I have a collection that stores information about articles. The collection is for archival purposes so it is read only. Only two fields are being used at the moment: "title" and "page_length". Because I am always interested in getting longer articles first, I have the following index in place: { title: 1, page_length: -1}.
I have found that sorts are still slow because the collection is very large and won't fit into memory.
Assuming that almost every query I use on this collection will require a sort({page_length:-1}), is there any way to simply have the records stored on disk in order of page_length descending? In other words, is there a simple way to make the first record in the collection the largest page_length value, the second record the second largest, and so on?
That way I could just grab the first n records using limit(n) without having to run a sort. Any ideas?
Updating with more information:
I'm using this for a search autocomplete feature so speed is critical. The query I've been using looks like this:
db.articles.find({"title": /^SomeKeyword/}).sort({page_length:-1})
I'm happy to create multiple indexes since inserts are not a concern, I just want to maximize read speed.
EDIT: For reference, I actually was able to reorganize the records in the collection by using a find().forEach() into a new collection. I then searched the collection and grabbed the first N results without the need for any sort, which worked very well. Note that this ONLY works because my dataset does not ever change.
Your index { title: 1, page_length: -1 } is not used for the a query that looks like this:
db.collection.find( {} ).sort( { page_length: -1 } );
MongoDB can only use compound indexes from left to right, so in order for the index to be used, you need to have the "title" as a find or sort argument:
db.collection.find( { title: 'foo' } ).sort( { page_length: -1 } );
db.collection.find().sort( { title: 1, page_length: -1 } );
Explain will tell you:
db.so.find( {} ).sort( { page_length: -1 } ).explain();
{
"cursor" : "BasicCursor",
…
If you change your index to:
db.so.ensureIndex({ page_length: -1, title: 1 } );
Then the index will be used for sorting, but you can't use the index for just doing a lookup by title and you will need an additional index for that. If you're really only interested in those two fields and making sure you use a covered index helps. You will have to have the compound index with { page_length: -1, title: 1 } and you can make sure it is used by using a projection:
db.collection.find( {}, { page_length: 1, title: 1, _id: 0 } ).sort( { page_length: -1 } );
But you can not decide or influence how MongoDB stores things on disk.
I can think of a solution that uses two queries.
First, you can do a covered query to get the list of documents you care about. Second you can use the list of documents retrieved and the $in operator to get the final result.
The covered query will operate within memory (or at least sequentially on disk), so it should be fast, and the $in can utilize the _id index and should be tolerably efficient with a reasonable number of documents.
Related
I'm having a problem with the time of my mongoDB query, from a node backend using mongoose. i have a collection called people that has 10M records, and every record is queried from the backend and inserted from another part of the system that's written in c++ and needs to be very fast.
this is my mongoose schema:
{
_id: {type: String, index: {unique: true}}, // We generate our own _id! Might it be related to the slowness?
age: { type: Number },
id_num: { type: String },
friends: { type: Object }
}
schema.index({'id_num': 1}, { unique: true, collation: { locale: 'en_US', strength: 2 } })
schema.index({'age': 1})
schema.index({'id_num': 'text'});
Friends is an object looking like that: {"Adam": true, "Eve": true... etc.}.
there's no meaning to the value, and we use dictionaries to avoid duplicates fast on C++.
also, we didn't encounter a set/unique-list type of field in mongoDB.
The Problem:
We display people in a table with pagination. the table has abilities of sort, search, and select number of results.
At first, I queried all people and searched, sorted and paged it on the js. but when there are a lot of documents, It's turning problematic (memory problems).
The next thing i did was to try to fit those manipulations (searching, sorting & paging) on my query.
I used mongo's text search- but it not matches a partial word. is there any way to search a partial insensitive string? (I prefer not to use regex, to avoid unexpected problems)
I have to sort before paging, so I tried to use mongo sort. the problem is, that when the user wants to sort by "Friends", we want to return the people sorted by their number of friends (number of entries in the object).
The only way i succeeded pulling it off was using $addFields in aggregation:
{$addFields: {$size: {$ifNull: [{$objectToArray: '$friends'}, [] ]}}}
this addition is taking forever! when sorting by friends, the query takes about 40s for 8M people, and without this part it takes less than a second.
