MongoDB aggregation: count appearances of each value of a field per id - mongodb

Question

Data example:
{ id: 1, field: a, .. }
{ id: 1, field: a, .. }
{ id: 1, field: b, .. }
{ id: 2, field: b, .. }
Desired result:
{ id: 1, countA: 2, countB: 1 }
{ id :2, countA:0, countB: 1 }
'field' is an enum, so I know all the values in advance and can give names to the counters.
I have a solution but it seems that there is a better one. My solution:
db.collection.aggregate([
{ $group: { _id: { id: "$id", field: "$field"}, count: { $sum : 1}}},
{ $project: {
_id: 1,
countA: { $cond: { if: { $eq: ["$_id.field", "a"] }, then: "$count", else: 0 }},
countB: { $cond: { if: { $eq: ["$_id.field", "b"] }, then: "$count", else: 0 }}
}
},
{ $group:
{_id: "$_id.id", countA: { $max: "$countA"}, countB: { $max :"$countB"}}
}
])
upd: I have a better solution - placing the project before grouping in some way and no need for 2 groupings, but it uses the same principle. But it still seems that there should be somehting more built-in for this purpose
Thanks!

Answer 1

I believe you just need one $group stage to acheive what you want.
Just use $sum to count the number of fields with value a and B with $cond.
Try this:
db.collection.aggregate([
{
$group: {
_id: "$id",
id: {
$first: "$id"
},
countA: {
$sum: {
$cond: {
if: { $eq: [ "$field","a"] },
then: 1,
else: 0
}
}
},
countB: {
$sum: {
$cond: {
if: { $eq: [ "$field", "b"] },
then: 1,
else: 0
}
}
}
}
}
])
Have a look at this Mongo Playground for working demo of the query.
I hope this is what you are looking for!