I have a mongo collection where each document has an array with multiple hashtags (a simple string). I would like to count how many times each hashtag has appeared and return something like this:
{hashtag: "hashtag1",
count: numOcurrences
}
{hashtag: "hashtag2",
count: numOcurrences
}
...
It seems similar to this problem, but since I don't want to filter by any parameter, just count the overall occurrences I think It has to be a cleaner way to solve this, sadly my mongo knowledge is very limited...
The collection which the hashtags are in, looks similar to this, being the field "hastag" the array of hashtags:
{"_id": ...,
"hashtag" : [
"hashtag1",
"hashtag2"
],
"likes" : ...
},
{"_id": ...,
...
}
Your case is a bit easier than the other problem you mentioned, and you could solve it using the aggregation below:
db.hashtags.aggregate([
{
"$unwind": "$hashtag"
},
{
"$group": {
"_id": "$hashtag",
"count": {
"$sum": 1
}
}
},
// you can skip this projection if it's okay for you to have the result like [{ _id: "hashtag1", count: 2 }]
{
"$project": {
"_id": 0,
"hashtag": "$_id",
"count": 1
}
}
])
You can see a working example in mongoplayground
Related
I have a large collection of documents with datetime fields in them, and I need to retrieve the most recent document for any given queried list.
Sample data:
[
{"_id": "42.abc",
"ts_utc": "2019-05-27T23:43:16.963Z"},
{"_id": "42.def",
"ts_utc": "2019-05-27T23:43:17.055Z"},
{"_id": "69.abc",
"ts_utc": "2019-05-27T23:43:17.147Z"},
{"_id": "69.def",
"ts_utc": "2019-05-27T23:44:02.427Z"}
]
Essentially, I need to get the most recent record for the "42" group as well as the most recent record for the "69" group. Using the sample data above, the desired result for the "42" group would be document "42.def".
My current solution is to query each group one at a time (looping with PyMongo), sort by the ts_utc field, and limit it to one, but this is really slow.
// Requires official MongoShell 3.6+
db = db.getSiblingDB("someDB");
db.getCollection("collectionName").find(
{
"_id" : /^42\..*/
}
).sort(
{
"ts_utc" : -1.0
}
).limit(1);
Is there a faster way to get the results I'm after?
Assuming all your documents have the format displayed above, you can split the id into two parts (using the dot character) and use aggregation to find the max element per each first array (numeric) element.
That way you can do it in a one shot, instead of iterating per each group.
db.foo.aggregate([
{ $project: { id_parts : { $split: ["$_id", "."] }, ts_utc : 1 }},
{ $group: {"_id" : { $arrayElemAt: [ "$id_parts", 0 ] }, max : {$max: "$ts_utc"}}}
])
As #danh mentioned in the comment, the best way you can do is probably adding an auxiliary field to indicate the grouping. You may further index the auxiliary field to boost the performance.
Here is an ad-hoc way to derive the field and get the latest result per grouping:
db.collection.aggregate([
{
"$addFields": {
"group": {
"$arrayElemAt": [
{
"$split": [
"$_id",
"."
]
},
0
]
}
}
},
{
$sort: {
ts_utc: -1
}
},
{
"$group": {
"_id": "$group",
"doc": {
"$first": "$$ROOT"
}
}
},
{
"$replaceRoot": {
"newRoot": "$doc"
}
}
])
Here is the Mongo playground for your reference.
I tried to make a $group aggregation with MongoDB, like the following example:
"$group": {
"_id": "$test_id",
"feeling": {
"$push": "$feeling"
},
"reference_id": {
"$push": "$_id"
},
"training_start": {
"$push": "$training_start"
},
"training_duration": {
"$push": "$duration_ms"
}
}
The aggregation works fine, but the created arrays are sorted different. That means, if I check the result of the aggregation by looking at reference_id[x] and training_start[x] then the value of training_start in the source collection is not equal to training_start[x].
