I have MongoDB collection with sequential integer field.
I need to find best approach to find "holes" in that sequence that happens due to records deletion.
For example if I have collection with these documents:
{ _id: "aab", seq: 1 ... }, { _id: "aac", seq: 2 ... }, { _id: "aad", seq: 4 ... }
The next insert I do, needs to be:
{ _id: "aae", seq: 3 ... }
May be you can create a MongoDB Stored Function to achieve this.
You can create a JavaScript Stored function which store's the "Sequence Numbers" of the document, getting deleted in some collection "StoreDeletedSequenceNumCollection".
Whenever you perform a Delete operation on the collection, make sure you call the "Stored Function " which actually stores the Sequence Number of the Document which is getting deleted in the collection "StoreDeletedSequenceNumCollection".
db.system.js.save( { _id: "DeletedSeqNum",
value: function (seqNum) {
db.StoreDeletedSequenceNumCollection.insert(
{ _id : seqNum}); return 1 ; }
}
);
When you are inserting a document, you can check if there are any sequence numbers present in the "StoreDeletedSequenceNumCollection", If present you can get the MINIMUM Sequence number from it and can use for insertion, Else if the "StoreDeletedSequenceNumCollection" is empty, then you can get the "MAXIMUM" sequence number form your actual collection and can use " MAXIMUM + 1" as a sequence number for insertion.
Related
It seems that mongodb allows the same syntax for deleting documents where the value is a single value, and whenever the value is present in a collection:
This
db.SomeCollection.deleteMany({UserId: 12345});
Can affect both {UserId: 12345} and {UserId: [12345, 67895, 87897]}.
I understand that this schema isn't ideal; we are using a singular UserId prop to denote both a single id int, and an array of ints. However, I've inherited a database that makes use of mongodb's dynamic nature, if you catch my drift.
What is the safest way to execute a deleteMany query, specifying I only want to delete documents where the value is a single integer?
You can use $type operator to check field type but it's not so obvious in this case. You can't check if UserId is of type double because MongoDB will indicate that array of doubles is a double as well (more here). So you need to invert your logic and check if documents you're trying to remove are not an array (type 4):
db.SomeCollection.deleteMany({ $and: [ { UserId: 12345 }, { UserId: { $not: { $type: 4 } } } ] })
I have a collection1 of documents with tags in MongoDB. The tags are an embedded array of strings:
{
name: 'someObj',
tags: ['tag1', 'tag2', ...]
}
I want to know the count of each tag in the collection. Therefore I have another collection2 with tag counts:
{
{
tag: 'tag1',
score: 2
}
{
tag: 'tag2',
score: 10
}
}
Now I have to keep both in sync. It is rather trivial when inserting to or removing from collection1. However when I update collection1 I do the following:
1.) get the old document
var oldObj = collection1.find({ _id: id });
2.) calculate the difference between old and new tag arrays
var removedTags = $(oldObj.tags).not(obj.tags).get();
var insertedTags = $(obj.tags).not(oldObj.tags).get();
3.) update the old document
collection1.update(
{ _id: id },
{ $set: obj }
);
4.) update the scores of inserted & removed tags
// increment score of each inserted tag
insertedTags.forEach(function(val, idx) {
// $inc will set score = 1 on insert
collection2.update(
{ tag: val },
{ $inc: { score: 1 } },
{ upsert: true }
)
});
// decrement score of each removed tag
removedTags.forEach(function(val, idx) {
// $inc will set score = -1 on insert
collection2.update(
{ tag: val },
{ $inc: { score: -1 } },
{ upsert: true }
)
});
My questions:
A) Is this approach of keeping book of scores separately efficient? Or is there a more efficient one-time query to get the scores from collection1?
B) Even if keeping book separately is the better choice: can that be done in less steps, e.g. letting mongoDB calculate what tags are new / removed?
The solution, as nickmilion correctly states, would be an aggregation. Though I would do it with a nack: we'll save its results in a collection. What will do is to trade real time results for an extreme speed boost.
