I have a collection which im aggregating and grouping by the field "type". The final result should be just maximum of five documents in each type. But if i limit before group only five first docs will be grouped. if i limit after the group the first five types will return.
is there a way to do this without doing find() for each type , limiting to 5 and merging all the results ?
If you can use C# (which according to my quick google-search about mongodb you do), you can do this with one of the GroupBy's which have an "ResultSelector-Function", like this:
var groups = Enumerable.Range(0, 1000).
GroupBy(
x => x/10,
(key, elements) => new { Key = key, Elements = elements.Take(5) }
);
About the speed of this code - I believe the group is completely build before the result-selector is instantiated - so a custom foreach over the input sequence and building the groups by hand might be faster (if you can somehow determine when you are done)
P.S.: On second thought - I doubt my answer is the one you want. I had a look at the mongo-DB documentation, and "map" in combination with a suitable "reduce" function might be exactly what you want.
Related
I am using Couchbase 3.0 and I have a use case where I want to perform sorting, but as i cam through couchbase does not provide Sorting on Values but it provides Sorting on Keys, But as I use descending(true) it is returning me empty list. And on the other hand If I just simply use it without descending then it is giving me all the docs related.
My Map function is :
function (doc, meta) {
emit([meta.id,doc.latest.sortData],null);
}
}
Now my use case is that I want to perform a match query on meta.id and then for all the matched cases sort the data and then find out the top values.
The code that i am using to so is :
ViewQuery.from(DesignDocName, viewName).startKey(JsonArray.from(write(List("something","")))).descending(true).limit(5).stale(Stale.FALSE))
If I remove the descending parameter then I get the related rows but they are not sorted. So could you please provide me a way in which this can be done.
Any Help is appreciated
Thanks!
I was reading Couchbase's docs and stumbled upon this line:
Because selection is made after sorting the view results, if you
configure the results to be sorted in descending order and you are
selecting information using a key range, then you must also reverse
the startkey and endkey parameters.
see here: http://docs.couchbase.com/admin/admin/Views/views-querying.html#ordering
Let's say I have two collections, A and B, and a single document in A is related to N documents in B. For example, the schemas could look like this:
Collection A:
{id: (int),
propA1: (int),
propA2: (boolean)
}
Collection B:
{idA: (int), # id for document in Collection A
propB1: (int),
propB2: (...),
...
propBN: (...)
}
I want to return properties propB2-BN and propA2 from my API, and only return information where (for example) propA2 = true, propB6 = 42, and propB1 = propA1.
This is normally fairly simple - I query Collection B to find documents where propB6 = 42, collect the idA values from the result, query Collection A with those values, and filter the results with the Collection A documents from the query.
However, adding skip and limit parameters to this seems impossible to do while keeping the behavior users would expect. Naively applying skip and limit to the first query means that, since filtering occurs after the query, less than limit documents could be returned. Worse, in some cases no documents could be returned when there are actually still documents in the collection to be read. For example, if the limit was 10 and the first 10 Collection B documents returned pointed to a document in Collection A where propA2 = false, the function would return nothing. Then the user would assume there's nothing left to read, which may not be the case.
A slightly less naive solution is to simply check if the return count is < limit, and if so, repeat the queries until the return count = limit. The problem here is that skip/limit queries where the user would expect exclusive sets of documents returned could actually return the same documents.
I want to apply skip and limit at the mongo query level, not at the API level, because the results of querying collection B could be very large.
MapReduce and the aggregation framework appear to only work on a single collection, so they don't appear to be alternatives.
This seems like something that'd come up a lot in Mongo use - any ideas/hints would be appreciated.
Note that these posts ask similar sounding questions but don't actually address the issues raised here.
Sounds like you already have a solution (2).
You cannot optimize/skip/limit on first query, depending on search you can perhaps do it on second query.
You will need a loop around it either way, like you write.
I suppose, the .skip will always be costly for you, since you will need to get all the results and then throw them away, to simulate the skip, to give the user consistent behavior.
All the logic would have to go to your loop - unless you can match in a clever way to second query (depending on requirements).
Out of curiosity: Given the time passed, you should have a solution by now?!
What's the easiest way to get all the documents from a collection that are unique based on a single field.
