Mongodb 3.2 (Aggregation) : Group by multiple values of the same attribute - mongodb

Question

I have the following count queries :
db.collection.count({code : {$in : ['delta', 'beta']}});
db.collection.count({code : 'alpha'});
for the given data set
[
{
"code": "detla",
"name": "delta1"
},
{
"code": "detla",
"name": "delta2"
},
{
"code": "detla",
"name": "delta3"
},
{
"code": "detla",
"name": "delta4"
},
{
"code": "beta",
"name": "beta1"
},
{
"code": "beta",
"name": "beta2"
},
{
"code": "beta",
"name": "beta3"
},
{
"code": "beta",
"name": "beta4"
},
{
"code": "beta",
"name": "beta5"
},
{
"code": "alpha",
"name": "alpha1"
},
{
"code": "alpha",
"name": "alpha2"
},
{
"code": "alpha",
"name": "alpha3"
}
]
Is there any way to achieve this using single aggregation() and nested $group in mongodb#3.2?
I understand that this is a classic usecase for mapReduce(), but I am out of option here as the $group by parameters (code in the above example), is a dynamically generated attribute, hence the aggregation stages building is also dynamic.
Expected output is same as the result of the two count queries.
-> First count query (delta, beta combined count) -> 9
-> Second count query (alpha count) -> 3

Answer 1

You can use below $group aggregation
db.collection.aggregate([
{ "$group": {
"_id": null,
"first": {
"$sum": {
"$cond": [{ "$in": ["$code", ["detla", "beta"]] }, 1, 0]
}
},
"second": {
"$sum": {
"$cond": [{ "$eq": ["alpha", "$code"] }, 1, 0]
}
}
}}
])
But I will prefer to go with the two count queries for the better performance
$in is also released in version 3.4. You can either use $setIsSubset for the prior versions
db.collection.aggregate([
{ "$group": {
"_id": null,
"first": {
"$sum": {
"$cond": [{ "$setIsSubset": [["$code"], ["detla", "beta"]] }, 1, 0]
}
},
"second": {
"$sum": {
"$cond": [{ "$eq": ["alpha", "$code"] }, 1, 0]
}
}
}}
])