Created
July 12, 2016 11:19
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Delete duplicates from mongodb
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`dropDups: true` option is not available in 3.0. | |
I have solution with aggregation framework for collecting duplicates and then removing in one go. | |
It might be somewhat slower than system level "index" changes. But it is good by considering way you want to remove duplicate documents. | |
a. Remove all documents in one go | |
var duplicates = []; | |
db.collectionName.aggregate([ | |
{ $match: { | |
name: { "$ne": '' } // discard selection criteria | |
}}, | |
{ $group: { | |
_id: { name: "$name"}, // can be grouped on multiple properties | |
dups: { "$addToSet": "$_id" }, | |
count: { "$sum": 1 } | |
}}, | |
{ $match: { | |
count: { "$gt": 1 } // Duplicates considered as count greater than one | |
}} | |
]) // You can display result until this and check duplicates | |
// If your result getting response in "result" then use else don't use ".result" in query | |
.result | |
.forEach(function(doc) { | |
doc.dups.shift(); // First element skipped for deleting | |
doc.dups.forEach( function(dupId){ | |
duplicates.push(dupId); // Getting all duplicate ids | |
} | |
) | |
}) | |
// If you want to Check all "_id" which you are deleting else print statement not needed | |
printjson(duplicates); | |
// Remove all duplicates in one go | |
db.collectionName.remove({_id:{$in:duplicates}}) | |
b. You can delete documents one by one. | |
db.collectionName.aggregate([ | |
// discard selection criteria, You can remove "$match" section if you want | |
{ $match: { | |
source_references.key: { "$ne": '' } | |
}}, | |
{ $group: { | |
_id: { source_references.key: "$source_references.key"}, // can be grouped on multiple properties | |
dups: { "$addToSet": "$_id" }, | |
count: { "$sum": 1 } | |
}}, | |
{ $match: { | |
count: { "$gt": 1 } // Duplicates considered as count greater than one | |
}} | |
]) // You can display result until this and check duplicates | |
// If your result getting response in "result" then use else don't use ".result" in query | |
.result | |
.forEach(function(doc) { | |
doc.dups.shift(); // First element skipped for deleting | |
db.collectionName.remove({_id : {$in: doc.dups }}); // Delete remaining duplicates | |
}) |
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From stackoverflow URL: http://stackoverflow.com/a/33364353