I know I can use db.collections.distrinct to get an array of all the distinct values of a field, but I want to get the first (or really any one) document for every distinct value of one field.
e.g. if the database contained:
{number:1, data:'Test 1'}
{number:1, data:'This is something else'}
{number:2, data:'I'm bad at examples'}
{number:3, data:'I guess there\'s room for one more'}
it would return (based on number being unique:
{number:1, data:'Test 1'}
{number:2, data:'I'm bad at examples'}
{number:3, data:'I guess there\'s room for one more'}
Edit: I should add that the server is running Mongo 2.0.8 so no aggregation and there's more results than group will support.
Update to 2.4 and use aggregation :)
When you really need to stick to the old version of MongoDB due to too much red tape involved, you could use MapReduce.
In MapReduce, the map function transforms each document of the collection into a new document and a distinctive key. The reduce function is used to merge documents with the same distincitve key into one.
Your map function would emit your documents as-is and with the number-field as unique key. It would look like this:
var mapFunction = function(document) {
emit(document.number, document);
}
Your reduce-function receives arrays of documents with the same key, and is supposed to somehow turn them into one document. In this case it would just discard all but the first document with the same key:
var reduceFunction = function(key, documents) {
return documents[0];
}
Unfortunately, MapReduce has some problems. It can't use indexes, so at least two javascript functions are executed for every single document in the collections (it can be limited by pre-excluding some documents with the query-argument to the mapReduce command). When you have a large collection, this can take a while. You also can't fully control how the docments created by MapReduce are formed. They always have two fields, _id with the key and value with the document you returned for the key.
MapReduce is also hard to debug an troubleshoot.
tl;dr: Update to 2.4
I am implementing a simple 'get max value' map reduce in MongoDB (c# driver).
For my tests I have 10 items in a collection with int _id = 1 to 10.
My map and reduce are as follows:
var map = "function() {emit('_id', this.Id);}";
var reduce = "function(key, values) {var max = 1; for (id in values) {if(id>max) {max=id;}} } return max;}";
When I run however I get the result 9, strange!!
I think that the map is outputting a string, and thus the compare is not working as desired.
Any help would be great
Reduce function won't run if the values contain only one item. If all the ids are unique and your key in the map is only that id, reduce phase won't work because of a design issue (for improving performance). If you need to change the format of your reduce output, you should use finalize method. Or just take a look at the aggregation framework which provides quite useful tools for playing with data.
Check the jira
jira.mongodb.org/browse/SERVER-5818
If you are just trying to get familiar with map reduce I would suggest to try different scenarios where using map-reduce really makes sense
Cheers
I have over 300k records in one collection in Mongo.
When I run this very simple query:
db.myCollection.find().limit(5);
It takes only few miliseconds.
But when I use skip in the query:
db.myCollection.find().skip(200000).limit(5)
It won't return anything... it runs for minutes and returns nothing.
How to make it better?
One approach to this problem, if you have large quantities of documents and you are displaying them in sorted order (I'm not sure how useful skip is if you're not) would be to use the key you're sorting on to select the next page of results.
So if you start with
db.myCollection.find().limit(100).sort({created_date:true});
and then extract the created date of the last document returned by the cursor into a variable max_created_date_from_last_result, you can get the next page with the far more efficient (presuming you have an index on created_date) query
db.myCollection.find({created_date : { $gt : max_created_date_from_last_result } }).limit(100).sort({created_date:true});
From MongoDB documentation:
Paging Costs
Unfortunately skip can be (very) costly and requires the server to walk from the beginning of the collection, or index, to get to the offset/skip position before it can start returning the page of data (limit). As the page number increases skip will become slower and more cpu intensive, and possibly IO bound, with larger collections.
Range based paging provides better use of indexes but does not allow you to easily jump to a specific page.
You have to ask yourself a question: how often do you need 40000th page? Also see this article;
I found it performant to combine the two concepts together (both a skip+limit and a find+limit). The problem with skip+limit is poor performance when you have a lot of docs (especially larger docs). The problem with find+limit is you can't jump to an arbitrary page. I want to be able to paginate without doing it sequentially.
The steps I take are:
Create an index based on how you want to sort your docs, or just use the default _id index (which is what I used)
Know the starting value, page size and the page you want to jump to
Project + skip + limit the value you should start from
Find + limit the page's results
It looks roughly like this if I want to get page 5432 of 16 records (in javascript):
let page = 5432;
let page_size = 16;
let skip_size = page * page_size;
let retval = await db.collection(...).find().sort({ "_id": 1 }).project({ "_id": 1 }).skip(skip_size).limit(1).toArray();
let start_id = retval[0].id;
retval = await db.collection(...).find({ "_id": { "$gte": new mongo.ObjectID(start_id) } }).sort({ "_id": 1 }).project(...).limit(page_size).toArray();
This works because a skip on a projected index is very fast even if you are skipping millions of records (which is what I'm doing). if you run explain("executionStats"), it still has a large number for totalDocsExamined but because of the projection on an index, it's extremely fast (essentially, the data blobs are never examined). Then with the value for the start of the page in hand, you can fetch the next page very quickly.
