Mongoose/Mongo faster to have separate collections? - node.js

Question

Currently I have a schema USERS with a sub doc VIEWED
As a 'user' views other users, their ID gets logged in the viewers sub doc
So when viewing, technically this gets all the users, then filters that through all the viewed users [for any given user]. So you get a list of unique/fresh users.
My method is currently fetching the list of users - Query 1
Then its fetching the list of viewed users (for a given user) - Query 2
Then using array.filter function to get a list of new users.
(using async parallel for those queries)
Question is, would it be faster to just have a separate document/collection that stores a list of viewed users for any given user. e.g:
{
userID: 1002,
viewedID: 9112
},
{
userID: 1002,
viewedID: 9222
},
{
userID: 1002,
viewedID: 9332
}
Is it possible for me to some how do a query that gets me a fresh list of users, so i don't have to do the computation myself. i.e let mongo do all the work.
edit, adding code to make it more clear
var ViewedSchema = new Schema({
coupleId: {type: Number, required: true}
});
var UserSchema = new Schema({
name : { type: String, trim: true, required: true}
, partnerId : { type: Number}
, viewed : [ViewedSchema]
});
code to view partners/users that have not been viewed before
async.parallel([
function(callback) {
//gets all the users/partners
User.find({}, function(err, users) {
var allPartners = [];
users.forEach(function(user){
if(allPartners.indexOf(user.partnerId) == -1) {
allPartners.push(user.partnerId);
}
});
callback(null, allPartners);
});
},
function(callback) {
//gets all the users/partners i have already viewed
var votedPartners = [];
User.findById(id, function(err, user) {
user.viewed.forEach(function(user){
votedPartners.push(user.coupleId);
});
callback(null, votedPartners);
});
}
],
function(err, result) {
//gets the differences between the 2 arrays
function exists(element) {
return (result[1].indexOf(element) == -1);
}
var showPartners = result[0].filter(exists);
User.find({partnerId: showPartners[0]}, function(err, user){
var json = {objects: user};
res.render('index', json);
});
});
};

Answer 1

I'm not sure what you mean by fresh or new users, exactly, but have you loked at the distinct() command? You can use it to get all the unique viewed user IDs for all the users in the collection, which is what it sounds like you want to do. See
http://docs.mongodb.org/manual/reference/method/db.collection.distinct/
From the documentation:
Return an array of the distinct values of the field sku in the subdocument item from all documents in the orders collection:
db.orders.distinct( 'item.sku' )
If you give an example of your current document schema, I could try to write the exact query for you.
Edit: You can use $nin to find the userIds that are not in a given list. Here is an example I set up in my local Mongo:
> db.dating.insert({"userId":100,"viewedId":["200","201"]})
> db.dating.findOne()
{
"_id" : ObjectId("5398d81799b228e88aef2441"),
"userId" : 100,
"viewedId" : [
"200",
"201"
]
}
> db.dating.insert({"userId":200,"viewedId":[""]})
> db.dating.insert({"userId":201,"viewedId":[""]})
> db.dating.insert({"userId":202,"viewedId":[""]})
> db.dating.insert({"userId":203,"viewedId":[""]})
> db.dating.insert({"userId":204,"viewedId":[""]})
> db.dating.insert({"userId":205,"viewedId":[""]})
> db.dating.find({"userId":{$nin: [200,201]}})
{ "_id" : ObjectId("5398d81799b228e88aef2441"), "userId" : 100, "viewedId" : [ "
200", "201" ] }
{ "_id" : ObjectId("5398d84099b228e88aef2444"), "userId" : 202, "viewedId" : [ "
" ] }
{ "_id" : ObjectId("5398d84799b228e88aef2445"), "userId" : 203, "viewedId" : [ "
" ] }
{ "_id" : ObjectId("5398d85699b228e88aef2446"), "userId" : 204, "viewedId" : [ "
" ] }
{ "_id" : ObjectId("5398d85c99b228e88aef2447"), "userId" : 205, "viewedId" : [ "
" ] }