Maybe an example shows my problem more precisely:
One document after the $group aggregation:
{
_id: "string_1",
reference_id: [1, 2, 3],
training_start: [01:00:00, 02:00:00, 03:00:00] (date times)
}
Documents from source collection:
{
_id:1,
training_start: 01:00:00,
test_id: "string_1"
},
{
_id:2,
training_start: 03:00:00,
test_id: "string_1"
},
{
_id:3,
training_start: 02:00:00,
test_id: "string_1"
}
The first elements in these arrays are always in the right order. So I checked if each grouped field has the same number of entries by using the code below. And the annoying result is, that the amount of entries in each array is equal. So there is no shift in the arrays caused by missing values.
"$group": {
"_id": "$test_id",
"sum": {
"$sum": {
"$cond": {
"if": {
"$lte": [
"$training_start", null
]
},
"then": 0,
"else": 1
}
}
}
Does anybody know, if there is an other way to create arrays (already tried $addToSet) which keep the order, the elements where pushed in? Or am I the problem?
Greetings Max
Ok, I have an issue I cannot seem to solve.
I have a document like this:
{
"playerId": "43345jhiuy3498jh4358yu345j",
"leaderboardId": "5b165ca15399c020e3f17a75",
"data": {
"type": "EclecticData",
"holeScores": [
{
"type": "RoundHoleData",
"xtraStrokes": 0,
"strokes": 3,
},
{
"type": "RoundHoleData",
"xtraStrokes": 1,
"strokes": 5,
},
{
"type": "RoundHoleData",
"xtraStrokes": 0,
"strokes": 4
}
]
}
}
Now, what I am trying to accomplish is using aggregate sum the strokes and then order it afterwards. I am trying this:
var sortedBoard = db.collection.aggregate(
{$match: {"leaderboardId": boardId}},
{$group: {
_id: "$playerId",
played: { $sum: 1 },
strokes: {$sum: '$data.holeScores.strokes'}
}
},
{$project:{
type: "$SortBoard",
avgPoints: '$played',
sumPoints: "$strokes",
played : '$played'
}}
);
The issue here is that I do net get the strokes sum correct, since this is inside another array.
Hope someone can help me with this and thanks in advance :-)
You need to say $sum twice:
var sortedBoard = db.collection.aggregate([
{ "$match": { "leaderboardId": boardId}},
{ "$group": {
"_id": "$playerId",
"SortBoard": { "$first": "$SortBoard" },
"played": { "$sum": 1 },
"strokes": { "$sum": { "$sum": "$data.holeScores.strokes"} }
}},
{ "$project": {
"type": "$SortBoard",
"avgPoints": "$playeyed",
"sumPoints": "$strokes",
"played": "$played"
}}
])
The reason is because you are using it both as a way to "sum array values" and also as an "accumulator" for $group.
The other thing you appear to be missing is that $group only outputs the fields you tell it to, therefore if you want to access other fields in other stages or output, you need to keep them with something like $first or another accumulator. We also appear to be missing a pipeline stage in the question anyway, but it's worth noting just to be sure.
Also note you really should wrap aggregation pipelines as an official array [], because the legacy usage is deprecated and can cause problems in some language implementations.
Returns the correct details of course:
{
"_id" : "43345jhiuy3498jh4358yu345j",
"avgPoints" : 1,
"sumPoints" : 12,
"played" : 1
}
Say I have an aggregation that returns the following:
[
{driverId: 21312asd12, cars: 2, totalMiles: 30000, family: 4},
{driverId: 55512a23a2, cars: 3, totalMiles: 55000, family: 2},
...
]
How would I go about running a summation of each data set on a groupId basis to return the following? Do I use an $unwind? Do another grouping?