i connected two answer.
the problem is when you using skip and limit, without sort, it just pagination by order of table in the same sequence as you write data to table so engine needs make first temporary index. is better using ready _id index :) You need use sort by _id. Than is very quickly with large tables like.
db.myCollection.find().skip(4000000).limit(1).sort({ "_id": 1 });
In PHP it will be
$manager = new \MongoDB\Driver\Manager("mongodb://localhost:27017", []);
$options = [
'sort' => array('_id' => 1),
'limit' => $limit,
'skip' => $skip,
];
$where = [];
$query = new \MongoDB\Driver\Query($where, $options );
$get = $manager->executeQuery("namedb.namecollection", $query);
I'm going to suggest a more radical approach. Combine skip/limit (as an edge case really) with sort range based buckets and base the pages not on a fixed number of documents, but a range of time (or whatever your sort is). So you have top-level pages that are each range of time and you have sub-pages within that range of time if you need to skip/limit, but I suspect the buckets can be made small enough to not need skip/limit at all. By using the sort index this avoids the cursor traversing the entire inventory to reach the final page.
My collection has around 1.3M documents (not that big), properly indexed, but still takes a big performance hit by the issue.
After reading other answers, the solution forward is clear; the paginated collection must be sorted by a counting integer similar to the auto-incremental value of SQL instead of the time-based value.
The problem is with skip; there is no other way around it; if you use skip, you are bound to hit with the issue when your collection grows.
Using a counting integer with an index allows you to jump using the index instead of skip. This won't work with time-based value because you can't calculate where to jump based on time, so skipping is the only option in the latter case.
On the other hand,
by assigning a counting number for each document, the write performance would take a hit; because all documents must be inserted sequentially. This is fine with my use case, but I know the solution is not for everyone.
The most upvoted answer doesn't seem applicable to my situation, but this one does. (I need to be able to seek forward by arbitrary page number, not just one at a time.)
Plus, it is also hard if you are dealing with delete, but still possible because MongoDB support $inc with a minus value for batch updating. Luckily I don't have to deal with the deletion in the app I am maintaining.
Just write this down as a note to my future self. It is probably too much hassle to fix this issue with the current application I am dealing with, but next time, I'll build a better one if I were to encounter a similar situation.
If you have mongos default id that is ObjectId, use it instead. This is probably the most viable option for most projects anyway.
As stated from the official mongo docs:
The skip() method requires the server to scan from the beginning of
the input results set before beginning to return results. As the
offset increases, skip() will become slower.
Range queries can use indexes to avoid scanning unwanted documents,
typically yielding better performance as the offset grows compared to
using skip() for pagination.
Descending order (example):
function printStudents(startValue, nPerPage) {
let endValue = null;
db.students.find( { _id: { $lt: startValue } } )
.sort( { _id: -1 } )
.limit( nPerPage )
.forEach( student => {
print( student.name );
endValue = student._id;
} );
return endValue;
}
Ascending order example here.
If you know the ID of the element from which you want to limit.
db.myCollection.find({_id: {$gt: id}}).limit(5)
This is a lil genious solution which works like charm
For faster pagination don't use the skip() function. Use limit() and find() where you query over the last id of the precedent page.
Here is an example where I'm querying over tons of documents using spring boot:
Long totalElements = mongockTemplate.count(new Query(),"product");
int page =0;
Long pageSize = 20L;
String lastId = "5f71a7fe1b961449094a30aa"; //this is the last id of the precedent page
for(int i=0; i<(totalElements/pageSize); i++) {
page +=1;
Aggregation aggregation = Aggregation.newAggregation(
Aggregation.match(Criteria.where("_id").gt(new ObjectId(lastId))),
Aggregation.sort(Sort.Direction.ASC,"_id"),
new CustomAggregationOperation(queryOffersByProduct),
Aggregation.limit((long)pageSize)
);
List<ProductGroupedOfferDTO> productGroupedOfferDTOS = mongockTemplate.aggregate(aggregation,"product",ProductGroupedOfferDTO.class).getMappedResults();
lastId = productGroupedOfferDTOS.get(productGroupedOfferDTOS.size()-1).getId();
}