I used limit and skip for pagination. it works ok, but we have to wait until the user requests the second page and make another very long query.
In the end, this is the the interesting code part:
const { sortBy, sortDesc, search, page, itemsPerPage } = req.query
// Search never matches partial string
const match = search ? {$text: {$search: search}} : {}
const sortByInDB = ['age', 'id_num']
let sort = {$sort : {}}
const aggregate = [{$match: match}]
// if sortBy is on a simple type, we just use mongos sort
// else, we sortBy friends, and add a friends_count field.
if(sortByInDB.includes(sortBy)){
sort.$sort[sortBy] = sortDesc === 'true' ? -1 : 1
} else {
sort.$sort[sortBy+'_count'] = sortDesc === 'true' ? -1 : 1
// The problematic part of the query:
aggregate.push({$addFields: {friends_count: {$size: {
$ifNull: [{$objectToArray: '$friends'},[]]
}}}})
}
const numItems = parseInt(itemsPerPage)
const numPage = parseInt(page)
aggregate.push(sort, {$skip: (numPage - 1)*numItems}, {$limit: numItems})
// Takes a long time (when sorting by "friends")
let users = await User.aggregate(aggregate)
I tried indexing all simple fields, but the time is still too much.
The only other solution i could think of, is making mongo calculate a field "friends_count" every time a document is created or updated- but i have no idea how to do it, without slowing our c++ that writes to the DB.
Do you have any creative idea to help me? I'm lost, and I have to shorten the time drastically.
Thank you!
P.S: some useful information- the C++ area is writing the people to the DB in a bulk once in a while. we can sync once in a while and mostly rely on the data to be true. So, if that gives any of you any idea for a performance boost, i'd love to hear it.
Thanks!
we have a collection with big amount of documents, lets say around 100k. We now want to count the number of documents which has the key x set.
If I try it with Collection.countDocuments({ x: { $exists: true }}) I get the result, but it creates instantly a warning in the console: Query Targeting: Scanned Objects / Returned has gone above 1000.
So, is there a better way to count the documents? There is a Index on the field, is it possible to get the length of the index?
Thanks
Theres no real way of viewing the index trees in Mongo, what other people have linked you just returns the size of the tree, I'm not sure how useful that information is in this context.
Now to your question is this the best way to count?.
The answer is Yes ... -ish.
countDocuments is a wrapper function, it just simulates the following pipeline:
db.collection.aggregate([
{ $match: <query> },
{ $group: { _id: null, n: { $sum: 1 } } } )
])
This pipeline is the most efficient way to go, but the difference between running this aggregation and using the wrapper function is about 100-200 milliseconds, depending on your machine spec.
Meaning if you're looking for "way" better performance you're not going to find it.
With that said this warning is stupid, it just means you have more than 1000 documents with that field. The true purpose of it is to alert you in the case you're trying to query 1-20 documents without a proper index.
You can use the indexSizes field returned by the stats() method.
The stats() method "Returns statistics about the collection".
See example here :
https://docs.mongodb.com/manual/reference/method/db.collection.stats/#basic-stats-lookup
{
...,
"indexSizes" : {
"_id_" : 237568,
"cuisine_1" : 143360,
"borough_1_cuisine_1" : 151552,
"borough_1_address.zipcode_1" : 151552
},
...
}
indexSize key return size as in space used in storing not count
Check With Explain if index getting used or not . (Update in question Also)
can use hint option to check the performance after specifying index
Or precalculate count by $inc operator might good option if possible in you use case
try cursor.count if its faster countDocument should been faster but no harm in checking
https://docs.mongodb.com/manual/reference/method/cursor.count/
On a collection with over 100k records, when I query with Mongoose options like so:
contact.find({}, {}, {
collation: {
locale: 'en_US',
strength: 1
},
skip: 90000,
limit: 10,
sort: {
email: 1
}
});
I get this error:
MongoError: Executor error during find command: OperationFailed: Sort operation used more than the maximum 33554432 bytes of RAM. Add an index, or specify a smaller limit.
