Part of the way there but you need to get the "distinct" values for "holiday_type" first, then you $group
again:
db.transactions.aggregate([
{ "$group": {
"_id": {
"employee" : "$employee",
"Month": { "$month" : "$date" },
"Year": { "$year" : "$date" },
"holiday_type" : "$holiday_type"
},
}},
{ "$group": {
"_id": {
"employee" : "$_id.employee",
"Month": "$_id.Month",
"Year": "$_id.Year"
},
"count": { "$sum": 1 }
}}
], { "allowDiskUse": true }
);
That is the general process as "distinct" in SQL is kind of a grouping operation in itself. So it is a double $group
operation in order to get your correct result.