Mongodb Aggregation to count element pairs and individual elements - mongodb

Question

I have following data:
{ "id" : 1, "lsPairs" :[{"location" : "L0", "service" : "S0" }]}
{ "id" : 2, "lsPairs" :[{"location" : "L0", "service" : "S0" },{"location" : "L1", "service" : "S1"}]}
{ "id" : 3, "lsPairs" :[{"location" : "L0", "service" : "S0" },{"location" : "L1", "service" : "S1"}, {"location" : "L2", "service" : "S2"}]}
{ "id" : 4, "lsPairs" :[{"location" : "L0", "service" : "S0" },{"location" : "L1", "service" : "S1"},{"location" : "L2", "service" : "S2"}, {"location" : "L3", "service" : "S3"}]}`
I want to get location count, service count and (location,service) pair count
{ "_id" : "L3" , "count" : 1}
{ "_id" : "L2" , "count" : 2}
{ "_id" : "L1" , "count" : 3}
{ "_id" : "L0" , "count" : 4}
{ "_id" : "S3" , "count" : 1}
{ "_id" : "S2" , "count" : 2}
{ "_id" : "S1" , "count" : 3}
{ "_id" : "S0" , "count" : 4}
{ "_id" : { "loc" : "L2" , "srv" : "S2"} , "count" : 2}
{ "_id" : { "loc" : "L1" , "srv" : "S1"} , "count" : 3}
{ "_id" : { "loc" : "L3" , "srv" : "S3"} , "count" : 1}
{ "_id" : { "loc" : "L0" , "srv" : "S0"} , "count" : 4}`
Now I run group function three times, group different id.
Any idea for using one group to get these result?

Answer 1

You will need to deconstruct the array with $unwind then $group the documents.
collection.aggregate([
{ $unwind: "$lsPairs" },
{ $group: {
_id: {
"loc": "$lsPairs.location",
"srv": "$lsPairs.service"
},
"count": { $sum: 1 }
}}
])
Output
{ "_id" : { "loc" : "L3", "srv" : "S3" }, "count" : 1 }
{ "_id" : { "loc" : "L2", "srv" : "S2" }, "count" : 2 }
{ "_id" : { "loc" : "L1", "srv" : "S1" }, "count" : 3 }
{ "_id" : { "loc" : "L0", "srv" : "S0" }, "count" : 4 }

Answer 2

Keep the first round location-service pair to a collection and reused it.
db.locservice.aggregate([ {$unwind:"$lsPairs"},
{$group:{_id:"$lsPairs",count: { $sum: 1}}},
{$sort:{_id:1}},
{$out:"lsp"} ])
Take location from temp collection and group it.
db.lsp.aggregate([{$project:{_id:0, loc:"$_id.location", count:1}},
{$group:{_id:"$loc", cnt:{$sum:"$count"}}}, {$sort:{_id:1}} ])
Take service from temp collection and group it.
db.lsp.aggregate([{$project:{_id:0, srv:"$_id.service", count:1}},
{$group:{_id:"$srv", cnt:{$sum:"$count"}}}, {$sort:{_id:1}} ])

Answer 3

The following I add location and service to array, can I group two array same time
db.locservice.aggregate([ {$unwind:"$lsPairs"},
{$group:{_id:"$lsPairs",count: { $sum: 1},
locs:{$push:{item:"$lsPairs.location"}},
srvs:{$push:{item:"$lsPairs.service"}}}},
{$project:{count:1, locs:1, srvs:1}} ])
{ "_id" : { "location" : "L3", "service" : "S3" }, "count" : 1, "locs" : [ { "item" : "L3" } ], "srvs" : [ { "item" : "S3" } ] }
{ "_id" : { "location" : "L2", "service" : "S2" }, "count" : 2, "locs" : [ { "item" : "L2" }, { "item" : "L2" } ], "srvs" : [ { "item" : "S2" }, { "item" : "S2" } ] }
{ "_id" : { "location" : "L1", "service" : "S1" }, "count" : 3, "locs" : [ { "item" : "L1" }, { "item" : "L1" }, { "item" : "L1" } ], "srvs" : [ { "item" : "S1" }, { "item" : "S1" }, { "item" : "S1" } ] }
{ "_id" : { "location" : "L0", "service" : "S0" }, "count" : 4, "locs" : [ { "item" : "L0" }, { "item" : "L0" }, { "item" : "L0" }, { "item" : "L0" } ], "srvs" : [ { "item" : "S0" }, { "item" : "S0" }, { "item" : "S0" }, { "item" : "S0" } ] }