But I do have an index on the email field:
{
"v" : 2,
"key" : {
"email" : 1
},
"name" : "email_1",
"ns" : "leadfox.contact",
"background" : true
}
On the other hand when I query in the Mongo shell it works fine:
db.contact.find().sort({email: 1}).skip(90000).limit(10)
What you are experiencing is because of skip. As you can see in documentation
The cursor.skip() method is often expensive because it requires the server to walk from the beginning of the collection or index to get the offset or skip position before beginning to return results. As the offset (e.g. pageNumber above) increases, cursor.skip() will become slower and more CPU intensive. With larger collections, cursor.skip() may become IO bound.
You should find a better approach instead of skip. As you are sorting documents with email field, you can write a range query with email field instead of skiplike that:
contact.find({ "email": { $gt: the_last_email_from_previous_query } }, {}, {
collation: {
locale: 'en_US',
strength: 1
},
limit: 10,
sort: {
email: 1
}
});
Update:
First of all. Like I said above, what you want is not possible. Mongodb says it, not me.
Secondly, I suggest you to search about modern pagination methods and people use cases. Your example in comment is absurd. No user should/would go directly to 790th page for any data. If they go directly a page like that, it most probably means , they covered data till 790th page and they want to continue. So even you are building a stateless system(like all modern systems these days) you should store some information about users last-point-view for your paginated data. This is an example approach(I am not saying best, it is just an example) based on user behavior.
Another approach, you can use (like most of modern pagination tables) you only allow user to navigate 5-6 pages forward or backward. So you can skip only 50-60 document in your query combined with $gtand $lton emailfield.
Another approach can be caching data in memory with some other tools.
I think you get the picture. Happy coding.
What's a good way to store a set of documents in MongoDB where order is important? I need to easily insert documents at an arbitrary position and possibly reorder them later.
I could assign each item an increasing number and sort by that, or I could sort by _id, but I don't know how I could then insert another document in between other documents. Say I want to insert something between an element with a sequence of 5 and an element with a sequence of 6?
My first guess would be to increment the sequence of all of the following elements so that there would be space for the new element using a query something like db.items.update({"sequence":{$gte:6}}, {$inc:{"sequence":1}}). My limited understanding of Database Administration tells me that a query like that would be slow and generally a bad idea, but I'm happy to be corrected.
I guess I could set the new element's sequence to 5.5, but I think that would get messy rather quickly. (Again, correct me if I'm wrong.)
I could use a capped collection, which has a guaranteed order, but then I'd run into issues if I needed to grow the collection. (Yet again, I might be wrong about that one too.)
I could have each document contain a reference to the next document, but that would require a query for each item in the list. (You'd get an item, push it onto the results array, and get another item based on the next field of the current item.) Aside from the obvious performance issues, I would also not be able to pass a sorted mongo cursor to my {#each} spacebars block expression and let it live update as the database changed. (I'm using the Meteor full-stack javascript framework.)
I know that everything has it's advantages and disadvantages, and I might just have to use one of the options listed above, but I'd like to know if there is a better way to do things.
Based on your requirement, one of the approaches could be to design your schema, in such a way that each document has the capability to hold more than one document and in itself act as a capped container.
{
"_id":Number,
"doc":Array
}
Each document in the collection will act as a capped container, and the documents will be stored as array in the doc field. The doc field being an array, will maintain the order of insertion.
You can limit the number of documents to n. So the _id field of each container document will be incremental by n, indicating the number of documents a container document can hold.
By doing these you avoid adding extra fields to the document, extra indices, unnecessary sorts.
Inserting the very first record
i.e when the collection is empty.
var record = {"name" : "first"};
db.col.insert({"_id":0,"doc":[record]});
Inserting subsequent records
Identify the last container document's _id, and the number of
documents it holds.
If the number of documents it holds is less than n, then update the
container document with the new document, else create a new container
document.
