MongoDB pagination using cursors without sorting - mongodb

When looking for pagination techniques on the internet, one usually finds two ways :
Offset-based pagination :
Collection.find(
{ where_this: "equals that" },
{ skip: 15, limit: 5 }
)
Cursor-based pagination :
Collection.find(
{ where_this: "equals that", _id: { $gt: cursor }},
{ sort: { _id: 1 }}
)
But is there a way to have cursor-based pagination without sorting the collection according to that cursor ? Like, telling Mongo "Alright, I want the 5 next items after that _id, no matter in which order the _ids are, just give me 5 items after you see that _id". Something along those lines :
Collection.find(
{ where_this: "equals that", _id: { $must_come_after: cursor }},
{ sort: { other_field: 1 }}
)
It is not always possible to use the field you're sorting with as the cursor. First of all, because these fields can be of different types and you might allow your app users to sort, for example, tables as they please. With a strongly-typed API framework like GraphQL, this would be a mess to handle. Secondly, you could have two or more equal values for that field following each other in the sorted collection. If your pages split in the middle, asking for the next page will either give you duplicates or ignore items.
Is there a way to do that ? How is it usually done to allow custom sorting fields without offset-based pagination ? Thanks.

When we talk about "paginating a result set", the implicit assumption is that the result set stays the same throughout the process. In order for the result set to stay the same, it must normally be sorted. Without specifying an order, the database is free to return the documents in any order, and this order may change from one retrieval to the next. Reordering of documents that a user is paginating through creates a poor user experience.
you could have two or more equal values for that field following each other in the sorted collection. If your pages split in the middle, asking for the next page will either give you duplicates or ignore items.
The database can return the documents which compare equal in any order, and this order can change between adjacent queries. This is why when sorting on a field that has a low cardinality, it is a good idea to add another field with a high cardinality to the sort expression to ensure the documents are returned in a stable order.
But is there a way to have cursor-based pagination without sorting the collection according to that cursor ?
You can encode offset in a cursor identifier and use skip/limit.

