I'm trying to figure out if what I want to do is even possible in Mongodb. I'm open to all sorts of suggestions regarding more appropriate ways to achieve what I need.
Currently, I have 2 collections:
vehicles (Contains vehicle data such as make and model. This data can be highly unstructured, which is why I turned to Mongodb for this)
views (Simply contains an IP, a date/time that the vehicle was viewed and the vehicle_id. There could be thousands of views)
I need to return a list of vehicles that have views between 2 dates. The list should include the number of views. I need to be able to sort by the number of views in addition to any of the usual vehicle fields. So, to be clear, if a vehicle has had 1000 views, but only 500 of those between the given dates, the count should return 500.
I'm pretty sure I could perform this query without any issues in MySQL - however, trying to store the vehicle data in MySQL has been a real headache in the past and it has been great moving to Mongo where I can add new data fields with ease and not worry about the structure of my database.
What do you all think?? TIA!
As it turns out, it's totally possible. It took me a long while to get my head around this, so I'm posting it up for future google searches...
db.statistics.aggregate({
$match: {
branch_id: { $in: [14] }
}
}, {
$lookup: {
from: 'vehicles', localField: 'vehicle_id', foreignField: '_id', as: 'vehicle'
}
}, {
$group: {
_id: "$vehicle_id",
count: { $sum: 1 },
vehicleObject: { $first: "$vehicle" }
}
}, { $unwind: "$vehicleObject" }, {
$project: {
daysInStock: { $subtract: [ new Date(), "$vehicleObject.date_assigned" ] },
vehicleObject: 1,
count: 1
}
}, { $sort: { count: -1 } }, { $limit: 10 });
To explain the above:
The Mongodb aggregate framework is the way forward for complex queries like this. Firstly, I run a $match to filter the records. Then, we use $lookup to grab the vehicle record. Worth mentioning here that this is a Many to One relationship here (lots of stats, each having a single vehicle). I can then group on the vehicle_id field, which will enable me to return one record per vehicle with a count of the number of stats in the group. As it is a group, we technically have lots of copies of that same vehicle document now in each group, so I then add just the first one into the vehicleObject variable. This would be fine, but $first tends to return an array with a single entry (pointless in my opinion), so I added the $unwind stage to pull the actual vehicle out. I then added a $project stage to calculate an additional field, sorted by the count descending and limited the results to 10.
And take a breath :)
I hope that helps someone. If you know of a better way to do what I did, then I'm open to suggestions to improve.
Related
I want the total number of cases in all my documents,
This is the query I tried to use:
db.coviddatajson.aggregate([
{ $group: { _id: null, total: { $sum: "$total_cases"} } }
])
For some reason the result is 0 which does not make sense, as it's supposed to be 1000+ at least and the expected result anything that is not zero will make sense but it's supposed to be a few thousands or something like that.
This is the dataset I am using:
https://covid.ourworldindata.org/data/owid-covid-data.json
What am I doing wrong here?
Any ideas on how to fix this query?
The total_cases field is inside data array, and $sum requires field type as number in $group stage, so before we need to do total($sum) of data.total_cases in current document and then pass it to $group stage and count total sum,
db.coviddatajson.aggregate([
{
$project: { total_cases: { $sum: "$data.total_cases" } }
},
{
$group: {
_id: null,
total: { $sum: "$total_cases" }
}
}
])
Playground
The data set has some issues.
The document size is bigger than 16MiB, you cannot load documents >16MiB into MongoDB. This in an internal limitation. You would need to split the document into sub-documents.
The document contains data for each country but also summarized data for "World". Do you have to exclude the "World" data? Can you use it, instead of manual summary?
The data is not consistent. For example some countries do not provide a number of male/female smokers or median age. Not all countries provide all data for each date, you may have missing values. How to deal with them?
Do you like a simple sum of all total_cases? If yes, the query would be easy, however the result would be pointless (15'773'189'214 total cases, twice population of the world).
Kind of a complex one here and i'm pretty new to Mongo, so hopefully someone can help. I have a db of users. Each user has a state/province listed. I'm trying to create another collection of the total users in each state/province. Because users sign up pretty regularly, this will be an growing total i'm trying to generate and display on a map.
I'm able to query the database to find total number of users in a specific state, but i want to do this for all users and come out with a list of totals in all states/provinces and have a separate collection in the DB with all states/provinces listed and the field TOTAL to be dynamically populated with the count query of the other collection. But i'm not sure how to have a query be the result of a field in another collection.
used this to get users totals:
db.users.aggregate([
{"$group" : {_id:"$state", count:{$sum:1}}}
])
My main question is how to make the results of a query the value of a field in each corresponding record in another collection. Or if that's even possible.