Say, that each container document can hold 5 documents at most,and we want to insert a new document.
var record = {"name" : "newlyAdded"};
// using aggregation, get the _id of the last inserted container, and the
// number of record it currently holds.
db.col.aggregate( [ {
$group : {
"_id" : null,
"max" : {
$max : "$_id"
},
"lastDocSize" : {
$last : "$doc"
}
}
}, {
$project : {
"currentMaxId" : "$max",
"capSize" : {
$size : "$lastDocSize"
},
"_id" : 0
}
// once obtained, check if you need to update the last container or
// create a new container and insert the document in it.
} ]).forEach( function(check) {
if (check.capSize < 5) {
print("updating");
// UPDATE
db.col.update( {
"_id" : check.currentMaxId
}, {
$push : {
"doc" : record
}
});
} else {
print("inserting");
//insert
db.col.insert( {
"_id" : check.currentMaxId + 5,
"doc" : [ record ]
});
}
})
Note that the aggregation, runs on the server side and is very efficient, also note that the aggregation would return you a document rather than a cursor in versions previous to 2.6. So you would need to modify the above code to just select from a single document rather than iterating a cursor.
Inserting a new document in between documents
Now, if you would like to insert a new document between documents 1 and 2, we know that the document should fall inside the container with _id=0 and should be placed in the second position in the doc array of that container.
so, we make use of the $each and $position operators for inserting into specific positions.
var record = {"name" : "insertInMiddle"};
db.col.update(
{
"_id" : 0
}, {
$push : {
"doc" : {
$each : [record],
$position : 1
}
}
}
);
Handling Over Flow
Now, we need to take care of documents overflowing in each container, say we insert a new document in between, in container with _id=0. If the container already has 5 documents, we need to move the last document to the next container and do so till all the containers hold documents within their capacity, if required at last we need to create a container to hold the overflowing documents.
This complex operation should be done on the server side. To handle this, we can create a script such as the one below and register it with mongodb.
db.system.js.save( {
"_id" : "handleOverFlow",
"value" : function handleOverFlow(id) {
var currDocArr = db.col.find( {
"_id" : id
})[0].doc;
print(currDocArr);
var count = currDocArr.length;
var nextColId = id + 5;
// check if the collection size has exceeded
if (count <= 5)
return;
else {
// need to take the last doc and push it to the next capped
// container's array
print("updating collection: " + id);
var record = currDocArr.splice(currDocArr.length - 1, 1);
// update the next collection
db.col.update( {
"_id" : nextColId
}, {
$push : {
"doc" : {
$each : record,
$position : 0
}
}
});
// remove from original collection
db.col.update( {
"_id" : id
}, {
"doc" : currDocArr
});
// check overflow for the subsequent containers, recursively.
handleOverFlow(nextColId);
}
}
So that after every insertion in between , we can invoke this function by passing the container id, handleOverFlow(containerId).
Fetching all the records in order
Just use the $unwind operator in the aggregate pipeline.
db.col.aggregate([{$unwind:"$doc"},{$project:{"_id":0,"doc":1}}]);
Re-Ordering Documents
You can store each document in a capped container with an "_id" field:
.."doc":[{"_id":0,","name":"xyz",...}..]..
Get hold of the "doc" array of the capped container of which you want
to reorder items.
var docArray = db.col.find({"_id":0})[0];
Update their ids so that after sorting the order of the item will change.
Sort the array based on their _ids.
docArray.sort( function(a, b) {
return a._id - b._id;
});
update the capped container back, with the new doc array.
But then again, everything boils down to which approach is feasible and suits your requirement best.
Coming to your questions:
What's a good way to store a set of documents in MongoDB where order is important?I need to easily insert documents at an arbitrary
position and possibly reorder them later.
Documents as Arrays.
Say I want to insert something between an element with a sequence of 5 and an element with a sequence of 6?
use the $each and $position operators in the db.collection.update() function as depicted in my answer.
My limited understanding of Database Administration tells me that a
query like that would be slow and generally a bad idea, but I'm happy
to be corrected.
Yes. It would impact the performance, unless the collection has very less data.
I could use a capped collection, which has a guaranteed order, but then I'd run into issues if I needed to grow the collection. (Yet
again, I might be wrong about that one too.)
