Mongodb how to reduce the array within the matching key and calculate avg - mongodb

{
"_id" : {
"state" : "NY",
"st" : "value"
},
"List" : [
{
"id" : "21",
"score" : 18.75,
"name" : "PU"
},
{
"id" : "21",
"score" : 25.0,
"name" : "PU"
},
{
"id" : "23",
"score" : 25.0,
"name" : "CL"
},
{
"id" : "23",
"score" : 56.25,
"name" : "CL"
}
]
}
Desired result:
Match the key with id within the array and calculate avg of score.
{
"_id" : {
"state" : "New York",
"st" : "value"
},
"List" : [
{
"id" : "21",
"score" : 21.875,
"name" : "PU"
},
{
"id" : "23",
"score" : 40.625,
"name" : "CL"
}
]
}
Thank you in advance.

Query
(returns the expected result)
unwind List
group with including the id, and find avg
fix the structure to be similar with the document you want
group back to restore the document structure (reverse the unwind)
if 2 sames ids have different name(if possible to happen)
query will make them seperated members in the array.
(alternativly it could make them same member and pack the names in an array, but that would produce different schema from the one you expect to see)
Test code here
db.collection.aggregate([
{
"$unwind": {
"path": "$List"
}
},
{
"$group": {
"_id": {
"state": "$_id.state",
"st": "$_id.st",
"id": "$List.id",
"name": "$List.name"
},
"avg": {
"$avg": "$List.score"
}
}
},
{
"$project": {
"_id": {
"state": "$_id.state",
"st": "$_id.st"
},
"List": {
"name": "$_id.name",
"id": "$_id.id",
"avg": "$avg"
}
}
},
{
"$group": {
"_id": "$_id",
"List": {
"$push": "$List"
}
}
}
])

Related

match element in the array with aggregation

i have mongo db collection the follwing structure
{
{
"_id" : ObjectId("63e37afe7a3453d5014c011b"),
"schemaVersion" : NumberInt(1),
"Id" : "ObjectId("63e37afe7a3453d5014c0112")",
"Id1" : "ObjectId("63e37afe7a3453d5014c0113")",
"Id2" : "ObjectId("63e37afe7a3453d5014c0114")",
"collectionName" : "Country",
"List" : [
{
"countryId" : NumberInt(1),
"name" : "Afghanistan",
},{
"countryId" : NumberInt(1),
"name" : "India",
},
{
"countryId" : NumberInt(1),
"name" : "USA",
}
}
i need to match the value with id, id1, id2, collectionName and name in the list to get country id for example if match the below value
"Id" : "ObjectId("63e37afe7a3453d5014c0112")",
"Id1" : "ObjectId("63e37afe7a3453d5014c0113")",
"Id2" : "ObjectId("63e37afe7a3453d5014c0114")",
"collectionName" : "Country",
"name" : "Afghanistan",
i need result
{
"countryId" : 1,
"name" : "Afghanistan",
}
i tried like below
db.country_admin.aggregate([
{ $match: { collectionName: "Country" } },
{ $unwind : '$countryList' },
{ $project : { _id : 0, 'countryList.name' : 1, 'countryList.countryId' : 1 } }
]).pretty()
and i have following output
[
{
"List" : {
"countryId" : 1.0,
"name" : "Afghanistan"
}
},
{
"List" : {
"countryId" : 2.0,
"name" : "india"
}
},
{
"List" : {
"countryId" : 3.0,
"name" : "USA"
}
}]```
You can try using $filter to avoid $unwind like this example:
First $match by your desired condition(s).
Then $filter and get the first element (as "List.name": "Afghanistan" is used into $match stage there will be at least one result).
And output only values you want using $project.
db.collection.aggregate([
{
"$match": {
"Id": ObjectId("63e37afe7a3453d5014c0112"),
"Id1": ObjectId("63e37afe7a3453d5014c0113"),
"Id2": ObjectId("63e37afe7a3453d5014c0114"),
"collectionName": "Country",
"List.name": "Afghanistan",
}
},
{
"$project": {
"country": {
"$arrayElemAt": [
{
"$filter": {
"input": "$List",
"cond": {
"$eq": [
"$$this.name",
"Afghanistan"
]
}
}
},
0
]
}
}
},
{
"$project": {
"_id": 0,
"countryId": "$country.countryId",
"name": "$country.name"
}
}
])
Example here
By the way, using $unwind is also possible and you can check this example

