MongoDB - Getting #DBRef values from list as separate objects - mongodb

Question

This is the current object
{
"chargeCodeIds" : [
1,
2
],
"customChargeCode" : [
{
"_id" : 1,
"chargeCodeType" : "Department",
"name" : "IT",
"isBillable" : true,
},
{
"_id" : 2,
"chargeCodeType" : "Task"
"name" : "Development",
"isBillable" : true
}
]
We have the query that searched the charge codes from their collection say "customChargeCode" and listed them in an object "customChargeCode" as array of objects. BUT I don't want it an in array.
Here is the current query:
db.authorizeChargeAssociation.aggregate([
{
$project: {
chargeCodeIds: {
$map: {
input: {
$map: {
input:"$chargeCodes",
in: {
$arrayElemAt: [{$objectToArray: "$$this"}, 1]
},
}
},
in: "$$this.v"
}
},
}
},
{
$lookup: {
from: "customChargeCode",
localField:"chargeCodeIds",
foreignField:"_id",
as:"customChargeCode"
}
},
{$unwind: {path: "$customChargeCodes", preserveNullAndEmptyArrays: true} },
],
{ collation: { locale: "en_US", numericOrdering: true}},
{ allowDiskUse: true}
)
What I want is to get a separate object of each "chargeCodeIds" from the "customChargeCode" collection and name them the field as "chargeCodeType" value.
Expected Result be like:
{
"chargeCodeIds" : [
1,
2
],
"Department" : {
"_id" : 1,
"chargeCodeType" : "Department",
"name" : "IT",
"isBillable" : true,
},
"Task" : {
"_id" : 2,
"chargeCodeType" : "Task"
"name" : "Development",
"isBillable" : true
}