I'm trying to get a list of current holders at specific times from a collection. My collection looks like this:
[
{
"time": 1,
"holdings": [
{ "owner": "A", "tokens": 2 },
{ "owner": "B", "tokens": 1 }
]
},
{
"time": 2,
"holdings": [
{ "owner": "B", "tokens": 2 }
]
},
{
"time": 3,
"holdings": [
{ "owner": "A", "tokens": 3 },
{ "owner": "B", "tokens": 1 },
{ "owner": "C", "tokens": 1 }
]
},
{
"time": 4,
"holdings": [
{ "owner": "C", "tokens": 0 }
]
}
]
tokens show the current holdings of an owner if the holdings have changed to the last document. I would like to change the collection so that holdings always includes the full current holdings for any point in time.
At time: 1, the holdings are: A: 2, B: 1.
At time: 2, the holdings are: A: 2, B: 2. The collections does not include A's holdings however, because they haven't changed. So what I'd like to get is:
[
{
"time": 1,
"holdings": [
{ "owner": "A", "tokens": 2 },
{ "owner": "B", "tokens": 1 }
]
},
{
"time": 2,
"holdings": [
{ "owner": "A", "tokens": 2 }, // merged from prev doc.
{ "owner": "B", "tokens": 2 }
]
},
{
"time": 3,
"holdings": [
{ "owner": "A", "tokens": 3 },
{ "owner": "B", "tokens": 1 },
{ "owner": "C", "tokens": 1 }
]
},
{
"time": 4,
"holdings": [
{ "owner": "A", "tokens": 3 }, // merged from prev
{ "owner": "B", "tokens": 1 }, // merged from prev
{ "owner": "C", "tokens": 0 }
]
}
]
From what I understand $mergeObjects does that, but I don't understand how I can merge all previous docs in order up to the current doc for each doc. So I'm looking for a way to combine setWindowFields with mergeObjects I think.
This is a nice challenge.
So far, I got this complicated solution:
Get all of our timestamps in all of our documents. This is the purpose of the first 4 steps. $setWindowFields is used to accumulate this data.
$group by owner and calculate the empty timestamps as wantedTimes- next 5 steps.
$set empty timestamps with tokens: null to be filled with actual data and $unwind to separate - next 3 steps
Use $setWindowFields to find the last known token for each owner at each timestamp.
Fill this last known state for documents with unknown token - 2 steps
$group and format answer:
db.collection.aggregate([
{
$setWindowFields: {
sortBy: {time: 1},
output: {
allTimes: {$addToSet: "$time", window: {documents: ["unbounded", "current"]}
}
}
}
},
{
$setWindowFields: {
sortBy: {time: -1},
output: {
allTimes: {$addToSet: "$allTimes", window: {documents: ["unbounded", "current"]}
}
}
}
},
{
$set: {
allTimes: {
$reduce: {
input: "$allTimes",
initialValue: [],
in: {"$concatArrays": ["$$value", "$$this"]}
}
}
}
},
{$set: {allTimes: {$setIntersection: "$allTimes"}}},
{$unwind: "$holdings"},
{$sort: {time: 1}},
{$group: { _id: "$holdings.owner",
tokens: {$push: {tokens: "$holdings.tokens", time: "$time"}},
times: {$push: "$time"}, firstTime: {$first: "$time"},
allTimes: {$first: "$allTimes"}}
},
{
$addFields: {
wantedTimes: {
$filter: {
input: "$allTimes",
as: "item",
cond: {$gte: ["$$item", "$firstTime"]}
}
}
}
},
{
$project: {
tokens: 1,
wantedTimes: {$setDifference: ["$wantedTimes", "$times"]}
}
},
{
$set: {
data: {
$map: {
input: "$wantedTimes",
as: "item",
in: {time: "$$item", tokens: null}
}
}
}
},
{$project: {tokens: {"$concatArrays": ["$tokens", "$data"]}}},
{$unwind: "$tokens"},
{
$setWindowFields: {
partitionBy: "$_id",
sortBy: {"tokens.time": 1},
output: {
lastTokens: {
$push: "$tokens.tokens",
window: {documents: ["unbounded", "current"]}
}
}
}
},
{
$set: {
lastTokens: {
$filter: {
input: "$lastTokens",
as: "item",
cond: {$ne: ["$$item", null]}
}
}
}
},
{
$set: {
"tokens.tokens": {$ifNull: ["$tokens.tokens", {$last: "$lastTokens"}]}
}
},
{
$group: {
_id: "$tokens.time",
holdings: {$push: {owner: "$_id", tokens: "$tokens.tokens" }}
}
},
{$project: {time: "$_id", holdings: 1, _id: 0}},
{$sort: {time: 1}}
])
Playground example
From a performance perspective I recommend you split it into 2 calls, the first will be a quick findOne just to get the maximum time value in the collection.
