Distinct array element with condition

Question

My documents look like this:

{
    "_id": "1",
    "tags": [ 
        { "code": "01-01", "type": "machine" },
        { "code": "04-06", "type": "gearbox" },
        { "code": "07-01", "type": "machine" } 
    ]
},
{
    "_id": "2",
    "tags": [
        { "code": "03-04","type": "gearbox" },
        { "code": "01-01", "type": "machine" },
        { "code": "04-11", "type": "machine" }
    ]
}

I want to get distinct codes only for tags whose type is "machine". so, for the example above, the result should be ["01-01", "07-01", "04-11"]. How do I do this?

Possible duplicate of How to get array from mongoDB collection? — Sede
– Sede, Commented Dec 8, 2016 at 21:00

Wake · Accepted Answer · 2016-09-01 04:11:17Z

1

Using $unwind and then $group with the tag as the key will give you each tag in a separate document in your result set:

db.collection_name.aggregate([
    {
        $unwind: "$tags"
    },
    {
        $match: { 
            "tags.type": "machine" 
        }
    },
    {
        $group: {
            _id: "$tags.code"
        }
    },
    {
        $project:{
            _id:false
            code: "$_id"
        }
    }
]);

Or, if you want them put into an array within a single document, you can use $push within a second $group stage:

db.collection_name.aggregate([
    {
        $unwind: "$tags"
    },
    {
        $match: { 
            "tags.type": "machine" 
        }
    },
    {
        $group: {
            _id: "$tags.code"
        }
    },
    {
        $group:{
            _id: null,
            codes: {$push: "$_id"}
        }
    }
]);

Another user suggested including an initial stage of { $match: { "tags.type": "machine" } }. This is a good idea if your data is likely to contain a significant number of documents that do not include "machine" tags. That way you will eliminate unnecessary processing of those documents. Your pipeline would look like this:

db.collection_name.aggregate([
    {
        $match: { 
            "tags.type": "machine" 
        }
    },
    {
        $unwind: "$tags"
    },
    {
        $match: { 
            "tags.type": "machine" 
        }
    },
    {
        $group: {
            _id: "$tags.code"
        }
    },
    {
        $group:{
            _id: null,
            codes: {$push: "$_id"}
        }
    }
]);

edited Sep 1, 2016 at 4:11

answered Sep 1, 2016 at 2:27

Wake

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2 Comments

chrisbajorin Over a year ago

I'd suggest to lead off with { $match: { "tags.type": "machine" } } to get rid of any documents that won't contribute.

Wake Over a year ago

Good suggestion. Updated the answer.

yahai wan · Accepted Answer · 2016-09-01 03:58:35Z

1

> db.foo.aggregate( [
... { $unwind : "$tags" },
... { $match : { "tags.type" : "machine" } },
... { $group : { "_id" : "$tags.code" } },
... { $group : { _id : null , "codes" : {$push : "$_id"} }}
... ] )
{ "_id" : null, "codes" : [ "04-11", "07-01", "01-01" ] }

answered Sep 1, 2016 at 3:58

yahai wan

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Comments

Oded · Accepted Answer · 2016-09-29 12:25:46Z

1

A better way would be to group directly on tags.type and use addToSet on tags.code.

Here's how we can achieve the same output in 3 stages of aggregation :

db.name.aggregate([
  {$unwind:"$tags"},

  {$match:{"tags.type":"machine"}},

  {$group:{_id:"$tags.type","codes":{$addToSet:"$tags.code"}}} 
])

Output : { "_id" : "machine", "codes" : [ "04-11", "07-01", "01-01" ] }

Also, if you wish to filter out tag.type codes, we just need to replace "machine" in match stage with desired tag.type.

edited Sep 29, 2016 at 12:25

Oded

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answered Sep 1, 2016 at 8:19

inaitgaJ

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