How I would do it
More often than not, the need for real time results is overestimated. Hence, I'd go with precalculated stats for the tags and renew it every 5 minutes or so. That should be well enough, since most of such calls are requested async by the client and hence some delay in case the calculation has to be made on a specific request is negligible.
db.tags.aggregate(
{$unwind:"$tags"},
{$group: { _id:"$tags", score:{"$sum":1} } },
{$out:"tagStats"}
)
db.tagStats.update(
{'lastRun':{$exists:true}},
{'lastRun':new Date()},
{upsert:true}
)
db.tagStats.ensureIndex({lastRun:1}, {sparse:true})
Ok, here is the deal. First, we unwind the tags array, group it by the individual tags and increment the score for each occurrence of the respective tag. Next, we upsert lastRun in the tagStats collection, which we can do since MongoDB is schemaless. Next, we create a sparse index, which only holds values for documents in which the indexed field exists. In case the index already exists, ensureIndex is an extremely cheap query; however, since we are going to use that query in our code, we don't need to create the index manually. With this procedure, the following query
db.tagStats.find(
{lastRun:{ $lte: new Date( ISODate().getTime() - 300000 ) } },
{_id:0, lastRun:1}
)
becomes a covered query: A query which is answered from the index, which tends to reside in RAM, making this query lightning fast (slightly less than 0.5 msecs median in my tests). So what does this query do? It will return a record when the last run of the aggregation was run more than 5 minutes ( 5*60*1000 = 300000 msecs) ago. Of course, you can adjust this to your needs.
Now, we can wrap it up:
var hasToRun = db.tagStats.find(
{lastRun:{ $lte: new Date( ISODate().getTime() - 300000 ) } },
{_id:0, lastRun:1}
);
if(hasToRun){
db.tags.aggregate(
{$unwind:"$tags"},
{$group: {_id:"$tags", score:{"$sum":1} } },
{$out:"tagStats"}
)
db.tagStats.update(
{'lastRun':{$exists:true}},
{'lastRun':new Date()},
{upsert:true}
);
db.tagStats.ensureIndex({lastRun:1},{sparse:true});
}
// For all stats
var tagsStats = db.tagStats.find({score:{$exists:true}});
// score for a specific tag
var scoreForTag = db.tagStats.find({score:{$exists:true},_id:"tag1"});
Alternative approach
If real time results really matter and you need the stats for all the tags, simply use the aggregation without saving it to another collection:
db.tags.aggregate(
{$unwind:"$tags"},
{$group: { _id:"$tags", score:{"$sum":1} } },
)
If you only need the results for one specific tag at a time, a real time approach could be to use a special index, create a covered query and simply count the results:
db.tags.ensureIndex({tags:1})
var numberOfOccurences = db.tags.find({tags:"tag1"},{_id:0,tags:1}).count();
answering your questions:
B): you don't have to calculate the dif yourself use $addToSet
A): you can get the counts via aggregation framework with a combination of $unwind and $count
I have users' collection containing many lists of sub documents. Schema is something like this:
{
_id: ObjectId(),
name: aaa,
age: 20,
transactions:[
{
trans_id: 1,
product: mobile,
price: 30,
},
{
trans_id: 2,
product: tv,
price: 10
},
...]
...
}
So I have one doubt. trans_id in transactions list is unique over all the products, but it may be possible that I may have copied the same transaction again with same trans_id (due to bad ETL programming). Now I want to drop those duplicate sub documents. I have indexed trans_id thought not unique. I read about dropDups option. But will it delete a particular duplicate exists in DB or it'll drop whole document (which I definitely don't want). If not how to do it?
PS: I am using MongoDB 2.6.6 version.
Nearest case to all we can see presented here it that now you need a way of defining the "distinct" items within the array where some items are in fact an "exact copy" of other items in the array.
The best case is to use $addToSet along with the $each modifier within a looping operation for the collection. Ideally you use the Bulk Operations API to take advantage of the reduced traffic when doing so:
var bulk = db.collection.initializeOrderedBulkOperation();
var count = 0;
// Read the docs
db.collection.find({}).forEach(function(doc) {
// Blank the array
bulk.find({ "_id": doc.id })
.updateOne({ "$set": { "transactions": [] } });
// Resend as a "set"
bulk.find({ "_id": doc.id })
.updateOne({
"$addToSet": {
"trasactions": { "$each": doc.transactions }
}
});
count++;
// Execute once every 500 statements ( actually 1000 )
if ( count % 500 == 0 ) {
bulk.execute()
bulk = db.collection.initializeOrderedBulkOperation();
}
});
// If a remainder then execute the remaining stack
if ( count % 500 != 0 )
bulk.execute();
So as long as the "duplicate" content is "entirely the same" then this approach will work. If the only thing that is actually "duplicated" is the "trans_id" field then you need an entirely different approach, since none of the "whole documents" are "duplicated" and this means you need more logic in place to do this.
