Remove duplicates from array without $unwind - mongodb

Question

Using query aggregation I want to create a new array by a filter of another array, so that the filtered result will be done by a specific field of the preliminary array.
In this case, I want to filter by the field "fieldName".
I will always want to filter out the last occur
example:
I have one document:
{
"fullyQualifiedName" : "MongoDB",
"items" : [
{
"fieldName" : "_id",
"fieldCount" : 7,
"confidence_level" : 1,
"fieldClassifications" : [
"LineageGuid"
],
},
{
"fieldName" : "_id",
"fieldCount" : 7,
"fieldClassifications" : [
{
"classificationName" : "LineageGuid",
}
]
},
{
"fieldName" : "details",
"fieldCount" : 7,
},
{
"fieldName" : "state",
"fieldCount" : 7,
}
]
}
I want to create a new array like:
"items" : [
{
"fieldName" : "_id",
"fieldCount" : 7,
"confidence_level" : 1,
"fieldClassifications" : [
"LineageGuid"
],
},
{
"fieldName" : "details",
"fieldCount" : 7,
},
{
"fieldName" : "state",
"fieldCount" : 7,
}
]
The simple solution is to $unwind and $group again but I can't do it because of performance issue.
I am using MongoDB 3.4

Answer 1

You can use below aggregation
db.collection.aggregate([
{ "$addFields": {
"items": {
"$map": {
"input": {
"$setUnion": [
{ "$map": {
"input": "$items",
"in": { "$indexOfArray": ["$items.fieldName", "$$this.fieldName"] }
}}
]
},
"as": "i",
"in": {
"fieldName": { "$arrayElemAt": ["$items.fieldName", "$$i"] },
"fieldCount": { "$arrayElemAt": ["$items.fieldCount", "$$i"] },
"confidence_level": { "$arrayElemAt": ["$items.confidence_level", "$$i"] },
"fieldClassifications": { "$arrayElemAt": ["$items.fieldClassifications", "$$i"] }
}
}
}
}}
])