Remove aggregate with conditions - mongodb

Question

I have the a collection of documents as follows an example document:
{
'publicacao' : { 'data': '2013-13-13', 'hora': '13:13:13'},
'conteudo' : 'https://docs.google.com/document/d/1EQynJTiBa6FNI2O8XfoV0clMPxS5uOAu0_jKyEwsTBE/edit?usp=sharing',
'titulo' : 'As histórias de ciclano',
'categoria' : 'Romance',
'autor' : 'Ciclano',
'avaliacoes': [
{
'leitor': 'Fulano',
'nota': 1
},
{
'leitor': 'Beltrano',
'nota': 0
}
],
'denuncias': [
{
'denunciante': 'Ciclano'
},
{
'denunciante': 'Beltrano'
}
]
}
then i made an aggregate to define some documents to be removed:
var cursor = db.livro.aggregate( [
{
$project: {
id:1,
remover: {
$gt: [
{ $size: "$denuncias" },
{
$divide: [
{ $size: "$avaliacoes" },
2
]
}
]
}
}
}
]);
this aggregation returns the following docs:
{ "_id" : ObjectId("5cf5a9be7d48c53504974439"), "remover" : false }
{ "_id" : ObjectId("5cf5a9be7d48c5350497443a"), "remover" : false }
{ "_id" : ObjectId("5cf5a9be7d48c5350497443b"), "remover" : true }
{ "_id" : ObjectId("5cfd746e40d53565ca52132b"), "remover" : false }
{ "_id" : ObjectId("5cfd746e40d53565ca52132c"), "remover" : false }
{ "_id" : ObjectId("5cfd746e40d53565ca52132d"), "remover" : true }
I need to remove all docs with "remove": true.
I don't know how to get those ones!

Answer 1

Resolved:
I used the cursor.forEach() to iterate the objects and find the objects with remove = true.
cursor.forEach(function (doc){
if(doc.remover == true) {
db.livro.remove({"_id": doc._id});
print("Doc removido: "+ doc._id)
}
});

Answer 2

Instead of doing this with code after the query it is better overall to do it in the actual aggregation. Just check with a $match if the array has an element:
db.collection.aggregate([
{
$match: {
"denuncias.0": {
$exists: true
}
}
},
{
$project: {
id: 1,
remover: {
$gt: [
{
$size: "$denuncias"
},
{
$divide: [
{
$size: "$avaliacoes"
},
2
]
}
]
}
}
}
])
You can see if working here