Once you have that value the pipeline can be much leaner:
const maxItem = await db.collection.findOne({}).sort({ time: -1 });
db.collection.aggregate([
{
$unwind: "$holdings"
},
{
$group: {
_id: "$holdings.owner",
times: {
$push: {
time: "$time",
tokens: "$holdings.tokens"
}
},
minTime: {
$min: "$time"
}
}
},
{
$addFields: {
times: {
$reduce: {
input: {
$range: [
"$minTime",
maxItem.time + 1 // this is max time
]
},
initialValue: {
values: [],
lastIndex: 0
},
in: {
values: {
"$concatArrays": [
"$$value.values",
[
{
$cond: [
{
$in: [
"$$this",
"$times.time"
]
},
{
"$arrayElemAt": [
"$times",
"$$value.lastIndex"
]
},
{
"$mergeObjects": [
{
tokens: 0
},
{
"$arrayElemAt": [
"$times",
{
$subtract: [
"$$value.lastIndex",
1
]
}
]
},
{
time: "$$this"
}
]
}
]
}
]
]
},
lastIndex: {
$cond: [
{
$in: [
"$$this",
"$times.time"
]
},
{
$sum: [
"$$value.lastIndex",
1
]
},
"$$value.lastIndex"
]
}
}
}
}
}
},
{
$unwind: "$times.values"
},
{
$group: {
_id: "$times.values.time",
holdings: {
$push: {
owner: "$_id",
tokens: "$times.values.tokens"
}
}
}
},
{
$project: {
_id: 0,
time: "$_id",
holdings: 1
}
},
{
$sort: {
time: 1
}
}
])
This is still quite a heavy query as it requires to $unwind and $group the entire collection, however there is no workaround this due to the requirements. if the collection is too big for this approach I recommend iteration owner by owner, or time by time and doing separate updates accordingly.
Mongo Playground
If you don't care about performance at all and want it in a single query you can still use the same pipeline, you will have to first extract the max time in the collection, this will require you to add an initial $group stage, like so:
db.collection.aggregate([
{
$group: {
_id: null,
maxTime: {
$max: "$time"
},
roots: {
$push: "$$ROOT"
}
}
},
{
$unwind: "$roots"
},
{
$replaceRoot: {
newRoot: {
"$mergeObjects": [
"$roots",
{
maxTime: "$maxTime"
}
]
}
}
},
... same pipeline ...
])
Related
I have 4 products. I want to know the count of product-4 for users who has product-1 or product-2
Sample data:
[
{
"user_id": 1,
"product_type": "product-1"
},
{
"user_id": 1,
"product_type": "product-4"
},
{
"user_id": 1,
"product_type": "product-4"
},
{
"user_id": 2,
"product_type": "product-1"
}
]
user-1 has two product-4 and one product-1 (that counts 2)
user-2 has only product-1, but no product-4 (hence that does not count)
This is how I tried
db.collection.aggregate([
{
$match: {
product_type: {
$in: [
"product-1",
"product-2",
],
},
},
},
{
$group: {
_id: "$user_id",
},
},
{
$match: {
user_id: { $in: "$_id"}, // I want to use $group's result in here
product_type: "product-4",
},
}
]);
Expected results are:
[
{
"_id": 1,
"count": 2
},
{
"_id": 2,
"count": 0
}
]
Note:
I dont have a backend, I have to this using mongodb only.
Does this answer your question?
db.collection.aggregate([
{$group: {_id: "$user_id", data: {$push: "$product_type"}}},
{$match: {$expr: {$or: [
{$in: ["product-1", "$data"]},
{$in: ["product-2", "$data"]}
]}}},
{$project: {
count: {
$size: {
$filter: {
input: "$data",
cond: {$eq: ["$$this", "product-4"]}
}
}
}
}}
])
See how it works on the playground example
I want to split the following array according to the group-value. I know I can do this using $unwind and $group. Is there any way to this in a single $project-stage?
