MongoDB - Aggregate muliple rows - mongodb

Question

Suppose the following aggregation query:
Machine.aggregate( [ { $match : { $and: [ {"idc": req.query.idc }, {"customer":req.query.customer} ] } } ,{"$group":{_id: {"cluster":"$cluster","idc":"$idc","type":"$type"},"SumCores":{"$sum":"$cores"},"SumMemory": { "$sum":"$memory" }}}, { $sort : { idc : -1, cluster: 1 } } ]);
Which returns:
[
{
"_id": {
"cluster": 1,
"idc": "LH5",
"type": "Virtual"
},
"SumCores": 112,
"SumMemory": 384
},
{
"_id": {
"cluster": 1,
"idc": "LH5",
"type": "Physical"
},
"SumCores": 192,
"SumMemory": 768
},
{
"_id": {
"cluster": 1,
"idc": "LH8",
"type": "Virtual"
},
"SumCores": 232,
"SumMemory": 469
},
{
"_id": {
"cluster": 1,
"idc": "LH8",
"type": "Physical"
},
"SumCores": 256,
"SumMemory": 1024
}
]
Is there a way to change the aggregation to retrieve this desired output:
[
{
"_id": {
"cluster": 1,
"idc": "LH5"
},
"Virtual": {
"SumCores": 112,
"SumMemory": 384
},
"Physical": {
"SumCores": 192,
"SumMemory": 768
}
},
{
"_id": {
"cluster": 1,
"idc": "LH8"
},
"Virtual": {
"SumCores": 232,
"SumMemory": 469
},
"Physical": {
"idc": "LH8",
"type": "Physical"
}
}
]
Assumptions:
There will always be a Physical and Virtual "pair" per IDC/Cluster
I am happy to receive solutions which:
a) Change the aggregation query
b) Receive the existing data and change it into this format by way of a library and/or an algorithm

Answer 1

You are doing all the right things in the query already as you need to $group at the level you have in order to get the correct sums. The only thing remaining is to bring it all together.
Personally I would stick with the "pair" in an array as the final output:
Machine.aggregate([
{ "$match": {
"idc": req.query.idc, "customer": req.query.customer}
} ,
{ "$group": {
"_id": {
"cluster": "$cluster",
"idc":"$idc",
"type": "$type"
},
"SumCores": { "$sum":"$cores" },
"SumMemory": { "$sum":"$memory" }
}},
{ "$group": {
"_id": {
"cluster": "$_id.cluster",
"idc": "$_id.idc"
},
"data": {
"$push": {
"type": "$_id.type",
"SumCores": "$SumCores",
"SumMemory": "$SumMemory"
}
}
}},
{ "$sort" : { "_id.idc": -1, "_id.cluster": 1 } }
]);
Which would give you:
{
"_id" : {
"cluster" : 1,
"idc" : "LH8"
},
"data" : [
{
"type" : "Virtual",
"SumCores" : 232,
"SumMemory" : 469
},
{
"type" : "Physical",
"SumCores" : 256,
"SumMemory" : 1024
}
]
}
{
"_id" : {
"cluster" : 1,
"idc" : "LH5"
},
"data" : [
{
"type" : "Virtual",
"SumCores" : 112,
"SumMemory" : 384
},
{
"type" : "Physical",
"SumCores" : 192,
"SumMemory" : 768
}
]
}
But if you really must, then you can filter out the matched elements from the array and put them in their own properties:
Machine.aggregate([
{ "$match": {
"idc": req.query.idc, "customer": req.query.customer}
} ,
{ "$group": {
"_id": {
"cluster": "$cluster",
"idc":"$idc",
"type": "$type"
},
"SumCores": { "$sum":"$cores" },
"SumMemory": { "$sum":"$memory" }
}},
{ "$group": {
"_id": {
"cluster": "$_id.cluster",
"idc": "$_id.idc"
},
"data": {
"$push": {
"type": "$_id.type",
"SumCores": "$SumCores",
"SumMemory": "$SumMemory"
}
}
}},
{ "$project": {
"Physical": {
"$setDifference": [
{ "$map": {
"input": "$data",
"as": "el",
"in": {
"$cond": [
{ "$eq": [ "$$el.type", "Physical" ] },
{
"SumCores": "$$el.SumCores",
"SumMemory": "$$el.SumMemory"
},
false
]
}
}},
[false]
]
},
"Virtual": {
"$setDifference": [
{ "$map": {
"input": "$data",
"as": "el",
"in": {
"$cond": [
{ "$eq": [ "$$el.type", "Virtual" ] },
{
"SumCores": "$$el.SumCores",
"SumMemory": "$$el.SumMemory"
},
false
]
}
}},
[false]
]
}
}},
{ "$unwind": "$Physical" },
{ "$unwind": "$Virtual"},
{ "$sort" : { "_id.idc": -1, "_id.cluster": 1 } }
]);
Which gives you your result:
{
"_id" : {
"cluster" : 1,
"idc" : "LH8"
},
"Physical" : {
"SumCores" : 256,
"SumMemory" : 1024
},
"Virtual" : {
"SumCores" : 232,
"SumMemory" : 469
}
}
{
"_id" : {
"cluster" : 1,
"idc" : "LH5"
},
"Physical" : {
"SumCores" : 192,
"SumMemory" : 768
},
"Virtual" : {
"SumCores" : 112,
"SumMemory" : 384
}
}
But the first is just going to give you the same essential data without needing an extra pass through the results.
At any rate it's really just one more $group to bring it all together and then the optional stages if you really must have that data format. But I would personally handle any accessing of the "pair" in the code that needs to deal with it.