For example I would like to return:
{
totalDrivers: 2,
totalCars: 5,
totalMiles: 85000,
totalFamily: 6
}
You seem to just be referring to the documents in the output as an "array", therefore just add another $group to the end of your pipeline:
{ "$group": {
"_id": null,
"totalDrivers": { "$sum": 1 },
"totalCars": { "$sum": "$cars" },
"totalMiles": { "$sum": "$totalMiles" },
"totalFamily": { "$sum": "$family" }
}}
Where null is essentially just a blank grouping key that is not a field present in the document to group on. The result should be a single document (albeit in an array, depending on the API method call used or server version).
Or if you actually mean that each document has a field with an array like this, then $unwind and process the group either per document or with a null as above:
{ "$unwind": "$someArray" },
{ "$group": {
"_id": "$_id",
"totalDrivers": { "$sum": 1 },
"totalCars": { "$sum": "$someArray.cars" },
"totalMiles": { "$sum": "$someArray.totalMiles" },
"totalFamily": { "$sum": "$someArray.family" }
}}
At any rate, you should really post the code you are using when asking questions like this. It is very likely that your pipeline may not be as efficient to get to your end goal as you think, and if you posted that it both gives a clear picture of what you are doing as well as leaves it open for suggested improvement.
and thank you in advance for your help.
I have a mongoDB database structured like this:
{
'_id' : objectID(...),
'userID' : id,
'movies' : [{
'movieID' : movieID,
'rating' : rating
}]
}
My question is:
I want to search for a specific user that has 'userID' : 3, for example, get all is movies, then i want to get all the other users that have at least, 15 or more movies with the same 'movieID', then with that group i wanna select only the users that have those 15 movies in similarity and have one extra 'movieID' that i choose.
I already tried aggregation, but failed, and if i do single queries like getting all the users movies from a user, the cycling every user movie and comparing it takes a bunch of time.
Any ideias?
Thank you
There are a couple of ways to do this using the aggregation framework
Just a simple set of data for example:
{
"_id" : ObjectId("538181738d6bd23253654690"),
"movies": [
{ "_id": 1, "rating": 5 },
{ "_id": 2, "rating": 6 },
{ "_id": 3, "rating": 7 }
]
},
{
"_id" : ObjectId("538181738d6bd23253654691"),
"movies": [
{ "_id": 1, "rating": 5 },
{ "_id": 4, "rating": 6 },
{ "_id": 2, "rating": 7 }
]
},
{
"_id" : ObjectId("538181738d6bd23253654692"),
"movies": [
{ "_id": 2, "rating": 5 },
{ "_id": 5, "rating": 6 },
{ "_id": 6, "rating": 7 }
]
}
Using the first "user" as an example, now you want to find if any of the other two users have at least two of the same movies.
For MongoDB 2.6 and upwards you can simply use the $setIntersection operator along with the $size operator:
db.users.aggregate([
// Match the possible documents to reduce the working set
{ "$match": {
"_id": { "$ne": ObjectId("538181738d6bd23253654690") },
"movies._id": { "$in": [ 1, 2, 3 ] },
"$and": [
{ "movies": { "$not": { "$size": 1 } } }
]
}},
// Project a copy of the document if you want to keep more than `_id`
{ "$project": {
"_id": {
"_id": "$_id",
"movies": "$movies"
},
"movies": 1,
}},
// Unwind the array
{ "$unwind": "$movies" },
// Build the array back with just `_id` values
{ "$group": {
"_id": "$_id",
"movies": { "$push": "$movies._id" }
}},
// Find the "set intersection" of the two arrays
{ "$project": {
"movies": {
"$size": {
"$setIntersection": [
[ 1, 2, 3 ],
"$movies"
]
}
}
}},
// Filter the results to those that actually match
{ "$match": { "movies": { "$gte": 2 } } }
])
This is still possible in earlier versions of MongoDB that do not have those operators, just using a few more steps:
db.users.aggregate([
// Match the possible documents to reduce the working set
{ "$match": {
"_id": { "$ne": ObjectId("538181738d6bd23253654690") },
"movies._id": { "$in": [ 1, 2, 3 ] },
"$and": [
{ "movies": { "$not": { "$size": 1 } } }
]
}},
// Project a copy of the document along with the "set" to match
{ "$project": {
"_id": {
"_id": "$_id",
"movies": "$movies"
},
"movies": 1,
"set": { "$cond": [ 1, [ 1, 2, 3 ], 0 ] }
}},
// Unwind both those arrays
{ "$unwind": "$movies" },
{ "$unwind": "$set" },
// Group back the count where both `_id` values are equal
{ "$group": {
"_id": "$_id",
"movies": {
"$sum": {
"$cond":[
{ "$eq": [ "$movies._id", "$set" ] },
1,
0
]
}
}
}},
// Filter the results to those that actually match
{ "$match": { "movies": { "$gte": 2 } } }
])
In Detail
That may be a bit to take in, so we can take a look at each stage and break those down to see what they are doing.
