How to remove duplicates based on a key in Mongodb?

Question

I have a collection in MongoDB where there are around (~3 million records). My sample record would look like,

 { \"_id\" = ObjectId(\"50731xxxxxxxxxxxxxxxxxx         


        

Answer 1

I had a similar requirement but I wanted to retain the latest entry. The following query worked withmy collections with millions of records and duplicates.

/** Create a array to store all duplicate records ids*/
var duplicates = [];

/** Start Aggregation pipeline*/
db.collection.aggregate([
  {
    $match: { /** Add any filter here. Add index for filter keys*/
      filterKey: {
        $exists: false
      }
    }
  },
  {
    $sort: { /** Sort it in such a way that you want to retain first element*/
      createdAt: -1
    }
  },
  {
    $group: {
      _id: {
        key1: "$key1", key2:"$key2" /** These are the keys which define the duplicate. Here document with same value for key1 and key2 will be considered duplicate*/
      },
      dups: {
        $push: {
          _id: "$_id"
        }
      },
      count: {
        $sum: 1
      }
    }
  },
  {
    $match: {
      count: {
        "$gt": 1
      }
    }
  }
],
{
  allowDiskUse: true
}).forEach(function(doc){
  doc.dups.shift();
  doc.dups.forEach(function(dupId){
    duplicates.push(dupId._id);
  })
})

/** Delete the duplicates*/
var i,j,temparray,chunk = 100000;
for (i=0,j=duplicates.length; i