i have following collection...
[{
"_id":"abcd1234",
"Date" :{"$date":1430418601000},
"Count":{
"hr1":0,
"hr2":20,
"hr3":1,
"hr4":4,
"hr5":0,
}
},
{
"_id":"abcd1234",
"Date" :{"$date":1430505001000},
"Count":{
"hr1":2,
"hr2":15,
"hr3":15,
"hr4":0,
"hr5":1,
}
}
]
I want to aggregate count for these dates so total count will be like...
{"hr1":2,
"hr2":35,
"hr3":16,//......so on
}
Is it possible in mongodb? I don't have much knowledge of mongo, if someone can guide me with this query will be helpful...
Running the following aggregation pipeline should give you the desired result:
db.collection.aggregate([
{
"$group": {
"_id": null,
"hr1": { "$sum": "$Count.hr1" },
"hr2": { "$sum": "$Count.hr2" },
"hr3": { "$sum": "$Count.hr3" },
"hr4": { "$sum": "$Count.hr4" },
"hr5": { "$sum": "$Count.hr5" }
}
}
])
Specifying an _id value of null will calculate accumulated values for all the input documents as a whole.
Related
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 $addToSet) which keep the order, the elements where pushed in? Or am I the problem?
Greetings Max
this is my document in the sensor collection
sensorId:1
sensorType:1
BuildingId:1
CalcItems:
0: Object
Ts:2019-08-30T18:30:00.000+00:00
Value:1
1:Object
Ts:2019-08-31T19:00:00.000+00:00
Value:2
i need to get sum of all value attributes with respect to same Ts date value
output like this
sensorType:1
BuildingId:1
CalcItems:
0: Object
Ts:2019-08-31T18:30:00.000+00:00
Value:23
1:Object
Ts:2019-08-31T19:00:00.000+00:00
Value:43
give me any suggestions
Try this..
Sample live demo
[
{
"$unwind": "$CalcItems"
},
{
$group: {
_id: {
"sensorType": "$sensorType",
"BuildingId": "$BuildingId",
"Ts": "$CalcItems.Ts"
},
"total": {
"$sum": "$CalcItems.Value"
}
}
},
{
"$project": {
"_id": 0,
"sensorType": "$_id.sensorType",
"BuildingIdVal": "$_id.BuildingId",
"CalcItems.Ts": "$_id.Ts",
"CalcItems.Value": "$total"
}
}
]
Reference
Mongodb $sum
Mongodb $group
Mongodb $project
When i do a find all in a mongodb collection, i want to get maintenanceList sorted by older maintenanceDate to newest.
The sort of maintenanceDate should not affect parents order in a find all query
{
"_id":"507f191e810c19729de860ea",
"color":"black",
"brand":"brandy",
"prixVenteUnitaire":200.5,
"maintenanceList":[
{
"cost":100.40,
"maintenanceDate":"2017-02-07T00:00:00.000+0000"
},
{
"cost":4000.40,
"maintenanceDate":"2019-08-07T00:00:00.000+0000"
},
{
"cost":300.80,
"maintenanceDate":"2018-08-07T00:00:00.000+0000"
}
]
}
Any guess how to do that ?
Thank you
Whatever order the fields are in with the previous pipeline stage, as operations like $project and $group effectively "copy" same position.So, it will not change the order of your fields in your aggregated result.
And the sort of maintenanceDate through aggregation will not affect parents order in a find all query.
So, simply doing this should work.
Assuming my collection name is example.
db.example.aggregate([
{
"$unwind": "$maintenanceList"
},
{
"$sort": {
"_id": 1,
"maintenanceList.maintenanceDate": 1
}
},
{
"$group": {
"_id": "$_id",
"color": {
$first: "$color"
},
"brand": {
$first: "$brand"
},
"prixVenteUnitaire": {
$first: "$prixVenteUnitaire"
},
"maintenanceList": {
"$push": "$maintenanceList"
}
}
}
])
Output:
Hi I want to change my sql query to mongo aggregation.
select c.year, c.minor_category, count(c.minor_category) from Crime as c
group by c.year, c.minor_category having c.minor_category = (
Select cc.minor_category from Crime as cc where cc.year=c.year group by
cc.minor_category order by count(*) desc, cc.minor_category limit 1)
I tried do something like this:
db.crimes.aggregate({
$group: {
"_id": {
year: "$year",
minor_category :"$minor_category",
count: {$sum: "$minor_category"}
}
},
},
{
$match : {
minor_category: ?
}
})
But i stuck in $match which is equivalent to having, but i dont know how to make subqueries in mongo like in my sql query.
Can anybody can help me ?
Ok based on the confirmation above , the below query should work.
db.crime.aggregate
([
{"$group":{"_id":{"year":"$year","minor":"$minor"},"count":{"$sum":1}}},
{"$project":{"year":"$_id.year","count":"$count","minor":"$_id.minor","document":"$$ROOT"}},
{"$sort":{"year":1,"count":-1}},
{"$group":{"_id":{"year":"$year"},"orig":{"$first":"$document"}}},
{"$project":{"_id":0,"year":"$orig._id.year","minor":"$orig._id.minor","count":"$orig.count"}}
)]
This translates into the following MongoDB query:
db.crime.aggregate({
$group: { // group by year and minor_catetory
_id: {
"year": "$year",
"minor_category": "$minor_category"
},
"count": { $sum: 1 }, // count all documents per group,
}
}, {
$sort: {
"count": -1, // sort descending by count
"minor_category": 1 // and ascending by minor_category
}
}, {
$group: { // now we get the highst element per year
_id: "$_id.year", // so group by year
"minor_category": { $first: "$_id.minor_category" }, // and get the first (we've sorted the data) value
"count": { $first: "$count" } // same here
}
}, {
$project: { // remove the _id field and add the others in the right order (if needed)
"_id": 0,
"year": "$_id",
"minor_category": "$minor_category",
"count": "$count"
}
})
Let's say that I want to aggregate and group by documents in MongoDb by the Description field.
Running the following (case-sensitive by default):
db['Products'].aggregate(
{ $group: {
_id: { 'Description': "$Description" },
count: { $sum: 1 },
docs: { $push: "$_id" }
}},
{ $match: {
count: { $gt : 1 }
}}
);
on my sample data gives me 1000 results, which is fine.
But now I expect that running a case-insensitive query (using $toLower) should give me less than or equal to 1000 results:
db['Products'].aggregate(
{ $group: {
_id: { 'Description': {$toLower: "$Description"} },
count: { $sum: 1 },
docs: { $push: "$_id" }
}},
{ $match: {
count: { $gt : 1 }
}}
);
But instead I get more than 1000 results. That can't be right, can it? More common entries should get grouped together to yield less number of total groupings ... I think.
So then probably my aggregation query is wrong! Which brings me to my question:
How should case-insensitive aggregation grouping in MongoDb be performed?
You approach to case-insensitive grouping is correct so perhaps your observation is not? ;)
Try this example:
// insert two documents
db.getCollection('test').insertOne({"name" : "Test"}) // uppercase 'T'
db.getCollection('test').insertOne({"name" : "test"}) // lowercase 't'
// perform the grouping
db.getCollection('test').aggregate({ $group: { "_id": { $toLower: "$name" }, "count": { $sum: 1 } } }) // case insensitive
db.getCollection('test').aggregate({ $group: { "_id": "$name", "count": { $sum: 1 } } }) // case sensitive
You may have a typo somewhere?
The documentation also states that
$toLower only has a well-defined behavior for strings of ASCII characters.
Perhaps that's what's biting you here?