Input
{
"_id": 1,
"some_field": "some_value",
"array": [
{
"group": "a",
"subgroup": "aa",
"value": 1
},
{
"group": "b",
"subgroup": "bb",
"value": 2
},
{
"group": "a",
"subgroup": "ab",
"value": 2
}
]
}
desired output:
{
"_id": 1,
"some_field": "some_value",
"array": [
{
"group": "a",
"values": [
{
"subgroup": "aa",
"value": 1
},
{
"subgroup": "ab",
"value": 2
}
]
},
{
"group": "b",
"values": [
{
"subgroup": "bb",
"value": 2
}
]
}
]
}
Try this: https://mongoplayground.net/p/pFn3tLtAG4D
$set: {
_id: "$_id",
some_field: "$some_field",
array: {
$map: {
input: {
$setUnion: [
"$array.group"
]
},
in: {
group: "$$this",
values: {
$map: {
input: {
$filter: {
input: "$array",
as: "elem",
cond: {
$eq: [
"$$elem.group",
"$$this"
]
}
}
},
as: "vals",
in: {
subgroup: "$$vals.subgroup",
value: "$$vals.value"
}
}
}
}
}
}
}
This is far from a single project stage, but it does produce the desired output from the given input.
db.collection.aggregate([
{'$match': {'_id': 1}},
{'$unwind': '$array'},
{'$project': {'array': {'group': '$array.group', 'values': '$array'},
'some_field': 1,
'my_id': '$_id'}},
{'$unset': 'array.values.group'},
{'$group': {'_id': '$array.group',
'values': {'$push': '$array.values'},
'some_field': {'$first': '$some_field'},
'my_id': {'$first': '$my_id'}}},
{'$set': {'array': {'group': '$_id', 'values': '$values'}}},
{'$unset': 'values'},
{'$group': {'_id': '$my_id',
'array': {'$push': '$array'},
'some_field': {'$first': '$some_field'}}}
])
Try it on mongoplayground.net.
It is doable, it's definitely not clean or sexy.
My approach is to use $reduce and $mergeObjects, we'll iterate over the array and keep reconstructing the result.
The main issue that plagues this approach is this feature that doesn't allow to $concatArrays expressions, so we have to use some very ugly workarounds.
Anyways here is how you can achieve this:
db.collection.aggregate([
{
$project: {
_id: 1,
some_field: 1,
array: {
$map: {
input: {
"$objectToArray": {
$reduce: {
input: "$array",
initialValue: {},
in: {
"$mergeObjects": [
"$$value",
{
"$arrayToObject": [
[
{
k: "$$this.group",
v: {
$map: {
input: {
"$concatArrays": [
[
"$$this"
],
{
$map: {
input: {
$filter: {
input: {
"$objectToArray": "$$value"
},
as: "filterItem",
cond: {
$eq: [
"$$filterItem.k",
"$$this.group"
]
}
}
},
as: "mapItem",
in: "$$mapItem.v"
}
},
]
},
as: "map2Item",
in: {
$cond: [
{
"$isArray": "$$map2Item"
},
{
$arrayElemAt: [
"$$map2Item",
0
]
},
"$$map2Item"
]
}
}
}
}
]
]
}
]
}
}
}
},
as: "item",
in: {
group: "$$item.k",
values: "$$item.v"
}
}
}
}
}
])
Mongo Playground
I have a similar collection where I have sort them by their startTime:
{"name": 'A', "startTime": '1634626355', "endTime": '1634631405'}
{"name": 'A', "startTime": '1634631406', "endTime": '1634631864'}
{"name": 'A', "startTime": '1634631865', "endTime": '1634656048'}
{"name": 'A', "startTime": '1634712642', "endTime": '1634718856'}
How can I compare the documents such that if the document endTime and the next document startTime duration is less than 5 minutes, merge it.
This is the result I'm trying to achieve (The 1st 3 documents are merged into 1 where it uses the startTime of the 1st document and the endTime of the 3rd document):
{"name": 'A', "startTime": '1634626355', "endTime": '1634656048'}
{"name": 'A', "startTime": '1634712642', "endTime": '1634718856'}
Thanks
First of all, you should never store date/time values as string, it's a design flaw. Store always proper Date object.