Mongodb aggregate with cond and query value

I'm new to mongodb. I need to know how it is possible to query item for set to the value with aggregate
Data
[
{
"_id" : "11111",
"parent_id" : "99",
"name" : "AAAA"
},
{
"_id" : "11112",
"parent_id" : "99",
"name" : "BBBB"
},
{
"_id" : "11113",
"parent_id" : "100",
"name" : "CCCC"
},
{
"_id" : "11114",
"parent_id" : "99",
"name" : "DDDD"
}
]
mongoshell
Assume $check is false
db.getCollection('test').aggregate(
[
{
"$group": {
"_id": "$id",
//...,
"item": {
"$last": {
"$cond": [
{"$eq": ["$check", true]},
"YES",
* * ANSWER **,
}
]
}
},
}
]
)
So i need the result for item is all the name contain with same parent_id as string of array
Expect result
[
{
"_id" : "11111",
"parent_id" : "99",
"name" : "AAAA",
"item" : ["AAAA","BBBB","DDDD"]
},
{
"_id" : "11112",
"parent_id" : "99",
"name" : "BBBB",
"item" : ["AAAA","BBBB","DDDD"]
},
{
"_id" : "11113",
"parent_id" : "100",
"name" : "CCCC",
"item" : ["CCCC"]
},
{
"_id" : "11114",
"parent_id" : "99",
"name" : "DDDD",
"item" : ["AAAA","BBBB","DDDD"]
}
]
Try this..
Sample live demo
db.collection.aggregate([
{
"$group": {
"_id": "$parent_id",
"item": {
"$push": "$name"
},
"data": {
"$push": {
"_id": "$_id",
"name": "$name"
}
}
}
},
{
"$unwind": "$data"
},
{
"$project": {
"_id": "$data._id",
"parent_id": "$_id",
"name": "$data.name",
"item": 1
}
}
])

