Find all duplicate documents in a MongoDB collection by a key field

Question

Suppose I have a collection with some set of documents. something like this.

{ \"_id\" : ObjectId(\"4f127fa55e7242718200002d\"), \"id\":1, \"name\" : \"foo\"         


        

Answer 1

The accepted answer is terribly slow on large collections, and doesn't return the _ids of the duplicate records.

Aggregation is much faster and can return the _ids:

db.collection.aggregate([
  { $group: {
    _id: { name: "$name" },   // replace `name` here twice
    uniqueIds: { $addToSet: "$_id" },
    count: { $sum: 1 } 
  } }, 
  { $match: { 
    count: { $gte: 2 } 
  } },
  { $sort : { count : -1} },
  { $limit : 10 }
]);

In the first stage of the aggregation pipeline, the $group operator aggregates documents by the name field and stores in uniqueIds each _id value of the grouped records. The $sum operator adds up the values of the fields passed to it, in this case the constant 1 - thereby counting the number of grouped records into the count field.

In the second stage of the pipeline, we use $match to filter documents with a count of at least 2, i.e. duplicates.

Then, we sort the most frequent duplicates first, and limit the results to the top 10.

This query will output up to $limit records with duplicate names, along with their _ids. For example:

{
  "_id" : {
    "name" : "Toothpick"
},
  "uniqueIds" : [
    "xzuzJd2qatfJCSvkN",
    "9bpewBsKbrGBQexv4",
    "fi3Gscg9M64BQdArv",
  ],
  "count" : 3
},
{
  "_id" : {
    "name" : "Broom"
  },
  "uniqueIds" : [
    "3vwny3YEj2qBsmmhA",
    "gJeWGcuX6Wk69oFYD"
  ],
  "count" : 2
}