This solution works without self-lookup, so it may perform better:
db.collection.aggregate([
{
$set: {
startDateTime: { $toDate: { $multiply: ["$startTime", 1000] } },
endDateTime: { $toDate: { $multiply: ["$endTime", 1000] } }
},
},
{ $sort: { startDateTime: 1 } },
{ $group: { _id: null, data: { $push: "$$ROOT" } } },
{
$set: {
data: {
$reduce: {
input: "$data",
initialValue: [],
in: {
$cond: {
if: {
$or: [
{ $eq: [{ $size: "$$value" }, 0] }, // for the initail element
{
$gt: [
{
$dateDiff: { // calculate difference
endDate: "$$this.startDateTime",
startDate: { $last: "$$value.endDateTime" },
unit: "minute"
}
},
5 // more than 5 Minutes
]
}
]
},
then: { $concatArrays: ["$$value", ["$$this"]] }, // append new element
else: {
$map: {
input: "$$value",
as: "data",
in: {
$cond: {
if: { $eq: ["$$data._id", { $last: "$$value._id" }] }, // find last element
then: { // update last element
$mergeObjects: [
"$$data",
{ endDateTime: "$$this.endDateTime" },
{ endTime: "$$this.endTime" }
]
},
else: "$$data"
}
}
}
}
}
}
}
}
}
},
// some cosmetic
{ $unwind: "$data" },
{ $replaceRoot: { newRoot: "$data" } }
])
Mongo Playground
You can use $lookup in an aggregation pipeline to find out the documents that you need to remove. Then, perform a forEach to remove them.
db.collection.aggregate([
{
$addFields: {
endDateTime: {
"$toDate": {
"$multiply": [
{
$toLong: "$endTime"
},
1000
]
}
}
},
},
{
"$lookup": {
"from": "collection",
let: {
end: "$endDateTime"
},
pipeline: [
{
"$addFields": {
startDateTime: {
"$toDate": {
"$multiply": [
{
$toLong: "$startTime"
},
1000
]
}
}
}
},
{
$match: {
$expr: {
$and: [
{
$lte: [
{
$subtract: [
"$startDateTime",
"$$end"
]
},
300000
]
},
{
$lte: [
"$$end",
"$startDateTime"
]
}
]
}
}
}
],
"as": "lessThan5min"
}
},
{
"$unwind": "$lessThan5min"
},
{
"$replaceRoot": {
"newRoot": "$lessThan5min"
}
}
]).forEach(function(doc){
db.collection.remove({ "_id": doc._id });
});
Here is the Mongo playground to find out the documents that you need to remove for your reference.
I'm trying to merge two arrays in my aggregation pipeline. After performing $facet, my MongoDB document has this format:
{
"final": [
{
"key": "TP-1",
"status_map": [
{ "status": "Closed", "final": [ "a", "b"]},
{ "status": "Done", "final": ["c","d" ] }
]
},
{
"key": "TP-2",
"status_map": [
{ "status": "Closed", "final": [ "x","y"] }
]
}
],
"start": [
{
"key": "TP-1",
"status_map": [
{ "status": "Closed", "start": [ "h"]},
{ "status": "Done", "start": ["a"]}
]
},
{
"key": "TP-2",
"status_map": [{ "status": "Done", "start": ["l","m"]}
]
}
]
}
Expected Output:
I need to merge final and start array corresponding to two groups:
Based on key and then
Based on status
{
"data": [
{
"key": "TP-1",
"status_map": [
{ "status": "Closed","final": ["a","b"],"start":["h"]},
{ "status": "Done","final": ["c","d"],"start":["a"]}
]
},
{
"key": "TP-2",
"status_map": [
{ "status": "Closed", "final":[ "x","y"],"start": []},
{ "status": "Done", "final": [ ],"start": [ "l","m"]}
]
}
]
}
How to achieve this use case?