Thanks for any help or guidance.
Looks like that On-Demand Materialized Views (just added on version 4.2 of MongoDB) should solve your problem!
You can create an On-Demand Materialized View using the $merge operator.
A possible definition of the Materialized View could be:
updateUsersLocationTotal = function() {
db.users.aggregate( [
{ $match: { <if you need to perform a match, like $state, otherwise remove it> } },
{ $group: { _id:"$state", users_quantity: { $sum: 1} } },
{ $merge: { into: "users_total", whenMatched: "replace" } }
] );
};
And then you perform updates just by calling updateUsersLocationTotal()
After that you can query the view just like a normal collection, using db.users_total.find() or db.users_total.aggregate().
I have to create a collection of document and I have a doubt about the right design.
Each document is an "identity"; each Identity has a list of "partner Data"; each partner data are defined by an ID and a set of Data.
One approach can be (1):
{
_id: ...
partners: [
{
id: partner1,
data: {
}
},
{
id: partner2,
data: {
}
},
]
}
Another approach can be (2)
{
_id: ...
partners: {
partner1: {
data: {
}
},
partner2: {
data: {
}
},
]
}
I prefer the first one, but considering that I could have million of these identities, which could be the most performed schema?
A typical query can be: "how many identities have partner with ID N".
With the second example, a query can be:
db.identities.find({partner.partnerName: {$exists:true}})
With first approach, how can I get this count?
The second solution is more easy to handle Server Side; each document will have a list where each KEY is the partner ID, so instead of scan all document, I can simply get partner data by key...
What do you think about these solutions? I prefer the first one but the second I think that is more "usable"...
Thanks
I prefer the first one, but considering that I could have million of these identities, wich could be the most performed schema?
If you going to have millions of identities, then both approaches
are not really scalable .
Each document in mongo has a size limit (16MB) (read about it here)
In case you are going to have really lot's of identities ,
the scalable approach would be to create a different collection,
only for the relations and partnership data.
Now , I also want you to consider how you treat "partnership",
if I'm a user and I got you on my partners list , will you see me as a partner on your list ?
In case we both see each other as partners , then mongo-db may not be the best solution. graph db's are more appropriate for dealing with relations of this type.
All solutions within mongo for two-ways relations will be built on double updates (Your id on my partner's list , My id on your partner's list).
(In SQL you could add an extra condition for joining but not in mongo) ,
so you don't need to save twice partnerships . (me and you , you and me)
just you and me.
Do you see where is going this way ?
If you need to go only in one way,
Then just create a second collection , "partnerships" ,
{
_id: should be uniqe,
user_id: 'your_id',
partner_id: 'his_id'
data: {} or just flatten the fields into the root object.
}
Please notice that you create a row for each partnership !
Then you could use $lookup in order to query for a user with all of his
partners .
something like:
db.getCollection('partners').aggregate([
{
$lookup: {
from: 'parterships',
localField: '_id',
foreignField: 'user_id',
as: 'partners'
}
},
{
$project: {
name: 1,
partners: 1,
num_partners: { $size: "$partners" }
}
}
])
Read more about the aggregation stages here.
In case you are not going to have lot's of partnership's Then please continue with
your first approach which is good .
The second approach will make most queries to this collection pretty weird and you will always have to write code in order to query this table .
It won't be "straight forward" mongo queries .
Currently I use the following find query to get the latest document of a certain ID
Conditions.find({
caveId: caveId
},
{
sort: {diveDate:-1},
limit: 1,
fields: {caveId: 1, "visibility.visibility":1, diveDate: 1}
});
How can I use the same using multiple ids with $in for example
I tried it with the following query. The problem is that it will limit the documents to 1 for all the found caveIds. But it should set the limit for each different caveId.
Conditions.find({
caveId: {$in: caveIds}
},
{
sort: {diveDate:-1},
limit: 1,
fields: {caveId: 1, "visibility.visibility":1, diveDate: 1}
});
One solution I came up with is using the aggregate functionality.
var conditionIds = Conditions.aggregate(
[
{"$match": { caveId: {"$in": caveIds}}},
{
$group:
{
_id: "$caveId",
conditionId: {$last: "$_id"},
diveDate: { $last: "$diveDate" }
}
}
]
).map(function(child) { return child.conditionId});
var conditions = Conditions.find({
_id: {$in: conditionIds}
},
{
fields: {caveId: 1, "visibility.visibility":1, diveDate: 1}
});
You don't want to use $in here as noted. You could solve this problem by looping through the caveIds and running the query on each caveId individually.
you're basically looking at a join query here: you need all caveIds and then lookup last for each.