Related

What is the correct way to Index in MongoDB when big combination of fields exist

Considering I have search pannel that inculude multiple options like in the picture below:
I'm working with mongo and create compound index on 3-4 properties with specific order.
But when i run a different combinations of searches i see every time different order in execution plan (explain()). Sometime i see it on Collection scan (bad) , and sometime it fit right to the index (IXSCAN).
The selective fields that should handle by mongo indexes are:(brand,Types,Status,Warehouse,Carries ,Search - only by id)
My question is:
Do I have to create all combination with all fields with different order , it can be 10-20 compound indexes. Or 1-3 big Compound Index , but again it will not solve the order.
What is the best strategy to deal with big various of fields combinations.
I use same structure queries with different combinations of pairs
// Example Query.
// fields could be different every time according to user select (and order) !!
db.getCollection("orders").find({
'$and': [
{
'status': {
'$in': [
'XXX',
'YYY'
]
}
},
{
'searchId': {
'$in': [
'3859447'
]
}
},
{
'origin.brand': {
'$in': [
'aaaa',
'bbbb',
'cccc',
'ddd',
'eee',
'bundle'
]
}
},
{
'$or': [
{
'origin.carries': 'YYY'
},
{
'origin.carries': 'ZZZ'
},
{
'origin.carries': 'WWWW'
}
]
}
]
}).sort({"timestamp":1})
// My compound index is:
{status:1 ,searchId:-1,origin.brand:1, origin.carries:1 , timestamp:1}
but it only 1 combination ...it could be plenty like
a. {status:1} {b.status:1 ,searchId:-1} {c. status:1 ,searchId:-1,origin.brand:1} {d.status:1 ,searchId:-1,origin.brand:1, origin.carries:1} ........
Additionally , What will happened with Performance write/read ? , I think write will decreased over reads ...
The queries pattern are :
1.find(...) with '$and'/'$or' + sort
2.Aggregation with Match/sort
thanks
Generally, indexes are only useful if they are over a selective field. This means the number of documents that have a particular value is small relative to the overall number of documents.
What "small" means varies on the data set and the query. A 1% selectivity is pretty safe when deciding whether an index makes sense. If an particular value exists in, say, 10% of documents, performing a table scan may be more efficient than using an index over the respective field.
With that in mind, some of your fields will be selective and some will not be. For example, I suspect filtering by "OK" will not be very selective. You can eliminate non-selective fields from indexing considerations - if someone wants all orders which are "OK" with no other conditions they'll end up doing a table scan. If someone wants orders which are "OK" and have other conditions, whatever index is applicable to other conditions will be used.
Now that you are left with selective (or at least somewhat selective) fields, consider what queries are both popular and selective. For example, perhaps brand+type would be such a combination. You could add compound indexes that match popular queries which you expect to be selective.
Now, what happens if someone filters by brand only? This could be selective or not depending on the data. If you already have a compound index on brand+type, you'd leave it up to the database to determine whether a brand only query is more efficient to fulfill via the brand+type index or via a collection scan.
Continue in this manner with other popular queries and fields.
So you have subdocuments, ranged queries, and sorting by 1 field only.
It can eliminate most of the possible permutations. Assuming there are no other surprises.
D. SM already covered selectivity - you should really listen what the man says and at least upvote.
The other things to consider is the order of the fields in the compound index:
fields that have direct match like $eq
fields you sort on
fields with ranged queries: $in, $lt, $or etc
These are common rules for all b-trees. Now things that are specific to mongo:
A compound index can have no more than 1 multikey index - the index by a field in subdocuments like "origin.brand". Again I assume origins are embedded docs, so the document's shape is like this:
{
_id: ...,
status: ...,
timestamp: ....,
origin: [
{brand: ..., carries: ...},
{brand: ..., carries: ...},
{brand: ..., carries: ...}
]
}
For your query the best index would be
{
searchId: 1,
timestamp: 1,
status: 1, /** only if it is selective enough **/
"origin.carries" : 1 /** or brand, depending on data **/
}
Regarding the number of indexes - it depends on data size. Ensure all indexes fit into RAM otherwise it will be really slow.
Last but not least - indexing is not a one off job but a lifestyle. Data change over time, so do queries. If you care about performance and have finite resources you should keep an eye on the database. Check slow queries to add new indexes, collect stats from user's queries to remove unused indexes and free up some room. Basically apply common sense.
I noticed this one-year-old topic, because I am more or less struggling with a similar issue: users can request queries with an unpredictable set of the fields, which makes it near to impossible to decide (or change) how indexes should be defined.
Even worse: the user should indicate some value (or range) for the fields that make up the sharding-key, otherwise we cannot help MongoDB to limit its search in only a few shards (or chunks, for that matter).
When the user needs the liberty to search on other fields that are not necessariy the ones which make up the sharding-key, then we're stuck with a full-database search. Our dbase is some 10's of TB size...
Indexes should fit in RAM ? This can only be achieved with small databases, meaning some 100's GB max. How about my 37 TB database ? Indexes won't fit in RAM.
So I am trying out a POC inspired by the UNIX filesystem structures where we have inodes pointing to data blocks:
we have a cluster with 108 shards, each contains 100 chunks
at insert time, we take some fields of which we know they yield a good cardinality of the data, and we compute the sharding-key with those fields; the document goes into the main collection (call it "Main_col") on that computed shard, so with a certain chunk-number (equals our computed sharding-key value)
from the original document, we take a few 'crucial' fields (the list of such fields can evolve as your needs change) and store a small extra document in another collection (call these "Crucial_col_A", Crucial_col_B", etc, one for each such field): that document contains the value of this crucial field, plus an array with the chunk-number where the original full document has been stored in the 'big' collection "Main_col"; consider this as a 'pointer' to the chunk in collecton "Main_col" where this full document exists. These "Crucial_col_X" collections are sharded based on the value of the 'crucial' field.
when we insert another document that has the same value for some 'crucial' field "A", then that array in "Crucial_col_A" with chunk-numbers with be updated (with 'merge') to contain the different or same chunk number of this next full document from "Main_col"
a user can now define queries with criteria for at least one of those 'crucial' fields, plus (optional) any other criteria on other fields in the documents; the first criterium for the crucial field (say field "B") will run very quickly (because sharded on the value of "B") and return the small document from "Crucial_col_B", in which we have the array of chunk-numbers in "Main_col" where any document exists that has field "B" equal to the given criterium. Then we run a second set of parallel queries, one for each shardkey-value=chunk-number (or one per shard, to be decided) that we find in the array from before. We combine the results of those parallel subqueries, and then apply further filtering if the user gave additional criteria.
Thus this involves 2 query-steps: first in the "Crucial_col_X" collection to obtain the array with chunk-numbers where the full documents exist, and then the second query on those specific chunks in "Main_col".
The first query is done with a precise value for the 'crucial' field, so the exact shard/chunk is known, thus this query goes very fast.
The second (set of) queries are done with precise values for the sharding-keys (= the chunk numbers), so these are expected to go also very fast.
This way of working would eliminate the burden of defining many index combinations.