There are several ways to approach this, not necessarily with $mergeObjects. But since you mentioned $mergeObjects this is one that uses it:
Note that, with this approach, we are merging objects of the same key and status, the values in the arrays will not get concatenated if the same key exists for multiple documents, The arrays will get replaced instead.
db.collection.aggregate([
{
$project: {
all: { $concatArrays: ["$final","$start"] }
}
},
{
$unwind: "$all"
},
{
$unwind: "$all.status_map"
},
{
$group: {
_id: {
_id: "$_id", // keep _id in $group to apply the group for each document, otherwise if you want to apply group on all documents, omit this
key: "$all.key",
status: "$all.status_map.status"
},
status_map: { $mergeObjects: "$$ROOT.all.status_map" }
}
},
{ // some data don't have start or end at all, we have to set a default empty array
$addFields: { // you can skip this stage if you allow data without start and final keys
"status_map.start": { $ifNull: ["$status_map.start", []] },
"status_map.final": { $ifNull: ["$status_map.final", []] }
}
},
{
$group: {
_id: { _id: "$_id._id", key: "$_id.key" },
key: { $first: "$_id.key" },
status_map: { $push: "$status_map" }
}
}
])
Mongo Playground
With no assumptions (for example for both keys to always appear) my strategy was to concat both arrays, unwind and finally group by the key.
db.collection.aggregate([
{
$project: {
concat: {
$concatArrays: [
"$final",
"$start"
]
}
}
},
{
$unwind: "$concat"
},
{
$unwind: "$concat.status_map"
},
{
$group: {
_id: {
k: "$concat.key",
status: "$concat.status_map.status"
},
final: {
$push: "$concat.status_map.final"
},
start: {
$push: "$concat.status_map.start"
}
}
},
{
$group: {
_id: "$_id.k",
status_map: {
$push: {
status: "$_id.status",
final: "$final",
start: "$start"
}
}
}
},
{
$project: {
key: "$_id",
status_map: 1,
_id: 0
}
}
])
Mongo Playground
Adding to #Tom Slabbaert's answer,
Mongo Playground
Here, final and start array is of format array of array. But It has to be simply an array.
It can be achieved by using $unwind on status_map and $reduce on status_map.final and status_map.start arrays.
Final query:
db.collection.aggregate([
{
$project: {
concat: {
$concatArrays: [
"$final",
"$start"
]
}
}
},
{
$unwind: "$concat"
},
{
$unwind: "$concat.status_map"
},
{
$group: {
_id: {
k: "$concat.key",
status: "$concat.status_map.status"
},
final: {
$push: "$concat.status_map.final"
},
start: {
$push: "$concat.status_map.start"
}
}
},
{
$group: {
_id: "$_id.k",
status_map: {
$push: {
status: "$_id.status",
final: "$final",
start: "$start"
}
}
}
},
{
$project: {
key: "$_id",
status_map: 1,
_id: 0
}
},
{
$unwind: "$status_map"
},
{
$project: {
key: 1,
"status_map.status": 1,
final: {
$reduce: {
input: "$status_map.final",
initialValue: [],
in: {
$concatArrays: [
"$$value",
"$$this"
]
}
}
},
start: {
$reduce: {
input: "$status_map.start",
initialValue: [],
in: {
$concatArrays: [
"$$value",
"$$this"
]
}
}
}
}
},
{
$group: {
_id: "$key",
status_map: {
$push: {
status: "$status_map.status",
final: "$final",
start: "$start"
}
}
}
}
])
Mongo Playground
How do I get counts data grouped by every hour in 24 hours even if data is not present i.e. IF 0 will select 0
MonogDB 3.6
Input
[
{
"_id": ObjectId("5ccbb96706d1d47a4b2ced4b"),
"date": "2019-05-03T10:39:53.108Z",
"id": 166,
"update_at": "2019-05-03T02:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2ced4c"),
"date": "2019-05-03T10:39:53.133Z",
"id": 166,
"update_at": "2019-05-03T02:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2ced4d"),
