In the following example, "Algorithms in C++" is present twice.
The $unset modifier can remove a particular field but how to remove an entry from a field?
{
"_id" : ObjectId("4f6cd3c47156522f4f45b26f"),
"favorites" : {
"books" : [
"Algorithms in C++",
"The Art of Computer Programming",
"Graph Theory",
"Algorithms in C++"
]
},
"name" : "robert"
}
As of MongoDB 2.2 you can use the aggregation framework with an $unwind, $group and $project stage to achieve this:
db.users.aggregate([{$unwind: '$favorites.books'},
{$group: {_id: '$_id',
books: {$addToSet: '$favorites.books'},
name: {$first: '$name'}}},
{$project: {'favorites.books': '$books', name: '$name'}}
])
Note the need for the $project to rename the favorites field, since $group aggregate fields cannot be nested.
$group stage why are you using name: {$first: '$name'}?
name, so you can take any in the $group stage, so I'm just taking the first.
The easiest solution is to use setUnion (Mongo 2.6+):
db.users.aggregate([
{'$addFields': {'favorites.books': {'$setUnion': ['$favorites.books', []]}}}
])
Another (more lengthy) version that is based on the idea from @kynan's answer, but preserves all the other fields without explicitly specifying them (Mongo 3.4+):
> db.users.aggregate([
{'$unwind': {
'path': '$favorites.books',
// output the document even if its list of books is empty
'preserveNullAndEmptyArrays': true
}},
{'$group': {
'_id': '$_id',
'books': {'$addToSet': '$favorites.books'},
// arbitrary name that doesn't exist on any document
'_other_fields': {'$first': '$$ROOT'},
}},
{
// the field, in the resulting document, has the value from the last document merged for the field. (c) docs
// so the new deduped array value will be used
'$replaceRoot': {'newRoot': {'$mergeObjects': ['$_other_fields', "$$ROOT"]}}
},
// this stage wouldn't be necessary if the field wasn't nested
{'$addFields': {'favorites.books': '$books'}},
{'$project': {'_other_fields': 0, 'books': 0}}
])
{ "_id" : ObjectId("4f6cd3c47156522f4f45b26f"), "name" : "robert", "favorites" :
{ "books" : [ "The Art of Computer Programmning", "Graph Theory", "Algorithms in C++" ] } }
addFields. Maybe something changed in the API recently.What you have to do is use map reduce to detect and count duplicate tags .. then use $set to replace the entire books based on { "_id" : ObjectId("4f6cd3c47156522f4f45b26f"),
This has been discussed sevel times here .. please seee
Removing duplicate records using MapReduce
Fast way to find duplicates on indexed column in mongodb
http://csanz.posterous.com/look-for-duplicates-using-mongodb-mapreduce
function unique(arr) {
var hash = {}, result = [];
for (var i = 0, l = arr.length; i < l; ++i) {
if (!hash.hasOwnProperty(arr[i])) {
hash[arr[i]] = true;
result.push(arr[i]);
}
}
return result;
}
db.collection.find({}).forEach(function (doc) {
db.collection.update({ _id: doc._id }, { $set: { "favorites.books": unique(doc.favorites.books) } });
})
Starting in Mongo 4.4, the $function aggregation operator allows applying a custom javascript function to implement behaviour not supported by the MongoDB Query Language.
For instance, in order to remove duplicates from an array:
// {
// "favorites" : { "books" : [
// "Algorithms in C++",
// "The Art of Computer Programming",
// "Graph Theory",
// "Algorithms in C++"
// ]},
// "name" : "robert"
// }
db.collection.aggregate(
{ $set:
{ "favorites.books":
{ $function: {
body: function(books) { return books.filter((v, i, a) => a.indexOf(v) === i) },
args: ["$favorites.books"],
lang: "js"
}}
}
}
)
// {
// "favorites" : { "books" : [
// "Algorithms in C++",
// "The Art of Computer Programming",
// "Graph Theory"
// ]},
// "name" : "robert"
// }
This has the advantages of:
$unwind and $group stages.$function takes 3 parameters:
body, which is the function to apply, whose parameter is the array to modify.args, which contains the fields from the record that the body function takes as parameter. In our case "$favorites.books".lang, which is the language in which the body function is written. Only js is currently available.
