Count by multiple fields - mongodb

Question

I work with mongodb and aggregation-framework. I have fallowing data in database I need to sort and group it by year
[{
"_id": "5df537d615a0cd001759f5e4",
"timeDoneA": {
"year":2020,
"day": 5,
"month": 12
},
"timeDoneB": {
"year": 2020,
"day": 4,
"month": 2
},
"timeDoneC": {
"year": 2020,
"day": 4,
"month": 2
},
},
{
"_id": "5df6595dab96a000174e29d7",
"timeDoneA": {
"year": 2020,
"day": 12,
"month": 12
},
"timeDoneB": {
"year": 2019,
"day": 15,
"month": 12
},
"timeDoneC": {
"year": 2019,
"day": 15,
"month": 12
},
}...etc],
I want to get something like this to sort data by years:
2019: {
timeDoneA: 0,
timeDoneB: 1,
timeDoneC: 1
},
2020: {
timeDoneA: 2,
timeDoneB: 1,
timeDoneC: 1
}
How can I do that with aggregation query?

Answer 1

You could use a $facet stage to collect the year values and remove unnecessary fields, then $unwind both of these, and finally use $group to collate the values.
A quick and dirty example:
db.aggtest.aggregate([
{$facet:{
years:[
{$group:{ _id:null,
listC:{$addToSet: "$timeDoneC.year"},
listB:{$addToSet:"$timeDoneB.year"},
listA:{$addToSet:"$timeDoneA.year"}}},
{$project:{ _id:0,
list:{$setUnion:["$listA","$listB","$listC"]}}}],
done:[{$project:{ _id:0,
timeDoneA:"$timeDoneA.year",
timeDoneB:"$timeDoneB.year",
timeDoneC:"$timeDoneC.year"}}]}},
{$unwind:"$done"},
{$unwind:"$years"},
{$unwind:"$years.list"},
{$group:{ _id:"$years.list",
timeDoneA:{$sum:{$cond:[{$eq:["$done.timeDoneA","$years.list"]},1,0]}},
timeDoneB:{$sum:{$cond:[{$eq:["$done.timeDoneB","$years.list"]},1,0]}},
timeDoneC:{$sum:{$cond:[{$eq:["$done.timeDoneC","$years.list"]},1,0]}}}}])
Testing this on your sample data gives:
{ "_id" : 2020, "timeDoneA" : 2, "timeDoneB" : 1, "timeDoneC" : 1 }
{ "_id" : 2019, "timeDoneA" : 0, "timeDoneB" : 1, "timeDoneC" : 1 }