Group by an optional field in mongodb - mongodb

Question

I would like to independently group the results of an or clause, including overlap. The data set is rather large so running 2 queries sequentially will result in an undesirable wait time. I am hoping I can somehow project which clause returned the corresponding data. Given this data set:
[
{
"_id": 1,
"item": "abc",
"name": "Michael",
"price": NumberDecimal("10"),
"quantity": NumberInt("2"),
"date": ISODate("2014-03-01T08:00:00Z")
},
{
"_id": 2,
"item": "jkl",
"name": "Toby",
"price": NumberDecimal("20"),
"quantity": NumberInt("1"),
"date": ISODate("2014-03-01T09:00:00Z")
},
{
"_id": 3,
"item": "xyz",
"name": "Keith",
"price": NumberDecimal("5"),
"quantity": NumberInt("10"),
"date": ISODate("2014-03-15T09:00:00Z")
},
{
"_id": 4,
"item": "abc",
"name": "Dwight",
"price": NumberDecimal("5"),
"quantity": NumberInt("20"),
"date": ISODate("2014-04-04T11:21:39.736Z")
},
{
"_id": 5,
"item": "abc",
"name": "Ryan",
"price": NumberDecimal("10"),
"quantity": NumberInt("10"),
"date": ISODate("2014-04-04T21:23:13.331Z")
},
{
"_id": 6,
"item": "def",
"name": "Jim",
"price": NumberDecimal("7.5"),
"quantity": NumberInt("5"),
"date": ISODate("2015-06-04T05:08:13Z")
},
{
"_id": 7,
"item": "abc",
"name": "Keith",
"price": NumberDecimal("7.5"),
"quantity": NumberInt("10"),
"date": ISODate("2015-09-10T08:43:00Z")
},
{
"_id": 8,
"item": "abc",
"name": "Michael",
"price": NumberDecimal("10"),
"quantity": NumberInt("5"),
"date": ISODate("2016-02-06T20:20:13Z")
},
]
I would like to receive this result:
[{
"_id": {
"name": "Keith"
},
"count": 2
},
{
"_id": {
"item": "abc",
},
"count": 5
}]
Here is what I have tried so far:
db.collection.aggregate([
{
$match: {
$or: [
{
item: "abc"
},
{
name: "Keith"
}
]
}
},
{
$group: {
_id: {
item: "$item",
name: "$name"
},
count: {
$sum: 1
}
}
}
])

Answer 1

You can use $facet to get multiple aggregation pipelines into the same stage in this way:
Using $facet there are two "outputs" one group by name and other by item.
In each one there are multiple stages:
First $match to process only documents you want.
Then $group with _id name or item, and $count to get the total.
db.collection.aggregate([
{
"$facet": {
"groupByName": [
{
"$match": {"name": "Keith"}
},
{
"$group": {"_id": "$name","count": {"$sum": 1}}
}
],
"groupByItem": [
{
"$match": {"item": "abc"}
},
{
"$group": {"_id": "$item","count": {"$sum": 1}}
}
]
}
}
])
Example here
The output is:
{
"groupByItem": [
{
"_id": "abc",
"count": 5
}
],
"groupByName": [
{
"_id": "Keith",
"count": 2
}
]
}

Answer 2

Here it is:
mongos> db.n.aggregate([ { $facet:{ names:[ {$match:{name:"Keith"}} , {$group:{_id:{name:"$name"}, count:{$sum:1}}} ] , items:[ {$match:{item:"abc"}},{ $group:{_id:{item:"$item"}, count:{$sum:1}} } ] } } , {$project:{ "namesANDitems":{$concatArrays:[ "$names","$items" ]} }} ,{$unwind:"$namesANDitems"} ,{$replaceRoot:{newRoot:"$namesANDitems"} } ]).pretty()
{ "_id" : { "name" : "Keith" }, "count" : 2 }
{ "_id" : { "item" : "abc" }, "count" : 5 }
mongos>
explained:
You create two pipes via $facet
Match in every facet pipe what you need to group pipe1=names , pipe2=items
Join the arrays from the two pipes in single array named "namesANDitems"
Convert the array to object with $unwind
Remove the temporary object name namesANDitems so you have only the two objects as requested