"date": "2019-05-03T10:39:53.180Z",
"id": 166,
"update_at": "2019-05-03T20:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2ced7a"),
"date": "2019-05-10T10:39:53.218Z",
"id": 166,
"update_at": "2019-12-04T10:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2ced7b"),
"date": "2019-05-03T10:39:53.108Z",
"id": 166,
"update_at": "2019-05-05T10:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2cedae"),
"date": "2019-05-03T10:39:53.133Z",
"id": 166,
"update_at": "2019-05-05T10:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2cedad"),
"date": "2019-05-03T10:39:53.180Z",
"id": 166,
"update_at": "2019-05-06T10:45:36.208Z",
"type": "image"
},
{
"_id": ObjectId("5ccbb96706d1d47a4b2cedab"),
"date": "2019-05-10T10:39:53.218Z",
"id": 166,
"update_at": "2019-12-06T10:45:36.208Z",
"type": "image"
}
]
Implementation
db.collection.aggregate({
$match: {
update_at: {
"$gte": "2019-05-03T00:00:00.0Z",
"$lt": "2019-05-05T00:00:00.0Z"
},
id: {
"$in": [
166
]
}
}
},
{
$group: {
_id: {
$substr: [
"$update_at",
11,
2
]
},
count: {
"$sum": 1
}
},
},
{
$project: {
_id: 0,
hour: "$_id",
count: "$count"
}
},
{
$sort: {
hour: 1
}
})
Actual Output:
{
"count": 2,
"hour": "02"
},
{
"count": 1,
"hour": "20"
}
My expectation code show 24 hours event data is 0 or null and convert from example "02" as "02 AM" , "13" as "01 PM":
Expected Output
{
"count": 0,
"hour": "01" // 01 AM
},
{
"count": 2,
"hour": "02"
},
{
"count": 0,
"hour": "03"
},
{
"count": 0,
"hour": "04"
},
{
"count": 0,
"hour": "05"
},
{
"count": 1,
"hour": "20" // to 08 pm
}
Try this solution:
Explanation
We group by hour to count how many images are uploaded.
Then, we add extra field hour to create time interval (if you had v4.x, there is a better solution).
We flattern hour field (will create new documents) and split first 2 digits to match count and split last 2 digits to put AM / PM periods.
db.collection.aggregate([
{
$match: {
update_at: {
"$gte": "2019-05-03T00:00:00.0Z",
"$lt": "2019-05-05T00:00:00.0Z"
},
id: {
"$in": [
166
]
}
}
},
{
$group: {
_id: {
$substr: [
"$update_at",
11,
2
]
},
count: {
"$sum": 1
}
}
},
{
$addFields: {
hour: [
"0000",
"0101",
"0202",
"0303",
"0404",
"0505",
"0606",
"0707",
"0808",
"0909",
"1010",
"1111",
"1212",
"1301",
"1402",
"1503",
"1604",
"1705",
"1806",
"1907",
"2008",
"2109",
"2210",
"2311"
]
}
},
{
$unwind: "$hour"
},
{
$project: {
_id: 0,
hour: 1,
count: {
$cond: [
{
$eq: [
{
$substr: [
"$hour",
0,
2
]
},
"$_id"
]
},
"$count",
0
]
}
}
},
{
$group: {
_id: "$hour",
count: {
"$sum": "$count"
}
}
},
{
$sort: {
_id: 1
}
},
{
$project: {
_id: 0,
hour: {
$concat: [
{
$substr: [
"$_id",
2,
2
]
},
{
$cond: [
{
$gt: [
{
$substr: [
"$_id",
0,
2
]
},
"12"
]
},
" PM",
" AM"
]
}
]
},
count: "$count"
}
}
])
MongoPlayground
There's no "magic" solution, you'll have to hardcode it into your aggregation:
Heres an example using Mongo v3.2+ syntax with some $map and $filter magic:
db.collection.aggregate([
{
$match: {
update_at: {
"$gte": "2019-05-03T00:00:00.0Z",
"$lt": "2019-05-05T00:00:00.0Z"
},
id: {"$in": [166]}
}
},
{
$group: {
_id: {$substr: ["$update_at", 11, 2]},
count: {"$sum": 1}
}
},
{
$group: {
_id: null,
hours: {$push: {hour: "$_id", count: "$count"}}
}
},
{
$addFields: {
hours: {
$map: {
input: {
$concatArrays: [
"$hours",
{
$map: {
input: {
$filter: {
input: ["00", "01", "02", "03", "04", "05", "06", "07", "08", "09", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23"],
as: "missingHour",
cond: {
$not: {
$in: [
"$$missingHour",
{
$map: {
input: "$hours",
as: "hourObj",