Aggregate group multiple fields

Given the following dataset:
{ "_id" : 1, "city" : "Yuma", "cat": "roads", "Q1" : 0, "Q2" : 25, "Q3" : 0, "Q4" : 0 }
{ "_id" : 2, "city" : "Reno", "cat": "roads", "Q1" : 30, "Q2" : 0, "Q3" : 0, "Q4" : 60 }
{ "_id" : 3, "city" : "Yuma", "cat": "parks", "Q1" : 0, "Q2" : 0, "Q3" : 45, "Q4" : 0 }
{ "_id" : 4, "city" : "Reno", "cat": "parks", "Q1" : 35, "Q2" : 0, "Q3" : 0, "Q4" : 0 }
{ "_id" : 5, "city" : "Yuma", "cat": "roads", "Q1" : 0, "Q2" : 15, "Q3" : 0, "Q4" : 20 }
I'm trying to achieve the following result. It would be great to just return the totals greater than zero, and also compress each city, cat and Qx total to a single record.
{
"city" : "Yuma",
"cat" : "roads",
"Q2total" : 40
},
{
"city" : "Reno",
"cat" : "roads",
"Q1total" : 30
},
{
"city" : "Reno",
"cat" : "roads",
"Q4total" : 60
},
{
"city" : "Yuma",
"cat" : "parks",
"Q3total" : 45
},
{
"city" : "Reno",
"cat" : "parks",
"Q1total" : 35
},
{
"city" : "Yuma",
"cat" : "roads",
"Q4total" : 20
}
Possible?
We could ask, to what end? Your documents already have a nice consistent Object structure which is recommended. Having objects with varying keys is not a great idea. Data is "data" and should not really be the name of the keys.
With that in mind, the aggregation framework actually follows this sense and does not allow for the generation of arbitrary key names from data contained in the document. But you could get a similar result with the output as data points:
db.junk.aggregate([
// Aggregate first to reduce the pipeline documents somewhat
{ "$group": {
"_id": {
"city": "$city",
"cat": "$cat"
},
"Q1": { "$sum": "$Q1" },
"Q2": { "$sum": "$Q2" },
"Q3": { "$sum": "$Q3" },
"Q4": { "$sum": "$Q4" }
}},
// Convert the "quarter" elements to array entries with the same keys
{ "$project": {
"totals": {
"$map": {
"input": { "$literal": [ "Q1", "Q2", "Q3", "Q4" ] },
"as": "el",
"in": { "$cond": [
{ "$eq": [ "$$el", "Q1" ] },
{ "quarter": "$$el", "total": "$Q1" },
{ "$cond": [
{ "$eq": [ "$$el", "Q2" ] },
{ "quarter": "$$el", "total": "$Q2" },
{ "$cond": [
{ "$eq": [ "$$el", "Q3" ] },
{ "quarter": "$$el", "total": "$Q3" },
{ "quarter": "$$el", "total": "$Q4" }
]}
]}
]}
}
}
}},
// Unwind the array produced
{ "$unwind": "$totals" },
// Filter any "0" resutls
{ "$match": { "totals.total": { "$ne": 0 } } },
// Maybe project a prettier "flatter" output
{ "$project": {
"_id": 0,
"city": "$_id.city",
"cat": "$_id.cat",
"quarter": "$totals.quarter",
"total": "$totals.total"
}}
])
Which gives you results like this:
{ "city" : "Reno", "cat" : "parks", "quarter" : "Q1", "total" : 35 }
{ "city" : "Yuma", "cat" : "parks", "quarter" : "Q3", "total" : 45 }
{ "city" : "Reno", "cat" : "roads", "quarter" : "Q1", "total" : 30 }
{ "city" : "Reno", "cat" : "roads", "quarter" : "Q4", "total" : 60 }
{ "city" : "Yuma", "cat" : "roads", "quarter" : "Q2", "total" : 40 }
{ "city" : "Yuma", "cat" : "roads", "quarter" : "Q4", "total" : 20 }
You could alternately use mapReduce which allows "some" flexibility with key names. The catch is though that your aggregation is still by "quarter", so you need that as part of the primary key, which cannot be changed once emitted.
Additionally, you cannot "filter" any aggregated results of "0" without a second pass after outputting to a collection, so it's not really of much use for what you want to do, unless you can live with a second mapReduce operation of "transform" query on the output collection.
Worth note is if you look at what is being done in the "second" pipeline stage here with $project and $map you will see that the document structure is essentially being altered to sometime like what you could alternately structure your documents like originally, like this:
{
"city" : "Reno",
"cat" : "parks"
"totals" : [
{ "quarter" : "Q1", "total" : 35 },
{ "quarter" : "Q2", "total" : 0 },
{ "quarter" : "Q3", "total" : 0 },
{ "quarter" : "Q4", "total" : 0 }
]
},
{
"city" : "Yuma",
"cat" : "parks"
"totals" : [
{ "quarter" : "Q1", "total" : 0 },
{ "quarter" : "Q2", "total" : 0 },
{ "quarter" : "Q3", "total" : 45 },
{ "quarter" : "Q4", "total" : 0 }
]
}
Then the aggregation operation becomes simple for your documents to the same results as shown above:
db.collection.aggregate([
{ "$unwind": "$totals" },
{ "$group": {
"_id": {
"city": "$city",
"cat": "$cat",
"quarter": "$totals.quarter"
},
"ttotal": { "$sum": "$totals.total" }
}},
{ "$match": { "ttotal": { "$ne": 0 } },
{ "$project": {
"_id": 0,
"city": "$_id.city",
"cat": "$_id.cat",
"quarter": "$_id.quarter",
"total": "$ttotal"
}}
])
So it might make more sense to consider structuring your documents in that way to begin with and avoid any overhead required by the document transformation.
I think you'll find that consistent key names makes a far better object model to program to, where you should be reading the data point from the key-value and not the key-name. If you really need to, then it's a simple matter of reading the data from the object and transforming the keys of each already aggregated result in post processing.

How to query a mongo collection to return the full document with virtual fields containing calculated values from the sub-document?

