Multiple mongodb aggregations in one operation

Question

I have a collection of items with the following documents.

{ "item" : "i1", "category" : "c1", "brand" : "b1" }  
{ "item" : "i2", "category" : "c2", "brand" : "b1" }  
{ "item" : "i3", "category" : "c1", "brand" : "b2" }  
{ "item" : "i4", "category" : "c2", "brand" : "b1" }  
{ "item" : "i5", "category" : "c1", "brand" : "b2" }  

I want to divide the aggregation results → count by category, count by brand. Please note that this does not count (category, brand)

I can do this using map-reduce using the following code.

map = function(){
    emit({type:"category",category:this.category},1);
    emit({type:"brand",brand:this.brand},1);
}
reduce = function(key, values){
    return Array.sum(values)
}
db.item.mapReduce(map,reduce,{out:{inline:1}})

And the result

{
        "results" : [
                {
                        "_id" : {
                                "type" : "brand",
                                "brand" : "b1"
                        },
                        "value" : 3
                },
                {
                        "_id" : {
                                "type" : "brand",
                                "brand" : "b2"
                        },
                        "value" : 2
                },
                {
                        "_id" : {
                                "type" : "category",
                                "category" : "c1"
                        },
                        "value" : 3
                },
                {
                        "_id" : {
                                "type" : "category",
                                "category" : "c2"
                        },
                        "value" : 2
                }
        ],
        "timeMillis" : 21,
        "counts" : {
                "input" : 5,
                "emit" : 10,
                "reduce" : 4,
                "output" : 4
        },
        "ok" : 1,
}

I can get the same results by running two different aggregation commands, as shown below.

db.item.aggregate({$group:{_id:"$category",count:{$sum:1}}})
db.item.aggregate({$group:{_id:"$brand",count:{$sum:1}}})

In any case, I can do the same using the aggregation structure using the single aggregation command.

I simplified my case here, but in reality I need this grouping of fields in an array of subdocuments. Suppose the above is the structure after I unwind.

(- ), , .

MongoDB 2.4.

Answer 1

, mapReduce , . , , , :

db.items.aggregate([
    { "$group": {
        "_id": null,
        "categories": { "$push": "$category" },
        "brands": { "$push": "$brand" }
    }},
    { "$project": {
        "_id": {
            "categories": "$categories",
            "brands": "$brands"
        },
        "categories": 1
    }},
    { "$unwind": "$categories" },
    { "$group": {
        "_id": {
            "brands": "$_id.brands",
            "category": "$categories"
        },
        "count": { "$sum": 1 }
    }},
    { "$group": {
        "_id": "$_id.brands",
        "categories": { "$push": {
            "category": "$_id.category",
            "count": "$count"
        }},
    }},
    { "$project": {
        "_id": "$categories",
        "brands": "$_id"
    }},
    { "$unwind": "$brands" },
    { "$group": {
        "_id": {
            "categories": "$_id",
            "brand": "$brands"
        },
        "count": { "$sum": 1 }
    }},
    { "$group": {
        "_id": null,
        "categories": { "$first": "$_id.categories" },
        "brands": { "$push": {
            "brand": "$_id.brand",
            "count": "$count"
        }}
    }}
])

, mapReduce, , :

{
    "_id" : null,
    "categories" : [
            {
                    "category" : "c2",
                    "count" : 2
            },
            {
                    "category" : "c1",
                    "count" : 3
            }
    ],
    "brands" : [
            {
                    "brand" : "b2",
                    "count" : 2
            },
            {
                    "brand" : "b1",
                    "count" : 3
            }
    ]
}

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