This is a problem of database schema/denormalization in my opinion: (but this is only an opinion!):
You could as mentioned here, lookup all caveIds and then run the single query for each, every single time you need to look up last dives.
However I think you are much better off recording/updating the last dive inside your cave document, and then lookup all caveIds of interest pulling only the lastDive field.
That will give you immediately what you need, rather than going through expensive search/sort queries. This is at the expense of maintaining that field in the document, but it sounds like it should be fairly trivial as you only need to update the one field when a new event occurs.
I am parsing Wikipedia dumps in order to play with the link-oriented metadata. One of the collections is named articles and it is in the following form:
{
_id : "Tree",
id: "18955875",
linksFrom: " [
{
name: "Forest",
count: 6
},
[...]
],
categories: [
"Trees",
"Forest_ecology"
[...]
]
}
The linksFrom field stores all articles this article points to, and how many times that happens. Next, I want to create another field linksTo with all the articles that point to this article. In the beginning, I went through the whole collection and updated every article, but since there's lots of them it takes too much time. I switched to aggregation for performance purposes and tried it on a smaller set - works like a charm and is super fast in comparison with the older method. The aggregation pipeline is as follows:
db.runCommand(
{
aggregate: "articles",
pipeline : [
{
$unwind: "$linksFrom"
},
{
$sort: { "linksFrom.count": -1 }
},
{
$project:
{
name: "$_id",
linksFrom: "$linksFrom"
}
},
{
$group:
{
_id: "$linksFrom.name",
linksTo: { $push: { name: "$name", count: { $sum : "$linksFrom.count" } } },
}
},
{
$out: "TEMPORARY"
}
] ,
allowDiskUse: true
}
)
However, on a large dataset being the english Wikipedia I get the following error after a few minutes:
{
"ok" : 0,
"errmsg" : "insert for $out failed: { connectionId: 24, err: \"BSONObj size: 24535193 (0x1766099) is invalid. Size must be between 0 and 16793600(16MB) First element: _id: \"United_States\"\", code: 10334, n: 0, ok: 1.0 }",
"code" : 16996
}
I understand that there are too many articles, which link to United_States article and the corresponding document's size grows above 16MB, currently almost 24MB. Unfortunately, I cannot even check if that's the case (error messages sometimes tend to lie)... Because of that, I'm trying to change the model so that the relationship between articles is stored with IDs rather than long names but I'm afraid that might not be enough - especially because my plan is to merge the two collections for every article later...
The question is: does anyone have a better idea? I don't want to try to increase the limit, I'm rather thinking about a different approach of storing this data in the database.
UPDATE after comment by Markus
Markus is correct, I am using a SAX parser and, as a matter of fact, I'm already storing all the links in a similar way. Apart from articles I have three more collections - one with links and two others, labels and stemmed-labels. The first one stores all links that occur in the dump in the following way:
{
_id : "tree",
stemmedName: "tree",
targetArticle: "Christmas_tree"
}
_id stores the text that is used to represent a given link, stemmedName represents stemmed _id and targetArticle marks what article this text pointed to. I'm in the middle of adding sourceArticle to this one, because it's obviously a good idea.
The second collection labels contains documents as follows:
{
_id : "tree",
targetArticles: [
{
name: "Christmas_tree",
count: 1
},
{
name: "Tree",
count: 166
}
[...]
]
}
The third stemmed-labels is analogous to the labels with its _id being a stemmed version of the root label.
So far, the first collection links serves as a baseline for the two other collections. I group the labels together by their name so that I only do one lookup for every phrase and then I can immiedately get all target articles with one query. Then I use the articles and labels collections in order to:
Look for label with a given name.
Get all articles it might
point to.
Compare the incoming and outcoming links for these
articles.
This is where the main question comes. I thought that it's better if I store all possible articles for a given phrase in one document rather than leave them scattered in the links collection. Only now did it occur to me, that - as long as the lookups are indexed - the overall performance might be the same for one big document or many smaller ones! Is this a correct assumption?
I think your data model is wrong. It may well be (albeit a bit theoretical) that individual articles (let's stick with the wikipedia example) are linked more often than you could store in a document. Embedding only works with One-To(-Very)-Few™ relationships.