in: "$$hourObj.hour"
}
}
]
}
}
}
},
as: "missingHour",
in: {hour: "$$missingHour", count: 0}
}
}
]
},
as: "hourObject",
in: {
count: "$$hourObject.count",
hour: {
$cond: [
{$eq: [{$substr: ["$$hourObject.hour", 0, 1]}, "0"]},
{$concat: ["$$hourObject.hour", " AM"]},
{
$concat: [{
$switch: {
branches: [
{case: {$eq: ["$$hourObject.hour", "13"]}, then: "1"},
{case: {$eq: ["$$hourObject.hour", "14"]}, then: "2"},
{case: {$eq: ["$$hourObject.hour", "15"]}, then: "3"},
{case: {$eq: ["$$hourObject.hour", "16"]}, then: "4"},
{case: {$eq: ["$$hourObject.hour", "17"]}, then: "5"},
{case: {$eq: ["$$hourObject.hour", "18"]}, then: "6"},
{case: {$eq: ["$$hourObject.hour", "19"]}, then: "7"},
{case: {$eq: ["$$hourObject.hour", "20"]}, then: "8"},
{case: {$eq: ["$$hourObject.hour", "21"]}, then: "9"},
{case: {$eq: ["$$hourObject.hour", "22"]}, then: "10"},
{case: {$eq: ["$$hourObject.hour", "23"]}, then: "11"},
],
default: "None"
}
}, " PM"]
}
]
}
}
}
}
}
},
{
$unwind: "$hours"
},
{
$project: {
_id: 0,
hour: "$hours.hour",
count: "$hours.count"
}
},
{
$sort: {
hour: 1
}
}
]);
A short explanation of the $addFields stage: we first add hours that we're missing, we then merge the two arrays (of the original found hours and the "new" missing hours), finally we convert to the required output ("01" to "01 AM").
If you're using Mongo v4+ I recommend you change the $group _id stage to use $dateFromString as its more consistent.
_id: {$hour: {$dateFromString: {dateString: "$update_at"}}}
If you do do that, you'll have to update the $filter and $map section to use numbers and not strings and eventually using $toString to cast into the format you want, hence the v4+ requirement.
You should store date values as Date objects instead of strings. I would do the formatting like this:
db.collection.aggregate(
[
{ $match: { ... } },
{
$group: {
_id: { h: { $hour: "$update_at" } },
count: { $sum: 1 }
}
},
{
$project: {
_id: 0,
hour: {
$switch: {
branches: [
{ case: { $lt: ["$_id.h", 10] }, then: { $concat: ["0", { $toString: "$_id.h" }, " AM"] } },
{ case: { $lt: ["$_id.h", 13] }, then: { $concat: [{ $toString: "$_id.h" }, " AM"] } },
{ case: { $lt: ["$_id.h", 22] }, then: { $concat: ["0", { $toString: { $subtract: ["$_id.h", 12] } }, " PM"] } },
{ case: { $lt: ["$_id.h", 24] }, then: { $concat: [{ $toString: { $subtract: ["$_id.h", 12] } }, " PM"] } }
]
}
},
hour24: "$_id.h",
count: 1
}
},
{ $sort: { hour24: 1 } }
])
As non-American I am not familiar with AM/PM rules, esp. for midnight and midday but I guess you get the principle.
Here is the query you can test it out, for MongoDB 4.0+
i will be improving query and update
const query = [{
$match: {
update_at: {
"$gte": ISODate("2019-05-03T00:00:00.0Z"),
"$lt": ISODate("2019-05-05T00:00:00.0Z")
},
id: {
"$in": [
166
]
}
}
},
{
$group: {
_id: { $hour: "$update_at" },
count: {
"$sum": 1
}
},
},
{
$addFields: {
hourStr: { $toString: { $cond: { if: { $gte: ["$_id", 12] }, then: { $subtract: [12, { $mod: [24, '$_id'] }] }, else: "$_id" } } },
}
},
{
$project: {
formated: { $concat: ["$hourStr", { $cond: { if: { $gt: ["$_id", 12] }, then: " PM", else: " AM" } }] },
count: "$count",
hour: 1,
}
}]
If you want to output in Indian Time formate. then below code work!
const query = [
{
$match: {
update_at: {
"$gte": ISODate("2019-05-03T00:00:00.0Z"),
"$lt": ISODate("2019-05-05T00:00:00.0Z")
},
id: {
"$in": [
166
]
}
}
},
{
$project: {
"h": { "$hour": { date: "$update_at", timezone: "+0530" } },
}
},
{
$group:
{
_id: { $hour: "$h" },
count: { $sum: 1 }
}
}
];