I have the following query to be executed on my MongoDB collection order_error. It has over 60 million documents. The main concern is I am having a $in operator within my query. I tried several possibilities of indices but none of them gave a high-performance improvement. The query is as follows
db.getCollection("order_error").find({
"$and":[
{
"type":"order"
},
{
"Origin.SN":{
"$in":[
"4095",
"4100",
"4509",
"4599",
"4510"
]
}
}
]
}).sort({"timestamp.milliseconds" : 1}).skip(1).limit(100).explain("executionStats")
One issue that needs to be noted is I am allowing sort on timestamp.milliseconds in both directions(ASC + DESC). I have limited the entries within the $in. Usually, it is more. SO what kind of index gives the performance improvement. I tried creating the following indices already
type_1_Origin.SN_1_timestamp.milliseconds_-1
type_1_timestamp.milliseconds_-1_Origin.SN
Is there any better way for index creation?
I am looking to a way to FindAndModify not more than 5 documents in MongoDB.
This is collection for queue which will be processed from multiple workers, so I want to put it into single query.
While I cannot control amount of updates in UpdateOptions parameter, is it possible to limit number of rows which will be found in filterDefinition?
Problem 1: findAndModify() can only update a single document at a time, as per the documentation. This is an inherent limit in MongoDB's implementation.
Problem 2: There is no way to update a specific number of arbitrary documents with a simple update() query of any kind. You can update one or all depending on the boolean value of your multi option, but that's it.
If you want to update up to 5 documents at a time, you're going to have to retrieve these documents first then update them, or update them individually in a foreach() call. Either way, you'll either be using something like:
db.collection.update(
{_id: {$in: [ doc1._id, doc2._id, ... ]}},
{ ... },
{multi: true}
);
Or you'll be using something like:
db.collection.find({ ... }).limit(5).forEach(function(doc) {
//do something to doc
db.collection.update({_id: doc._id}, doc);
});
Whichever approach you choose to take, it's going to be a workaround. Again, this is an inherent limitation.
There are 30,000,000 records in one collection.
when I use distinct command on this collection by java, it takes about 4 minutes, the result's count is about 40,000.
Is mongodb's distinct operation so inefficiency?
and how can I make it more efficient?
Is mongodb's distinct operation so inefficiency?
At 30m records? I would say 4 minutes is actually quite good, I think that's just as fast, maybe a little faster than SQL does it.
I would probably test this in other databases before saying it is inefficient.
However, one way of looking at performance is to see if the field is indexed first and if that index is in RAM or can be loaded without page thrashing. Distinct() can use an index so long as the field has an index.
and how can I make it more efficient?
You could use a couple of methods:
Incremental map reduce to distinct the main collection once every, say, 5 mins to a unique collection
And Pre-aggregate the unique collection on save by saving to two collections, one detail and one unique
Those are the two most viable methods of getting around this performantly.
Edit
Distinct() is not outdated and if it fits your needs is actually more performant than $group since it can use an index.
The .distinct() operation is an old one, as is .group(). In general these have been superseded by .aggregate() which should be generally used in preference to these actions:
db.collection.aggregate([
{ "$group": {
"_id": "$field",
"count": { "$sum": 1 }
}
)
Substituting "$field" with whatever field you wish to get a distinct count from. The $ prefixes the field name to assign the value.
Look at the documentation and especially $group for more information.
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 $gt for the other.
This strategy should give you the distinct values of a string field in alphabetical order, or distinct numerical values sorted ascendingly.
Document structure looks like this,
{
blacklists:[] // elements should be unique
blacklistsLength:0 // length of blacklists
}
Adding sets of value to blacklists is easy.
db.posts.update({_id:...}, {$addtoSet:{blacklists:{$each:['peter', 'bob', 'steven']}}});
But How can I update blacklistLength at the same time to reflect the changes?
This is not possible. Either you have
Update the length seperately using a subsequent findAndModify
command or
You can do it per name and rewrite the query using a negation in
your criteria and $push rather than $addToSet (not necessarily
needed but a lot faster with large blacklists since addToSet is
always o(n) regardless of indexes) :
db.posts.update({_id:..., blacklists:{$ne:'peter'}}, {$push:{blacklists:{'peter'}},$inc:{blacklistsLength: 1}});
The latter being perfectly safe since the list and the length are adjusted atomically but obviously has slightly degraded performance. Since it also has the benefit of better overall performance due to the $push versus $addToSet performance issue on large arrays (and blacklists tend to become huge and remember that the $push version of the update uses an index on blacklist in the update criteria while $addToSet will NOT use an index during it's set scan) it is generally the best solution.
Would the following not work?
db.posts.update({_id:...}, {
$addtoSet:{blacklists:{$each:['peter', 'bob', 'steven']}},
$set: {blacklistsLength: ['peter', 'bob', 'steven'].length}
});
I had a similar problem, please see the discussion here: google groups mongo
As you can notice, following to this discussion, a bug was open:
Mongo Jira
As you upsert items into the database, simply query the item to see if it's in your embedded array. That way, you're avoiding pushing duplicate items, and only incrementing the counter as you add new items.
q = {'blacklists': {'$nin': ['blacklist_to_insert'] }}
u = {
'$push' : {'blacklists': { 'blacklist_to_insert' } },
'$inc' : {'total_blacklists': 1 }
}
o = { 'upsert' : true }
db.posts.update(q,u,o)