Is it possible to index some documents of the collection "only if" one of the fields to be indexed has a particular value?
Let me explain with an example:
The collection "posts" has millions of documents, ALL defined as follows:
{
"network": "network_1",
"blogname": "blogname_1",
"post_id": 1234,
"post_slug": "abcdefg"
}
Let's assume that the distribution of the post is equally split on network_1 and network_2
My application OFTEN select the type of query based on the value of "network" (although sometimes I need the data from both networks):
For example:
www.test.it/network_1/blog_1/**postid**/1234/
-> db.posts.find ({network: "network_1" blogname "blog_1", post_id: 1234})
www.test.it/network_2/blog_4/**slug**/aaaa/
-> db.posts.find ({network: "network_2" blogname "blog_4" post_slug: "yyyy"})
I could create two separate indexes (network / blogname / post_id and network / blogname / post_slug) but I would get a huge waste of RAM, since 50% of the data in the index will never be used.
Is there a way to create an index "filtered"?
Example:
(Note the WHERE parameter)
db.posts.ensureIndex ({network: 1 blogname: 1, post_id: 1}, {where: {network: "network_1"}})
db.posts.ensureIndex ({network: 1 blogname: 1, post_slug: 1}, {where: {network: "network_2"}})
Indeed it's possible in MongoDB 3.2+ They call it partialFilterExpression where you can set a condition based on which index will be created.
Example
db.users.createIndex({ "userId": 1, "project": 1 },
{ unique: true, partialFilterExpression:{
userId: { $exists: true, $gt : { $type : 10 } } } })
Please see Partial Index documentation
As of MongoDB v3.2, partial indexes are supported. Documentation: https://docs.mongodb.org/manual/core/index-partial/
It's possible, but it requires a workaround which creates redundancy in your documents, requires you to rewrite your find-queries and limits find-queries to exact matches.
MongoDB supports sparse indexes which only index the documents where the given field exists. You can use this feature to only index a part of the collection by adding this field only to those documents you want to index.
The bad news is that sparse indexes can only include a single field. But the good news is, that this field can also contain an object with multiple fields, so you can still store all the data you want to search for in this field.
To do this, add a new field to the included documents which includes an object with the fields you search for:
{
"network": "network_1",
"blogname": "blogname_1",
"post_id": 1234,
"post_slug": "abcdefg"
"network_1_index_key": {
"blogname": "blogname_1",
"post_id": 1234
}
}
Your ensureIndex command would index the field network_1_index_key:
db.posts.ensureIndex( { network_1_index_key: 1 }, { sparse: true } )
A find-query which is supposed to use this index, must now query for the exact object of the field network_1_index_key:
db.posts.find ({
network_1_index_key: {
blogname: "blogname_1",
post_id: 1234
}
})
Doing this would likely only make sense when the documents you want to index are a very small part of the collection. When its about half, I would just create a regular index and live with it because the larger document-size could mitigate the gains from the reduced index size.
You can try create index on all field (network / blogname / post_id / post_slug)
I have a collection in MongoDB where there are around (~3 million records). My sample record would look like,
{ "_id" = ObjectId("50731xxxxxxxxxxxxxxxxxxxx"),
"source_references" : [
"_id" : ObjectId("5045xxxxxxxxxxxxxx"),
"name" : "xxx",
"key" : 123
]
}
I am having a lot of duplicate records in the collection having same source_references.key. (By Duplicate I mean, source_references.key not the _id).
I want to remove duplicate records based on source_references.key, I'm thinking of writing some PHP code to traverse each record and remove the record if exists.
Is there a way to remove the duplicates in Mongo Internal command line?
This answer is obsolete : the dropDups option was removed in MongoDB 3.0, so a different approach will be required in most cases. For example, you could use aggregation as suggested on: MongoDB duplicate documents even after adding unique key.
