MongoDB - Aggregation - To get unique items in array

Question

Here's my MongoDB collection:

{
    "_id" : ObjectId("515d8f53175b8ecb053425c2"),
    "category" : "Batteries",
    "products" : [
        {
            "brand" : "Duracell",
            "item" : [
                "AA",
                "AAA"
            ]
        },
        {
            "brand" : "Everyday",
            "item" : [
                "9V",
                "AA",
                "12V"
            ]
        }
    ]
}

The output that I need is

1) Unique list of all items

{["AA", "AAA", "9V", "12V"]}

and 2. unique list of items per product

{
    "category" : "Batteries",
    "item": ["AA", "AAA", "9V", "12V"]
}

I'm very new to MongoDB, and I tried different aggregations functions and nothing seems to work. Please help.

score 34 · Accepted Answer · edited Aug 04 '13 at 15:42

34

After few more tries, I had solved this. Here's the commands:

db.xyz.aggregate( {$project: {a: '$products.item'}}, 
    {$unwind: '$a'}, 
    {$unwind: '$a'}, 
    {$group: {_id: 'a', items: {$addToSet: '$a'}}});

and

db.xyz.aggregate( {$project: {category: 1, a: '$products.item'}}, 
    {$unwind: '$a'}, 
    {$unwind: '$a'}, 
    {$group: {_id: '$category', items: {$addToSet: '$a'}}});

edited Aug 04 '13 at 15:42

Yasin Okumuş

2,299
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answered Apr 04 '13 at 23:32

Ananth

767
1
7
15

7

Why are you $unwinding twice ? – Devesh Apr 05 '13 at 03:08
9

because 'item' is nested in an array. – Ananth Apr 06 '13 at 01:38
If you only have 1 item in an array, you can use `$first` instead of `unwind`. It returns the first item from an array and is much cheaper than $unwind. Use it in combination with `$addFields`. E.g {$addFields: {item: {$first: 'products.item'}}} Now you can unwind "item" and use it in the "$group" stage. – Emanuel Lindström Nov 24 '22 at 12:43

score 9 · Answer 2 · answered May 12 '20 at 03:40

After mongodb3.4, there is a $reduce operator, so we can flat a array without extra stage.

1.

col.aggregate([
  {
    $project: {
      items: {
        $reduce: {
          input: "$products.items",
          initialValue: [],
          in: { $concatArrays: ["$$value", "$$this"] },
        },
      },
    },
  },
  { $unwind: "$items" },
  { $group: { _id: null, items: { $addToSet: "$items" } } },
]);

2.

col.aggregate([
  {
    $project: {
      category: 1,
      items: {
        $setUnion: {
          $reduce: {
            input: "$products.items",
            initialValue: [],
            in: { $concatArrays: ["$$value", "$$this"] },
          },
        },
      },
    },
  },
]);

score 2 · Answer 3 · answered Aug 03 '20 at 12:06

2

I know it is an old question and you've solved it several years ago! But there is a small problem in the answer you've marked as correct and it may not suitable for all cases. The $unwind is an expensive operator and may affect latency and memory consumption for large datasets. I think the $reduce operator is more performant in this case.

answered Aug 03 '20 at 12:06

Mostafa Lavaei

1,960
1
18
27

Perhaps you could add a demonstration of how to use reduce to get a _unique set_ of items in this case? – ProgrammingLlama May 17 '22 at 09:06

Devesh · Answer 4 · 2013-04-04T15:03:09.877

0

I am not sure what you all you have tried in the aggregation function but i thought unwind will help you to do the same , assuming you are not able to get it done , we have a map-reduce which will allow you to easily do this one . You can look into the http://docs.mongodb.org/manual/applications/map-reduce/ . It allow you to get the data in a manner you want and you can easily get the list . I think $unwind on the tags column and then $group them will always give the us the list of distinct tags as required by you in 1 and for 2nd case create $group on two key category and item which was $unwind earlier.

edited Apr 04 '13 at 15:03

answered Apr 04 '13 at 14:57

Devesh

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1
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Devesh, thanks for your response. I was able to solve this with just aggregation. Posted my answer as well. – Ananth Apr 04 '13 at 23:33

score 0 · Answer 5 · answered Jan 17 '23 at 09:03

I know this is an old question but I would like to show an easier way of doing it! A setDifference function takes two sets and returns an array containing the elements that only exist in the first set. It ignores duplicate entries while doing it.

Therefore, I trick this by using an empty array at second variables.

Full Code

db.xyz.aggregate([
{
    $match: { 
        _id: ObjectId("515d8f53175b8ecb053425c2"),
        category: "Batteries"
    }
},
{
    $set: { 
        item: { $setDifference: ["$products.item", []] }
    }
}
])

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). — Community, Jan 18 '23 at 10:16

MongoDB - Aggregation - To get unique items in array

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