custom sort for a mongodb collection in meteor

I have this collection of products and i want to display a top 10 products based on a custom sort function
[{ _id: 1, title, tags:['a'], createdAt:ISODate("2016-01-28T00:00:00Z") } ,
{ _id: 2, title, tags:['d','a','e'], createdAt:ISODate("2016-01-24T00:00:00Z") }]
What i want to do is to sort it based on a "magic score" that can be calculated. For example, based on this formula: tag_count*5 - number_of_days_since_it_was_created.
If the first one is 1 day old, this makes the score:
[{_id:1 , score: 4}, {_id:2, score: 10}]
I have a few ideas on how i can achieve this, but i'm not sure how good they are, especially since i'm new to both mongo and meteor:
start an observer (Meteor.observe) and every time a document is
modified (or a new one created), recalculate the score and update it
on the collection itself. If i do this, i could just use $orderBy
where i need it.
after some reading i discovered that mongo aggregate or map_reduce
could help me achieve the same result, but as far as i found out,
meteor doesn't support it directly
sort the collection on the client side as an array, but using this
method i'm not sure how it will behave with pagination (considering that i subscribe to a limited number of documents)
Thank you for any information you can share with me!
Literal function sorting is just being implemented in meteor, so you should be able to do something like
Products.find({}, {sort: scoreComparator});
in an upcoming release.
You can use the transform property when creating collection. In this transform, store the magic operation as a function.
score=function(){
// return some score
};
transformer=function(product){
product.score=score;
// one could also use prototypal inheritance
};
Products=new Meteor.Collection('products',{transform:transformer});
Unfortunately, you cannot yet use the sort operator on virtual fields, because minimongo does not support it.
So the ultimate fall-back as you mentioned while nor the virtual field sorting nor the literate function sorting are supported in minimongo is client side sorting :
// Later, within some template
scoreComparator=function(prd1,prd2){
return prd1.score()-prd2.score();
}
Template.myTemplate.helpers({
products:function(){
return Products.find().fetch().sort(scoreComparator);
}
});
i'm not sure how it will behave with pagination (considering that i subscribe to a limited number of documents)
EDIT : the score will be computed among the subscribed documents, indeed.