Yes. With Capped Collections, you may lose data.
An _id field in MongoDB is a unique, indexed key similar to a primary key in relational databases. If there is an inherent order in your documents, ideally you should be able to associate a unique key to each document, with the key value reflecting the order. So while preparing your document for insertion, explicitly add an _id field as this key (if you do not, mongo creates it automatically with a BSON objectid).
As far as retrieving the results are concerned, MongoDB does not guarantee the order of return documents unless you explicitly use .sort() . If you do not use .sort(), the results are usually returned in natural order (order of insertion).Again, there is no guarantee on this behavior.
I'd advise you to override _id with your order while inserting, and use a sort while retrieving. Since _id is a necessary and auto-indexed entity, you will not be wasting any space defining a sort key, and storing the index for it.
For abitrary sorting of any collection, you'll need a field to sort it on. I call mine "sequence".
schema:
{
_id: ObjectID,
sequence: Number,
...
}
db.items.ensureIndex({sequence:1});
db.items.find().sort({sequence:1})
Here is a link to some general sorting database answers that may be relevant:
https://softwareengineering.stackexchange.com/questions/195308/storing-a-re-orderable-list-in-a-database/369754
I suggest going with Floating point solution - adding a position column:
Use a floating-point number for the position column.
You can then reorder the list changing only the position column in the "moved" row.
If your user wants to position "red" after "blue" but before "yellow" Then you just need to calculate
red.position = ((yellow.position - blue.position) / 2) + blue.position
After a few re-positions in the same place (Cuttin in half every time) - you might reach a wall - it's better that if you reach a certain threshold - to resort the list.
When retrieving it you can simply say col.sort() to get it sorted and no need for any client-side code (Like in the case of a Linked list solution)
I have a collection of md5 in mongodb. I'd like to find all duplicates. The md5 column is indexed. Do you know any fast way to do that using map reduce.
Or should I just iterate over all records and check for duplicates manually?
My current approach using map reduce iterates over the collection almost twice (assuming that there is very small amount of duplicates):
res = db.files.mapReduce(
function () {
emit(this.md5, 1);
},
function (key, vals) {
return Array.sum(vals);
}
)
db[res.result].find({value: {$gte:1}}).forEach(
function (obj) {
out.duplicates.insert(obj)
});
I personally found that on big databases (1TB and more) accepted answer is terribly slow. Aggregation is much faster. Example is below:
db.places.aggregate(
{ $group : {_id : "$extra_info.id", total : { $sum : 1 } } },
{ $match : { total : { $gte : 2 } } },
{ $sort : {total : -1} },
{ $limit : 5 }
);
It searches for documents whose extra_info.id is used twice or more times, sorts results in descending order of given field and prints first 5 values of it.
The easiest way to do it in one pass is to sort by md5 and then process appropriately.
Something like:
var previous_md5;
db.files.find( {"md5" : {$exists:true} }, {"md5" : 1} ).sort( { "md5" : 1} ).forEach( function(current) {
if(current.md5 == previous_md5){
db.duplicates.update( {"_id" : current.md5}, { "$inc" : {count:1} }, true);
}
previous_md5 = current.md5;
});
That little script sorts the md5 entries and loops through them in order. If an md5 is repeated, then they will be "back-to-back" after sorting. So we just keep a pointer to previous_md5 and compare it current.md5. If we find a duplicate, I'm dropping it into the duplicates collection (and using $inc to count the number of duplicates).
This script means that you only have to loop through the primary data set once. Then you can loop through the duplicates collection and perform clean-up.
You can do a group by that field and then query to get the duplicated (having a count > 1). http://www.mongodb.org/display/DOCS/Aggregation#Aggregation-Group
Although, the fastest thing might be to just do a query which only returns that field and then to do the aggregation in the client. Group/Map-Reduce need to provide access to the whole document which is much more costly than just providing the data from the index (which is now covered in 1.7.3+).
If this is a general problem you need to run periodically, you might want to keep a collection which is just {md5:value, count:value} so you can skip the aggregation, and it will be extremely fast when you need to cull duplicates.