solution would be great too.
Note: I have to omit some details about the query, so the store lookup aggregate is not there. However, I want to remove the duplicates later in the query so it hopefully shouldn't matter.
Please let me know if I need to provide more information.
Thanks.
EDIT: 31/ 07/ 2019: 13:47
I have tried the following:
val getItems = Seq (
Aggregates.lookup(Store...)...
Aggregates.lookup(Store.STORE_NAME, "relationship.itemID", "uniqueID", "item"),
Aggregates.unwind("$item"),
Aggregates.group("$item.itemID,
Accumulators.first("ID", "$ID"),
Accumulators.first("itemName", "$itemName"),
Accumulators.addToSet("item", "$item")
Aggregates.unwind("$items"),
Aggregates.lookup(Store.STORE_NAME, "item.content", "ID", "content"),
Aggregates.unwind("$content"),
Aggregates.project(Projections.fields(Projections.include("store", "items", "content")))
)
But my query now returns zero results instead of the duplicate result.

Answer 1

You can use $first to remove the duplicates.
Suppose I have the following data:
[
{"_id": 1,"item": "ABC","sizes": ["S","M","L"]},
{"_id": 2,"item": "EFG","sizes": []},
{"_id": 3, "item": "IJK","sizes": "M" },
{"_id": 4,"item": "LMN"},
{"_id": 5,"item": "XYZ","sizes": null
}
]
Now, let's aggregate it using $first and $unwind and see the difference:
First let's aggregate it using $first
db.collection.aggregate([
{ $sort: {
item: 1
}
},
{ $group: {
_id: "$item",firstSize: {$first: "$sizes"}}}
])
Output
[
{"_id": "XYZ","firstSize": null},
{"_id": "ABC","firstSize": ["S","M","L" ]},
{"_id": "IJK","firstSize": "M"},
{"_id": "EFG","firstSize": []},
{"_id": "LMN","firstSize": null}
]
Now, Let's aggregate it using $unwind
db.collection.aggregate([
{
$unwind: "$sizes"
}
])
Output
[
{"_id": 1,"item": "ABC","sizes": "S"},
{"_id": 1,"item": "ABC","sizes": "M"},
{"_id": 1,"item": "ABC","sizes": "L},
{"_id": 3,"item": "IJK","sizes": "M"}
]
You can see $first removes the duplicates where as $unwind keeps the duplicates.
Using $unwind and $first together.
db.collection.aggregate([
{ $unwind: "$sizes"},
{
$group: {
_id: "$item",firstSize: {$first: "$sizes"}}
}
])
Output
[
{"_id": "IJK", "firstSize": "M"},
{"_id": "ABC","firstSize": "S"}
]

Answer 2

group then addToSet is an effective way to deal with your problem !
it looks like this in mongoshell
db.sales.aggregate(
[
{
$group:
{
_id: { day: { $dayOfYear: "$date"}, year: { $year: "$date" } },
itemsSold: { $addToSet: "$item" }
}
}
]
)
in scala you can do it like
Aggregates.group("$groupfield", Accumulators.addToSet("fieldName","$expression"))
if you have multiple field to group
Aggregates.group(new BasicDBObject().append("fieldAname","$fieldA").append("fieldBname","$fieldB")), Accumulators.addToSet("fieldName","expression"))
then unwind