So basically, I think you should change your model. I will show you how I would do it.
I will use the mongoshell and JavaScript in this example, since it is the lingua franca. You might need to translate accordingly.
The questions
Lets begin with the questions you want to have answered:
For a given article, which other articles link to that article?
For a given article, to which other articles does that article link to?
For a given article, how many articles link to it?
Optional: For a given article, to how many articles does it link to?
The crawling
What I would do basically is to implement a SAX parser on the articles, creating a new document for each article link you encounter. The document itself should be rather simple:
{
"_id": new ObjectId(),
// optional, for recrawling or pointing out a given state
"date": new ISODate(),
"article": wikiUrl,
"linksTo": otherWikiUrl
}
Note that you should not do an insert, but an upsert. The reason for this is that we do not want to document the number of links, but the articles linked to. If we did an insert, the same combination of article and linksTocould occur multiple times.
So our statement when encountering a link would look like this for example:
db.links.update(
{ "article":"HMS_Warrior_(1860)", "linksTo":"Royal_Navy" },
{ "date": new ISODate(), "article":"HMS_Warrior_(1860)", "linksTo":"Royal_Navy" },
{ upsert:true }
)
Answering the questions
As you might already guess, answering the questions becomes pretty straightforward now. I have use the following statements for creating a few documents:
db.links.update(
{ "article":"HMS_Warrior_(1860)", "linksTo":"Royal_Navy" },
{ "date": new ISODate(), "article":"HMS_Warrior_(1860)", "linksTo":"Royal_Navy" },
{ upsert:true }
)
db.links.update(
{ "article":"Royal_Navy", "linksTo":"Mutiny_on_the_Bounty" },
{ "date":new ISODate(), "article":"Royal_Navy", "linksTo":"Mutiny_on_the_Bounty" },
{ upsert:true }
)
db.links.update(
{ "article":"Mutiny_on_the_Bounty", "linksTo":"Royal_Navy"},
{ "date":new ISODate(), "article":"Mutiny_on_the_Bounty", "linksTo":"Royal_Navy" },
{ upsert:true }
)
For a given article, which other articles link to that article?
We found out that we should not use an aggregation, since that might exceed the size limit. But we don't have to. We simply use a cursor and gather the results:
var toLinks =[]
var cursor = db.links.find({"linksTo":"Royal_Navy"},{"_id":0,"article":1})
cursor.forEach(
function(doc){
toLinks.push(doc.article);
}
)
printjson(toLinks)
// Output: [ "HMS_Warrior_(1860)", "Mutiny_on_the_Bounty" ]
For a given article, to which other articles does that article link to?
This works pretty much like the first question – we basically only change the query:
var fromLinks = []
var cursor = db.links.find({"article":"Royal_Navy"},{"_id":0,"linksTo":1})
cursor.forEach(
function(doc){
fromLinks.push(doc.linksTo)
}
)
printjson(fromLinks)
// Output: [ "Mutiny_on_the_Bounty" ]
For a given article, how many articles link to it?
It should be obvious that in case you already have answered question 1, you could simply check toLinks.length. But let's assume you haven't. There are two other ways of doing this
Using .count()
You can use this method on replica sets. On sharded clusters, this doesn't work well. But it is easy:
db.links.find({ "linksTo":"Royal_Navy" }).count()
// Output: 2
Using an aggregation
This works on any environment and isn't much more complicated:
db.links.aggregate([
{ "$match":{ "linksTo":"Royal_Navy" }},
{ "$group":{ "_id":"$linksTo", "isLinkedFrom":{ "$sum":1 }}}
])
// Output: { "_id" : "Royal_Navy", "isLinkedFrom" : 2 }
Optional: For a given article, to how many articles does it link to?
Again, you can answer this question by reading the length of the array from question 2 of use the .count()method. The aggregation again is simple
db.links.aggregate([
{ "$match":{ "article":"Royal_Navy" }},
{ "$group":{ "_id":"$article", "linksTo":{ "$sum":1 }}}
])
// Output: { "_id" : "Royal_Navy", "linksTo" : 1 }
Indices
As for the indices, I haven't really checked them, but individual indices on the fields is probably what you want:
db.links.createIndex({"article":1})
db.links.createIndex({"linksTo":1})
A compound index will not help much, since order matters and we do no always ask for the first field. So this is probably as optimized as it can get.
Conclusion
We are using an extremely simple, scalable model and rather simple queries and aggregations to get the questions answered you have to the data.