MongoDB: Find duplicate docs where a field has lowest values - mongodb

Question

so I have this problem
I have this duplicate collection that goes like:
{name: "a", otherField: 1, _id: "id1"},
{name: "a", otherField: 2, _id: "id2"},
{name: "a", otherField: 3, _id: "id3"},
{name: "b", otherField: 1, _id: "id4"}
{name: "b", otherField: 2, _id: "id5"}
My goal is to get id of with less otherField that will look like:
{"name": "a", _id: "id1"},
{"name": "a", _id: "id2"},
{"name": "b", _id: "id4"}
Since highest otherField from a and b is "id3" and "id5", I want id other than the highest otherField
How to achieve this through query in mongodb?
Thank you

Answer 1

You can try below query :
db.collection.aggregate([
/** group all docs based on name & push docs to data field & find max value for otherField field */
{
$group: {
_id: "$name",
data: {
$push: "$$ROOT"
},
maxOtherField: {
$max: "$otherField"
}
}
},
/** Recreate data field array with removing doc which has max otherField value */
{
$addFields: {
data: {
$filter: {
input: "$data",
cond: {
$ne: [
"$$this.otherField",
"$maxOtherField"
]
}
}
}
}
},
/** unwind data array */
{
$unwind: "$data"
},
/** Replace data field as new root for each doc in coll */
{
$replaceRoot: {
newRoot: "$data"
}
}
])
Test : MongoDB-Playground
Note : We might lean towards sorting docs on field otherField, but it's not preferable on huge datasets.