Iterating over distinct items in one field in MongoDB

I have a very large collection (~7M items) in MongoDB, primarily consisting of documents with three fields.
I'd like to be able to iterate over all the unique values for one of the fields, in an expedient manner.
Currently, I'm querying for just that field, and then processing the returned results by iterating on the cursor for uniqueness. This works, but it's rather slow, and I suspect there must be a better way.
I know mongo has the db.collection.distinct() function, but this is limited by the maximum BSON size (16 MB), which my dataset exceeds.
Is there any way to iterate over something similar to the db.collection.distinct(), but using a cursor or some other method, so the record-size limit isn't as much of an issue?
I think maybe something like the map/reduce functionality would possibly be suited for this kind of thing, but I don't really understand the map-reduce paradigm in the first place, so I have no idea what I'm doing. The project I'm working on is partially to learn about working with different database tools, so I'm rather inexperienced.
I'm using PyMongo if it's relevant (I don't think it is). This should be mostly dependent on MongoDB alone.
Example:
For this dataset:
{"basePath" : "foo", "internalPath" : "Neque", "itemhash": "49f4c6804be2523e2a5e74b1ffbf7e05"}
{"basePath" : "foo", "internalPath" : "porro", "itemhash": "ffc8fd5ef8a4515a0b743d5f52b444bf"}
{"basePath" : "bar", "internalPath" : "quisquam", "itemhash": "cf34a8047defea9a51b4a75e9c28f9e7"}
{"basePath" : "baz", "internalPath" : "est", "itemhash": "c07bc6f51234205efcdeedb7153fdb04"}
{"basePath" : "foo", "internalPath" : "qui", "itemhash": "5aa8cfe2f0fe08ee8b796e70662bfb42"}
What I'd like to do is iterate over just the basePath field. For the above dataset, this means I'd iterate over foo, bar, and baz just once each.
I'm not sure if it's relevant, but the DB I have is structured so that while each field is not unique, the aggregate of all three is unique (this is enforced with an index).
The query and filter operation I'm currently using (note: I'm restricting the query to a subset of the items to reduce processing time):
self.log.info("Running path query")
itemCursor = self.dbInt.coll.find({"basePath": pathRE}, fields={'_id': False, 'internalPath': False, 'itemhash': False}, exhaust=True)
self.log.info("Query complete. Processing")
self.log.info("Query returned %d items", itemCursor.count())
self.log.info("Filtering returned items to require uniqueness.")
items = set()
for item in itemCursor:
# print item
items.add(item["basePath"])
self.log.info("total unique items = %s", len(items))
Running the same query with self.dbInt.coll.distinct("basePath") results in OperationFailure: command SON([('distinct', u'deduper_collection'), ('key', 'basePath')]) failed: exception: distinct too big, 16mb cap
Ok, here is the solution I wound up using. I'd add it as an answer, but I don't want to detract from the actual answers that got me here.
reStr = "^%s" % fqPathBase
pathRE = re.compile(reStr)
self.log.info("Running path query")
pipeline = [
{ "$match" :
{
"basePath" : pathRE
}
},
# Group the keys
{"$group":
{
"_id": "$basePath"
}
},
# Output to a collection "tmp_unique_coll"
{"$out": "tmp_unique_coll"}
]
itemCursor = self.dbInt.coll.aggregate(pipeline, allowDiskUse=True)
itemCursor = self.dbInt.db.tmp_unique_coll.find(exhaust=True)
self.log.info("Query complete. Processing")
self.log.info("Query returned %d items", itemCursor.count())
self.log.info("Filtering returned items to require uniqueness.")
items = set()
retItems = 0
for item in itemCursor:
retItems += 1
items.add(item["_id"])
self.log.info("Recieved items = %d", retItems)
self.log.info("total unique items = %s", len(items))
General performance compared to my previous solution is about 2X in terms of wall-clock time. On a query that returns 834273 items, with 11467 uniques:
Original method(retreive, stuff into a python set to enforce uniqueness):
real 0m22.538s
user 0m17.136s
sys 0m0.324s
Aggregate pipeline method :
real 0m9.881s
user 0m0.548s
sys 0m0.096s
So while the overall execution time is only ~2X better, the aggregation pipeline is massively more performant in terms of actual CPU time.
Update:
I revisited this project recently, and rewrote the DB layer to use a SQL database, and everything was much easier. A complex processing pipeline is now a simple SELECT DISTINCT(colName) WHERE xxx operation.
Realistically, MongoDB and NoSQL databases in general are vary much the wrong database type for what I'm trying to do here.
From the discussion points so far I'm going to take a stab at this. And I'm also noting that as of writing, the 2.6 release for MongoDB should be just around the corner, good weather permitting, so I am going to make some references there.
Oh and the FYI that didn't come up in chat, .distinct() is an entirely different animal that pre-dates the methods used in the responses here, and as such is subject to many limitations.
And this soltion is finally a solution for 2.6 up, or any current dev release over 2.5.3
The alternative for now is use mapReduce because the only restriction is the output size
Without going into the inner workings of distinct, I'm going to go on the presumption that aggregate is doing this more efficiently [and even more so in upcoming release].
db.collection.aggregate([
// Group the key and increment the count per match
{$group: { _id: "$basePath", count: {$sum: 1} }},
// Hey you can even sort it without breaking things
{$sort: { count: 1 }},
// Output to a collection "output"
{$out: "output"}
])
So we are using the $out pipeline stage to get the final result that is over 16MB into a collection of it's own. There you can do what you want with it.
As 2.6 is "just around the corner" there is one more tweak that can be added.
Use allowDiskUse from the runCommand form, where each stage can use disk and not be subject to memory restrictions.
The main point here, is that this is nearly live for production. And the performance will be better than the same operation in mapReduce. So go ahead and play. Install 2.5.5 for you own use now.
A MapReduce, in the current version of Mongo would avoid the problems of the results exceeding 16MB.
map = function() {
if(this['basePath']) {
emit(this['basePath'], 1);
}
// if basePath always exists you can just call the emit:
// emit(this.basePath);
};
reduce = function(key, values) {
return Array.sum(values);
};
For each document the basePath is emitted with a single value representing the count of that value. The reduce simply creates the sum of all the values. The resulting collection would have all unique values for basePath along with the total number of occurrences.
And, as you'll need to store the results to prevent an error using the out option which specifies a destination collection.
db.yourCollectionName.mapReduce(
map,
reduce,
{ out: "distinctMR" }
)
#Neil Lunn 's answer could be simplified:
field = 'basePath' # Field I want
db.collection.aggregate( [{'$project': {field: 1, '_id': 0}}])
$project filters fields for you. In particular, '_id': 0 filters out the _id field.
Result still too large? Batch it with $limit and $skip:
field = 'basePath' # Field I want
db.collection.aggregate( [{'$project': {field: 1, '_id': 0}}, {'$limit': X}, {'$skip': Y}])
I think the most scalable solution is to perform a query for each unique value. The queries must be executed one after the other, and each query will give you the "next" unique value based on the previous query result. The idea is that the query will return you one single document, that will contain the unique value that you are looking for. If you use the proper projection, mongo will just use the index loaded into memory without having to read from disk.
You can define this strategy using $gt operator in mongo, but you must take into account values like null or empty strings, and potentially discard them using the $ne or $nin operator. You can also extend this strategy using multiple keys, using operators like $gte for one key and $gt for the other.
This strategy should give you the distinct values of a string field in alphabetical order, or distinct numerical values sorted ascendingly.

