Remove duplicate in MongoDB - mongodb

Question

I have a collection with the field called "contact_id".
In my collection I have duplicate registers with this key.
How can I remove duplicates, resulting in just one register?
I already tried:
db.PersonDuplicate.ensureIndex({"contact_id": 1}, {unique: true, dropDups: true})
But did not work, because the function dropDups is no longer available in MongoDB 3.x
I'm using 3.2

Answer 1

Yes, dropDups is gone for good. But you can definitely achieve your goal with little bit effort.
You need to first find all duplicate rows and then remove all except first.
db.dups.aggregate([{$group:{_id:"$contact_id", dups:{$push:"$_id"}, count: {$sum: 1}}},
{$match:{count: {$gt: 1}}}
]).forEach(function(doc){
doc.dups.shift();
db.dups.remove({_id : {$in: doc.dups}});
});
As you see doc.dups.shift() will remove first _id from array and then remove all documents with remaining _ids in dups array.
script above will remove all duplicate documents.

Answer 2

this is a good pattern for mongod 3+ that also ensures that you will not run our of memory which can happen with really big collections. You can save this to a dedup.js file, customize it, and run it against your desired database with: mongo localhost:27017/YOURDB dedup.js
var duplicates = [];
db.runCommand(
{aggregate: "YOURCOLLECTION",
pipeline: [
{ $group: { _id: { DUPEFIELD: "$DUPEFIELD"}, dups: { "$addToSet": "$_id" }, count: { "$sum": 1 } }},
{ $match: { count: { "$gt": 1 }}}
],
allowDiskUse: true }
)
.result
.forEach(function(doc) {
doc.dups.shift();
doc.dups.forEach(function(dupId){ duplicates.push(dupId); })
})
printjson(duplicates); //optional print the list of duplicates to be removed
db.YOURCOLLECTION.remove({_id:{$in:duplicates}});

Answer 3

We can also use an $out stage to remove duplicates from a collection by replacing the content of the collection with only one occurrence per duplicate.
For instance, to only keep one element per value of x:
// > db.collection.find()
// { "x" : "a", "y" : 27 }
// { "x" : "a", "y" : 4 }
// { "x" : "b", "y" : 12 }
db.collection.aggregate(
{ $group: { _id: "$x", onlyOne: { $first: "$$ROOT" } } },
{ $replaceWith: "$onlyOne" }, // prior to 4.2: { $replaceRoot: { newRoot: "$onlyOne" } }
{ $out: "collection" }
)
// > db.collection.find()
// { "x" : "a", "y" : 27 }
// { "x" : "b", "y" : 12 }
This:
$groups documents by the field defining what a duplicate is (here x) and accumulates grouped documents by only keeping one (the $first found) and giving it the value $$ROOT, which is the document itself. At the end of this stage, we have something like:
{ "_id" : "a", "onlyOne" : { "x" : "a", "y" : 27 } }
{ "_id" : "b", "onlyOne" : { "x" : "b", "y" : 12 } }
$replaceWith all existing fields in the input document with the content of the onlyOne field we've created in the $group stage, in order to find the original format back. At the end of this stage, we have something like:
{ "x" : "a", "y" : 27 }
{ "x" : "b", "y" : 12 }
$replaceWith is only available starting in Mongo 4.2. With prior versions, we can use $replaceRoot instead:
{ $replaceRoot: { newRoot: "$onlyOne" } }
$out inserts the result of the aggregation pipeline in the same collection. Note that $out conveniently replaces the content of the specified collection, making this solution possible.

Answer 4

maybe it be a good try to create a tmpColection, create unique index, then copy data from source, and last step will be swap names?
Other idea, I had is to get doubled indexes into array (using aggregation) and then loop thru calling the remove() method with the justOne parameter set to true or 1.
var itemsToDelete = db.PersonDuplicate.aggregate([
{$group: { _id:"$_id", count:{$sum:1}}},
{$match: {count: {$gt:1}}},
{$group: { _id:1, ids:{$addToSet:"$_id"}}}
])
and make a loop thru ids array
makes this sense for you?

Answer 5

I have used this approach:
Take the mongo dump of the particular collection.
Clear that collection
Add a unique key index
Restore the dump using mongorestore.