How to use findOne() when you want the few fields in one object and other fields in the other object?

I am querying the database using findOne which will return just one document. Now, I want a few of the fields of that document in one object and the other fields in the other object, both of which are wrapped in a single object. For example I have a table called Bus with the following fields –

_id,
busNo,
city,
agency,
date,
ticketPrice,
helper,
driver,
totalCollection

My find one query returns one document but I want it in the below format –

existingAggr -  {
  "result": [
    {
      "_id": "630dcd0c652489bca1b319f7",
      "busNo": "123",
      "city": "32",
      "agency": "58",
      "date": "2022-08-29T00:00:00.000Z",
    }
  ],
  "aggregates": {
    "ticketPrice": 8,
    "totalCollection": 402,
    "helper": 0,
    "driver": 23,
  }
}

I want this either with single access of database, or we can do some javascript calculations to further bring my result in this manner, but I cant seem to reach the solution. Currently, I am using the below code –

const res = await Bus.findOne(
    { busNo, date },
    { 
     _id :1,
     busNo:1, 
     city:1,
     agency:1,
     date:1,
     ticketPrice:1,
     helper:1,
     driver:1,
     totalCollection:1
   }
  );

This would return all the fields in one.

>Solution :

Given a result, you can straightfowardly create new objects from your result.

const res = await BusDayWise.findOne(
    { ...filter },
    { session: mongoSession }
  );

const result = [
    {
      "_id": res._id,
      "busNo": res.busNo,
      "city": res.city,
      "agency": res/agency,
      "date": res.date,
    }
],
const aggregates = 
{
    "ticketPrice": res.ticketPrice,
    "totalCollection": res.totalCollection,
    "helper": res.helper,
    "driver": res.driver,
}

More advanced answer

You can have a function that retrieve only certain key from a dictionnary

function subDict(dict, keys){
   const newDict = {};
   keys.forEach(key => newDict[key] = dict[key]);
   return newDict 
}

test = {"a": 1, "b": 2, "c": 3}
keys = ["a", "c"];
newTest = subDict(test, keys); // {"a": 1; "c": 3}

So in your case

const result = subDict(res, ["_id", "busNo", "city","agency", "date"]);
const aggregates = subDict(res, ["ticketPrice", "totalCollection", "helper", "driver"]);