Delete all but one duplicate from a mongo db - mongodb

Question

So I mad the mistake and saved a lot of doduments twice because I messed up my document id. Because I did a Insert, i multiplied my documents everytime I saved them. So I want to delete all duplicates except the first one, that i wrote. Luckilly the documents have an implicit unique key (match._id) and I should be able to tell what the first one was, because I am using the object id.
The documents look like this:
{
_id: "5e8e2d28ca6e660006f263e6"
match : {
_id: 2345
...
}
...
}
So, right now I have a aggregation that tells me what elements are duplicated and stores them in a collection. There is for sure a more elegant way, but I am still learning.
[{$sort: {"$_id": 1},
{$group: {
_id: "$match._id",
duplicateIds: {$push: "$_id"},
count: {$sum: 1}
}},
{$match: {
count: { $gt: 1 }
}}, {$addFields: {
deletableIds: { $slice: ["$duplicateIds", 1, 1000 ] }
}},
{$out: 'DeleteableIds'}]
Now I do not know how to proceed further, as it does not seem to have a "delete" operation in aggregations and I do not want to write those temp data to a db just so I can write a delete command with that, as I want to delete them in one go. Is there any other way to do this? I am still learning with mongodb and feel a little bit overwhelmed :/

Answer 1

Rather than doing all of those you can just pick first document in group for each _id: "$match._id" & make it as root document. Also, I don't think you need to do sorting in your case :
db.collection.aggregate([
{
$group: {
_id: "$match._id",
doc: {
$first: "$$ROOT"
}
}
},
{
$replaceRoot: {
newRoot: "$doc"
}
}, {$out: 'DeleteableIds'}
])
Test : MongoDB-Playground

Answer 2

I think you're on the right track, however, to delete the duplicates you've found you can use a bulk write on the collection.
So if we imagine you aggregation query saved the following in the the DeleteableIds collection
> db.DeleteableIds.insertMany([
... {deletableIds: [1,2,3,4]},
... {deletableIds: [103,35,12]},
... {deletableIds: [345,311,232,500]}
... ]);
We can now take them and write a bulk write command:
const bulkwrite = db.DeleteableIds.find().map(x => ({ deleteMany : { filter: { _id: { $in: x.deletableIds } } } }))
then we can execute that against the database.
> db.collection1.bulkWrite(bulkwrite)
this will then delete all the duplicates.