I'm trying to query a collection for a specific document that contains a sub-document. The sub-document contains values for which I'd like to obtain
the highest and lowest scores from that sub-document and return that result as virtual fields to the original document.
I have the following dataset:
{
"_id" : "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e",
"name" : "Addison Hunt",
"tests" : [
{
"name" : "lorem",
"score" : 79
},
{
"name" : "vallum",
"score" : 100
},
{
"name" : "ipsum",
"score" : 65
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU"
}
In mongo 2.4, how can I query mongo once to return the following result:
{
"_id" : "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e",
"name" : "Addison Hunt",
"tests" : [
{
"name" : "lorem",
"score" : 79
},
{
"name" : "vallum",
"score" : 100
},
{
"name" : "ipsum",
"score" : 65
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
"worst_test": {
"name" : "ipsum",
"score" : 65
},
"best_test": {
"name" : "vallum",
"score" : 100
}
}
Where "best_test" and "worst_test" are virtual fields representing the tests with the highest and lowest scores, respectively.
I've tried with many different ways and the closest I've gotten is with this query:
db.students.aggregate([
{ $match: {
'_id': 'd0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e'
}},
{ $unwind: '$tests' },
{ $sort: {'tests.score': 1} },
{ $group: {
_id: '$_id',
student_tests: {$push: "$$ROOT"},
worst_test: {$first: '$tests'},
best_test: { $last: '$tests' }
}}
]);
Which yields this result:
{
"_id" : "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e",
"student_tests" : [
{
"name" : "Addison Hunt",
"tests" : [
{
"name" : "ipsum",
"score" : 65
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
},
{
"name" : "Addison Hunt",
"tests" : [
{
"name" : "lorem",
"score" : 79
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
},
{
"name" : "Addison Hunt",
"tests" : [
{
"name" : "vallum",
"score" : 100
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
},
],
"worst_test": {
"name" : "ipsum",
"score" : 65
},
"best_test": {
"name" : "vallum",
"score" : 100
}
}
If you are using $$ROOT then in fact you are using MongoDB 2.6 as this is an aggregation variable only introduced in that version.
But while handy for various things, all it does is represent the entire document at the present stage of the pipeline where used. To do what you want and return the original document unmodified but with additional fields, you could use it in $project stage before the $unwind to assign to the _id field, but really you don't have exactly the same document as you would still need to $project at the end in order to get the correct document shape out of those elements.
You best bet is just projecting the fields, but keeping an un-altered copy of the array before any $sort is applied:
db.students.aggregate([
{ "$match": {
"_id": "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e"
}},
{ "$project": {
"name": 1,
"tests": 1,
"created_at": 1,
"class": 1,
"user_id": 1,
"testCopy": "$tests"
}},
{ "$unwind": "$testCopy" },
{ "$sort": { "testCopy.score": 1 } },
{ "$group": {
"_id: "$_id",
"tests": { "$first": "$tests" },
"created_at": { "$first": "$created_at" },
"class": { "$first": "$class" },
"user_id": { "$first": "$user_id" },
"worst_test": { "$first": "$testCopy" },
"best_test": { "$last": "$testCopy" }
}}
]);
Or using $$ROOT as mentioned before, alternately just placing the fields under the _id individually in the $project:
db.students.aggregate([
{ "$match": {
"_id": "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e"
}},
{ "$project": {
"_id": "$$ROOT",
"tests": 1
}},
{ "$unwind": "$tests" },
{ "$sort": { "tests.score": 1 } },
{ "$group": {
"_id": "$_id",
"aworst_test": { "$first": "$tests" },
"abest_test": { "$last": "$tests" }
}},
{ "$project": {
"_id": "$_id._id",
"tests": "$_id.tests",
"created_at": "$_id.created_at",
"class": "$_id.class",
"user_id": "$_id.user_id",
"worst_test": "$aworst_test",
"best_test": "$abest_test"
}}
]);
But as you see, you are still doing the $project work somewhere in order to get the structure you want, as well as the "renamed fields" to maintain the field order you want as the $project will otherwise "optimize" and "keep" any fields that have not been renamed and "append" new fields after the existing ones.
There really is no simple way to "get all fields" in the same way as you originally found them. Operations like $project and $group are an "all or nothing" affair, where they only explicitly produce what you tell them to.