$match : You do not want to operate on every document in the collection so this is an opportunity to remove the items that are not possibly matches even if there still is more work to do to find the exact ones. So the obvious things are to exclude the same "user" and then only match the documents that have at least one of the same movies as was found for that "user".
The next thing that makes sense is to consider that when you want to match n entries then only documents that have a "movies" array that is larger than n-1 can possibly actually contain matches. The use of $and here looks funny and is not required specifically, but if the required matches were 4 then that actual part of the statement would look like this:
"$and": [
{ "movies": { "$not": { "$size": 1 } } },
{ "movies": { "$not": { "$size": 2 } } },
{ "movies": { "$not": { "$size": 3 } } }
]
So you basically "rule out" arrays that are not possibly long enough to have n matches. Noting here that this $size operator in the query form is different to $size for the aggregation framework. There is no way for example to use this with an inequality operator such as $gt is it's purpose is to specifically match the requested "size". Hence this query form to specify all of the possible sizes that are less than.
$project : There are a few purposes in this statement, of which some differ depending on the MongoDB version you have. Firstly, and optionally, a document copy is being kept under the _id value so that these fields are not modified by the rest of the steps. The other part here is keeping the "movies" array at the top of the document as a copy for the next stage.
What is also happening in the version presented for pre 2.6 versions is there is an additional array representing the _id values for the "movies" to match. The usage of the $cond operator here is just a way of creating a "literal" representation of the array. Funny enough, MongoDB 2.6 introduces an operator known as $literal to do exactly this without the funny way we are using $cond right here.
$unwind : To do anything further the movies array needs to be unwound as in either case it is the only way to isolate the existing _id values for the entries that need to be matched against the "set". So for the pre 2.6 version you need to "unwind" both of the arrays that are present.
$group : For MongoDB 2.6 and greater you are just grouping back to an array that only contains the _id values of the movies with the "ratings" removed.
Pre 2.6 since all values are presented "side by side" ( and with lots of duplication ) you are doing a comparison of the two values to see if they are the same. Where that is true, this tells the $cond operator statement to return a value of 1 or 0 where the condition is false. This is directly passed back through $sum to total up the number of matching elements in the array to the required "set".
$project: Where this is the different part for MongoDB 2.6 and greater is that since you have pushed back an array of the "movies" _id values you are then using $setIntersection to directly compare those arrays. As the result of this is an array containing the elements that are the same, this is then wrapped in a $size operator in order to determine how many elements were returned in that matching set.
$match: Is the final stage that has been implemented here which does the clear step of matching only those documents whose count of intersecting elements was greater than or equal to the required number.
Final
That is basically how you do it. Prior to 2.6 is a bit clunkier and will require a bit more memory due to the expansion that is done by duplicating each array member that is found by all of the possible values of the set, but it still is a valid way to do this.
All you need to do is apply this with the greater n matching values to meet your conditions, and of course make sure your original user match has the required n possibilities. Otherwise just generate this on n-1 from the length of the "user's" array of "movies".