Mongo convert embedded document to array

Question

Is there a way to convert a nested document structure into an array? Below is an example:

Input

"experience" : {
        "0" : {
            "duration" : "3 months",
            "end" : "August 2012",
            "organization" : {
                "0" : {
                    "name" : "Bank of China",
                    "profile_url" : "http://www.linkedin.com/company/13801"
                }
            },
            "start" : "June 2012",
            "title" : "Intern Analyst"
        }
    },

Expected Output:

"experience" : [
           {
            "duration" : "3 months",
            "end" : "August 2012",
            "organization" : {
                "0" : {
                    "name" : "Bank of China",
                    "profile_url" : "http://www.linkedin.com/company/13801"
                }
            },
            "start" : "June 2012",
            "title" : "Intern Analyst"
        }
    ],

Currently I am using a script to iterate over each element, convert them to an array & finally update the document. But it is taking a lot of time, is there a better way of doing this?

Please mention your mongodb version? Also mention if you want to transform your document permanently? — Saleem
– Saleem, Commented Mar 22, 2016 at 20:53

Blakes Seven · Accepted Answer · 2016-03-26 09:02:27Z

You still need to iterate over the content, but instead you should be writing back using bulk operations:

Either for MongoDB 2.6 and greater:

var bulk = db.collection.initializeUnorderedBulkOp(),
    count = 0;

db.collection.find({ 
   "$where": "return !Array.isArray(this.experience)"
}).forEach(function(doc) {
    bulk.find({ "_id": doc._id }).updateOne({
        "$set": { "experience": [doc.experience["0"]] }
    });
    count++;

    // Write once in 1000 entries
    if ( count % 1000 == 0 ) {
        bulk.execute();    
        bulk = db.collection.initializeUnorderedBulkOp();
    }
})

// Write the remaining
if ( count % 1000 != 0 )
    bulk.execute();

Or in modern releases of MongoDB 3.2 and greater, the bulkWrite() method is preferred:

var ops = [];

db.collection.find({ 
   "$where": "return !Array.isArray(this.experience)"
}).forEach(function(doc) {
   ops.push({
       "updateOne": {
           "filter": { "_id": doc._id },
           "update": { "$set": { "experience": [doc.experience["0"]] } }
       }
   });

   if ( ops.length == 1000 ) {
       db.collection.bulkWrite(ops,{ "ordered": false })
       ops = [];
   }
})

if ( ops.length > 0 )
    db.collection.bulkWrite(ops,{ "ordered": false });

So when writing back to the database over a cursor, then bulk write operations with "unordered" set is the way to go. It's only one write/response per batch of 1000 requests, which reduces a lot of overhead, and "unordered" means that writes can happen in parallel rather than in a serial order. It all makes it faster.

whoami - fakeFaceTrueSoul · Accepted Answer · 2019-10-25 16:34:36Z

1

For mongoDB version >4.2 :

db.doc.aggregate([{ $match: {'experience.0': { $exists: false } } },
    {$project:{experience:["$experience.0"]}}, { $merge: { into: "doc", on: "_id" }
])

Note : Here we're merging the updated field/document with existing, but not replacing/updating entire document, default behavior of $merge is merge whenMatched document is found, You can pass other options like replace/keepExisting etc.

Ref: $merge

answered Oct 25, 2019 at 16:34

whoami - fakeFaceTrueSoul

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1 Comment

whoami - fakeFaceTrueSoul Over a year ago

This answer is based on @Saleem's answer, as my edit on his answer to include these new changes(features in mongoDB) was rejected & suggestion is to make this as a new answer !!

Ravi Soni · Accepted Answer · 2022-03-05 16:31:48Z

1

I am not sure, why aren't there any good answers yet.

It's super easy with aggregation "$set", set is used to add a new field. here you can add a new field with same name into an array. So it will override the older field.

Refer below example:

db.collectionName.aggregate[
   // match/other aggregations
   {$set: { "experience": ["$experience"] } }
];

answered Mar 5, 2022 at 16:31

Ravi Soni

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1 Comment

gscaparrotti Over a year ago

This should be the accepted answer. Dead simple yet effective.

Cerbrus · Accepted Answer · 2016-03-23 13:20:04Z

0

See if this query works with your MongoDB version

For MongoDB version 3.2+:

db.doc.aggregate([
    {$project:{experience:["$experience.0"]}}
])

MongoDB < 3.2:

db.doc.aggregate([
    {$group: {_id:"$_id", experience:{$push:"$experience.0"}}}
])

It should transform your document into:

{ 
    "_id" : ObjectId("56f1b046a65ea8a72c34839c"), 
    "experience" : [
        {
            "duration" : "3 months", 
            "end" : "August 2012", 
            "organization" : {
                "0" : {
                    "name" : "Bank of China", 
                    "profile_url" : "http://www.linkedin.com/company/13801"
                }
            }, 
            "start" : "June 2012", 
            "title" : "Intern Analyst"
        }
    ]
}

A better approach if you want to alter documents in collection permanently using aggregation framework.

Lets assume your collection name is doc

db.doc.aggregate([
    {$group: {_id:"$_id", experience:{$push:"$experience.0"}}},
    {$out: "doc"}
])

Query above will transform all of your documents in place.

edited Mar 23, 2016 at 13:20

Cerbrus

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answered Mar 22, 2016 at 20:57

Saleem

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1 Comment

Blakes Seven Over a year ago

It's still an incorrect approach regardless of the version. You keep repeating the same thing over and over without understanding what is wrong with it.

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