mongodb multiple aggregations in single operation

I have an item collection with following documents.
{ "item" : "i1", "category" : "c1", "brand" : "b1" }
{ "item" : "i2", "category" : "c2", "brand" : "b1" }
{ "item" : "i3", "category" : "c1", "brand" : "b2" }
{ "item" : "i4", "category" : "c2", "brand" : "b1" }
{ "item" : "i5", "category" : "c1", "brand" : "b2" }
I want to separate aggregation results --> count by category, count by brand. Please note, it is not count by (category,brand)
I am able to do this using map-reduce using following code.
map = function(){
emit({type:"category",category:this.category},1);
emit({type:"brand",brand:this.brand},1);
}
reduce = function(key, values){
return Array.sum(values)
}
db.item.mapReduce(map,reduce,{out:{inline:1}})
And the result is
{
"results" : [
{
"_id" : {
"type" : "brand",
"brand" : "b1"
},
"value" : 3
},
{
"_id" : {
"type" : "brand",
"brand" : "b2"
},
"value" : 2
},
{
"_id" : {
"type" : "category",
"category" : "c1"
},
"value" : 3
},
{
"_id" : {
"type" : "category",
"category" : "c2"
},
"value" : 2
}
],
"timeMillis" : 21,
"counts" : {
"input" : 5,
"emit" : 10,
"reduce" : 4,
"output" : 4
},
"ok" : 1,
}
I can get same results by firing two different aggregation commands as below.
db.item.aggregate({$group:{_id:"$category",count:{$sum:1}}})
db.item.aggregate({$group:{_id:"$brand",count:{$sum:1}}})
Is there anyway I can do the same using aggregation framework by single aggregation command.
I have simplified my case here, but in actual I need this grouping from fields in array of subdocuments. Assume the above is structure after I do unwind.
It is a real-time query (someone waiting for response), though on smaller dataset, so execution time is important.
I am using MongoDB 2.4.
Starting in Mongo 3.4, the $facet aggregation stage greatly simplifies this type of use case by processing multiple aggregation pipelines within a single stage on the same set of input documents:
// { "item" : "i1", "category" : "c1", "brand" : "b1" }
// { "item" : "i2", "category" : "c2", "brand" : "b1" }
// { "item" : "i3", "category" : "c1", "brand" : "b2" }
// { "item" : "i4", "category" : "c2", "brand" : "b1" }
// { "item" : "i5", "category" : "c1", "brand" : "b2" }
db.collection.aggregate(
{ $facet: {
categories: [{ $group: { _id: "$category", count: { "$sum": 1 } } }],
brands: [{ $group: { _id: "$brand", count: { "$sum": 1 } } }]
}}
)
// {
// "categories" : [
// { "_id" : "c1", "count" : 3 },
// { "_id" : "c2", "count" : 2 }
// ],
// "brands" : [
// { "_id" : "b1", "count" : 3 },
// { "_id" : "b2", "count" : 2 }
// ]
// }
Over a large data set I would say that your current mapReduce approach would be the best one, because the aggregation technique for this would not work well with large data. But possibly over a reasonably small size it might just be what you need:
db.items.aggregate([
{ "$group": {
"_id": null,
"categories": { "$push": "$category" },
"brands": { "$push": "$brand" }
}},
{ "$project": {
"_id": {
"categories": "$categories",
"brands": "$brands"
},
"categories": 1
}},
{ "$unwind": "$categories" },
{ "$group": {
"_id": {
"brands": "$_id.brands",
"category": "$categories"
},
"count": { "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.brands",
"categories": { "$push": {
"category": "$_id.category",
"count": "$count"
}},
}},
{ "$project": {
"_id": "$categories",
"brands": "$_id"
}},
{ "$unwind": "$brands" },
{ "$group": {
"_id": {
"categories": "$_id",
"brand": "$brands"
},
"count": { "$sum": 1 }
}},
{ "$group": {
"_id": null,
"categories": { "$first": "$_id.categories" },
"brands": { "$push": {
"brand": "$_id.brand",
"count": "$count"
}}
}}
])
Not really the same as the mapReduce output, you could throw in some more stages to change the output format, but this should be usable:
{
"_id" : null,
"categories" : [
{
"category" : "c2",
"count" : 2
},
{
"category" : "c1",
"count" : 3
}
],
"brands" : [
{
"brand" : "b2",
"count" : 2
},
{
"brand" : "b1",
"count" : 3
}
]
}
As you can see, this involves a fair bit of shuffling between arrays in order to group each set of either "category" or "brand" within the same pipeline process. Again I will say, this will not do well for large data, but for something like "items in an order" it would probably do nicely.
Of course as you say, you have simplified somewhat, so the first grouping key on null is either going to be something else or either narrowed down to do that null case by an earlier $match stage, which is probably what you want to do.