Delete all _id field from subdocuments - mongodb

Question

I have been using Mongoose to insert a large amount of data into a mongodb database. I noticed that by default, Mongoose adds _id fields to all subdocuments, leaving me with documents which look like this (I've removed many fields for brevity - I've also shrunken each array to one entry, they generally have more)
{
"start_time" : ISODate("2013-04-05T02:30:28Z"),
"match_id" : 165816931,
"players" : [
{
"account_id" : 4294967295,
"_id" : ObjectId("51daffdaa78cee5c36e29fba"),
"additional_units" : [ ],
"ability_upgrades" : [
{
"ability" : 5155,
"time" : 141,
"level" : 1,
"_id" : ObjectId("51daffdaa78cee5c36e29fca")
},
]
},
],
"_id" : ObjectId("51daffdca78cee5c36e2a02e")
}
I have found how to prevent Mongoose adding these by default (http://mongoosejs.com/docs/guide.html, see option: id), however I now have 95 million records with these extraneous _id fields on all subdocuments. I am interested in finding the best way of deleting all of these fields (leaving the _id on the top level document). My initial thoughts are to use a bunch of for...in loops on each object but this seems very inefficient.

Answer 1

Given Derick's answer, I have created a function to do this:
var deleteIdFromSubdocs = function (obj, isRoot) {
for (var key in obj) {
if (isRoot == false && key == "_id") {
delete obj[key];
} else if (typeof obj[key] == "object") {
deleteIdFromSubdocs(obj[key], false);
}
}
return obj;
And run it against a test collection using:
db.testobjects.find().forEach(function (x){ y = deleteIdFromSubdocs(x, true); db.testobjects.save(y); } )
This appears to work for my test collection. I'd like to see if anyone has any opinions on how this could be done better/any risks involved before I run it against the 95 million document collection.

Answer 2

The players._id could be removed using an update operation, like he following:
db.collection.update({'players._id': {$exists : 1}}, { $unset : { 'players.$._id' : 1 } }, false, true)
However, it's not possible use positional operator in nested arrays. So, one solution is run a script directly on our database:
var cursor = db.collection.find({'players.ability_upgrades._id': {$exists : 1}});
cursor.forEach(function(doc) {
for (var i = 0; i < doc.players.length; i++) {
var player = doc.players[i];
delete player['_id'];
for (var j = 0; j < player.ability_upgrades.length; j++) {
delete player.ability_upgrades[j]['_id'];
}
}
db.collection.save(doc);
});
Save the script to a file and call mongo with the file as parameter:
> mongo remove_oid.js --shell

Answer 3

The only solution is to do this one by one, exactly with a for...in loop as you described.

Answer 4

Just another version, try this with AngularJS and MongoDB ;-)
function removeIds (obj, isRoot) {
for (var key in obj._doc) {
if (isRoot == false && key == "_id") {
delete obj._doc._id;
} else if ((Object.prototype.toString.call( obj[key] ) === '[object Array]' )) {
for (var i=0; i<obj[key].length; i++)
removeIds(obj[key][i], false);
}
}
return obj;
}
Usage:
var newObj = removeIds(oldObj, true);
delete newObj._id;