If you are certain that the source_references.key identifies duplicate records, you can ensure a unique index with the dropDups:true index creation option in MongoDB 2.6 or older:
db.things.ensureIndex({'source_references.key' : 1}, {unique : true, dropDups : true})
This will keep the first unique document for each source_references.key value, and drop any subsequent documents that would otherwise cause a duplicate key violation.
Important Note: Any documents missing the source_references.key field will be considered as having a null value, so subsequent documents missing the key field will be deleted. You can add the sparse:true index creation option so the index only applies to documents with a source_references.key field.
Obvious caution: Take a backup of your database, and try this in a staging environment first if you are concerned about unintended data loss.
This is the easiest query I used on my MongoDB 3.2
db.myCollection.find({}, {myCustomKey:1}).sort({_id:1}).forEach(function(doc){
db.myCollection.remove({_id:{$gt:doc._id}, myCustomKey:doc.myCustomKey});
})
Index your customKey before running this to increase speed
While #Stennie's is a valid answer, it is not the only way. Infact the MongoDB manual asks you to be very cautious while doing that. There are two other options
Let the MongoDB do that for you using Map Reduce
Another way
You do programatically which is less efficient.
Here is a slightly more 'manual' way of doing it:
Essentially, first, get a list of all the unique keys you are interested.
Then perform a search using each of those keys and delete if that search returns bigger than one.
db.collection.distinct("key").forEach((num)=>{
var i = 0;
db.collection.find({key: num}).forEach((doc)=>{
if (i) db.collection.remove({key: num}, { justOne: true })
i++
})
});
I had a similar requirement but I wanted to retain the latest entry. The following query worked with my collection which had millions of records and duplicates.
/** Create a array to store all duplicate records ids*/
var duplicates = [];
/** Start Aggregation pipeline*/
db.collection.aggregate([
{
$match: { /** Add any filter here. Add index for filter keys*/
filterKey: {
$exists: false
}
}
},
{
$sort: { /** Sort it in such a way that you want to retain first element*/
createdAt: -1
}
},
{
$group: {
_id: {
key1: "$key1", key2:"$key2" /** These are the keys which define the duplicate. Here document with same value for key1 and key2 will be considered duplicate*/
},
dups: {
$push: {
_id: "$_id"
}
},
count: {
$sum: 1
}
}
},
{
$match: {
count: {
"$gt": 1
}
}
}
],
{
allowDiskUse: true
}).forEach(function(doc){
doc.dups.shift();
doc.dups.forEach(function(dupId){
duplicates.push(dupId._id);
})
})
/** Delete the duplicates*/
var i,j,temparray,chunk = 100000;
for (i=0,j=duplicates.length; i<j; i+=chunk) {
temparray = duplicates.slice(i,i+chunk);
db.collection.bulkWrite([{deleteMany:{"filter":{"_id":{"$in":temparray}}}}])
}
Expanding on Fernando's answer, I found that it was taking too long, so I modified it.
var x = 0;
db.collection.distinct("field").forEach(fieldValue => {
var i = 0;
db.collection.find({ "field": fieldValue }).forEach(doc => {
if (i) {
db.collection.remove({ _id: doc._id });
}
i++;
x += 1;
if (x % 100 === 0) {
print(x); // Every time we process 100 docs.
}
});
});
The improvement is basically using the document id for removing, which should be faster, and also adding the progress of the operation, you can change the iteration value to your desired amount.
Also, indexing the field before the operation helps.
pip install mongo_remove_duplicate_indexes
create a script in any language
iterate over your collection
create new collection and create new index in this collection with unique set to true ,remember this index has to be same as index u wish to remove duplicates from in ur original collection with same name
for ex-u have a collection gaming,and in this collection u have field genre which contains duplicates,which u wish to remove,so just create new collection
db.createCollection("cname")
create new index
db.cname.createIndex({'genre':1},unique:1)
now when u will insert document with similar genre only first will be accepted,other will be rejected with duplicae key error
now just insert the json format values u received into new collection and handle exception using exception handling
for ex pymongo.errors.DuplicateKeyError
check out the package source code for the mongo_remove_duplicate_indexes for better understanding
If you have enough memory, you can in scala do something like that:
cole.find().groupBy(_.customField).filter(_._2.size>1).map(_._2.tail).flatten.map(_.id)
.foreach(x=>cole.remove({id $eq x})