mongodb, make increment several times in single update

Having very simple 2 mongo documents:
{_id:1, v:1}
{_id:2, v:1}
Now, basing on array of _id I need increase field v as many times how _id appears. For example [1, 2, 1] should produce
{_id:1, v:3} //increased 2 times
{_id:2, v:2} //increased 1 times
Of course simple update eliminates duplicate in $in:
db.r.update({_id:{$in:[1,2,1]}}, {$inc:{v:1}}, {multi:true})
Is there a way to do it without for-loop? /Thank you in advance/
No there isn't a way to do this in a single update statement.
The reason why the $in operator "removes the duplicate" is a simple matter of the fact that th 1 was already matched, no point in matching again. So you can't make the document "match twice" as it were.
Also there is no current way to batch update operations. But that feature is coming.
You could look at your "batch" and make a decision to group together occurrences of the same document to be updated and then issue your increment to the appropriate number of units. However just like looping the array items, the operation would be programitic, albeit a little more efficient.
That isn't possible directly. You'll have to do that in your client, where you can at least try to minimize the number of batch updates required.
First, find the counts. This depends on your programming language, but what you want is something like [1, 2, 1] => [ { 1 : 2 }, { 2 : 1} ] (these are the counts for the respective ids, i.e. id 1 appears twice, etc.) Something like linq oder underscore.js is helpful here.
Next, since you can't perform different updates in a single operation, group them by their count, and update all objects whose count must be incremented by a common fixed value in one batch:
Pseudocode:
var groups = data.groupBy(p => p.Value);
foreach(var group in groups)
db.update({"_id" : { $in : group.values.asArray }},
// increase by the number of times those ids were present
{$inc : { v : group.key } })
That is better than individual updates only if there are many documents that must be increased by the same value.

