MongoDB - Aggregation - To get unique items in array

Question

Here\'s my MongoDB collection:

{
    \"_id\" : ObjectId(\"515d8f53175b8ecb053425c2\"),
    \"category\" : \"Batteries\",
    \"products\" : [
        {
              


        

Answer 1

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.

Answer 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.

Answer 3

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"] },
          },
        },
      },
    },
  },
]);

Answer 4

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'}}});