Sorting on Multiple fields mongo DB

I have a query in mongo such that I want to give preference to the first field and then the second field.
Say I have to query such that
db.col.find({category: A}).sort({updated: -1, rating: -1}).limit(10).explain()
So I created the following index
db.col.ensureIndex({category: 1, rating: -1, updated: -1})
It worked just fined scanning as many objects as needed, i.e. 10.
But now I need to query
db.col.find({category: { $ne: A}}).sort({updated: -1, rating: -1}).limit(10)
So I created the following index:
db.col.ensureIndex({rating: -1, updated: -1})
but this leads to scanning of the whole document and when I create
db.col.ensureIndex({ updated: -1 ,rating: -1})
It scans less number of documents:
I just want to ask to be clear about sorting on multiple fields and what is the order to be preserved when doing so. By reading the MongoDB documents, it's clear that the field on which we need to perform sorting should be the last field. So that is the case I assumed in my $ne query above. Am I doing anything wrong?
The MongoDB query optimizer works by trying different plans to determine which approach works best for a given query. The winning plan for that query pattern is then cached for the next ~1,000 queries or until you do an explain().
To understand which query plans were considered, you should use explain(1), eg:
db.col.find({category:'A'}).sort({updated: -1}).explain(1)
The allPlans detail will show all plans that were compared.
If you run a query which is not very selective (for example, if many records match your criteria of {category: { $ne:'A'}}), it may be faster for MongoDB to find results using a BasicCursor (table scan) rather than matching against an index.
The order of fields in the query generally does not make a difference for the index selection (there are a few exceptions with range queries). The order of fields in a sort does affect the index selection. If your sort() criteria does not match the index order, the result data has to be re-sorted after the index is used (you should see scanAndOrder:true in the explain output if this happens).
It's also worth noting that MongoDB will only use one index per query (with the exception of $ors).
So if you are trying to optimize the query:
db.col.find({category:'A'}).sort({updated: -1, rating: -1})
You will want to include all three fields in the index:
db.col.ensureIndex({category: 1, updated: -1, rating: -1})
FYI, if you want to force a particular query to use an index (generally not needed or recommended), there is a hint() option you can try.
That is true but there are two layers of ordering you have here since you are sorting on a compound index.
As you noticed when the first field of the index matches the first field of sort it worked and the index was seen. However when working the other way around it does not.
As such by your own obersvations the order needed to be preserved is query order of fields from first to last. The mongo analyser can sometimes move around fields to match an index but normally it will just try and match the first field, if it cannot it will skip it.
try this code it will sort data first based on name then keeping the 'name' in key holder it will sort 'filter'
var cursor = db.collection('vc').find({ "name" : { $in: [ /cpu/, /memo/ ] } }, { _id: 0, }).sort( { "name":1 , "filter": 1 } );
Sort and Index Use
MongoDB can obtain the results of a sort operation from an index which
includes the sort fields. MongoDB may use multiple indexes to support
a sort operation if the sort uses the same indexes as the query
predicate. ... Sort operations that use an index often have better
performance than blocking sorts.
db.restaurants.find().sort( { "borough": 1, "_id": 1 } )
more information :
https://docs.mongodb.com/manual/reference/method/cursor.sort/