MongoDB - Aggregate by distinct field then count per day - mongodb

Question

I have a mongodb database that collects device data.
Example document is
{
"_id" : ObjectId("5c125a185dea1b0252c5352"),
"time" : ISODate("2018-12-13T15:09:42.536Z"),
"mac" : "10:06:21:3e:0a:ff",
}
The goal would be to count the unique mac values per day, from the first document in the db to the last document in the db.
I've been playing around and came to the conclusion that I would need to have multiple groups as well as projects during my aggregations.
This is what I tried - not sure if it's in the right direction or not or just completely messed up.
pipeline = [
{"$project": {
"_id": 1,
"mac": 1,
"day": {
"$dayOfMonth":"$time"
},
"month": {
"$month":"$time"
},
"year": {
"$year":"$time"
}
}
},
{
"$project": {
"_id": 1,
"mac": 1,
"time": {
"$concat": [{
"$substr":["$year", 0, 4]
},
"-", {
"$substr": ["$month", 0, 2]
},
"-",
{
"$substr":["$day", 0, 2]
}]
}
}
},
{
"$group": {
"_id": {
"time": "$time",
"mac": "$mac"
}
},
"$group": {
"_id": "$_id.time",
"count":{"$sum": 1},
}
}
]
data = list(collection.aggregate(pipeline, allowDiskUse=True))
The output now doesn't look like it did any aggregation,
[{"_id": null, "count": 751050}]
I'm using Pymongo as my driver and using Mongodb 4.
Ideally it should just show the date and count (eg { "_id" : "2018-12-13", "count" : 2 }.
I would love some feedback and advice.
Thanks in advance.

Answer 1

I prefer to minimize the number of stages, and especially to avoid unnecessary $group stages. So I would do it with the following pipeline:
pipeline = [
{ '$group' : {
'_id': { '$dateToString': { 'format': "%Y-%m-%d", 'date': "$time" } },
'macs':{ '$addToSet': '$mac' }
} },
{$addFields:{ 'macs':{'$size':'$macs'}}}
]

Answer 2

There's an operator called "$dateToString", which would solve most of your problems.
Edit: Didn't read the question carefully, #Asya Kamsky, thank you for pointing out. Here' the new answer.
pipeline = [
{
"$group": {
"_id": {
"date": {
$dateToString: {
format: "%Y-%m-%d",
date: "$time"
}
},
"mac": "$mac"
}
}
},
{
"$group": {
"_id": "$_id.date",
"count": {
"$sum": 1
}
}
}
]

Answer 3

[
{
"$project": {
"_id": 1,
"mac": 1,
"time": { "$dateToString": { "format": "%Y-%m-%d", "date": "$time", "timezone": "Africa/Johannesburg"}}
},
},
{
"$group": {
"_id":{
"time": "$time",
"mac": "$mac",
}}},{
"$group": {
"_id": "$_id.time",
"count":{"$sum": 1}
}},
{"$sort": SON([("_id", -1)])}
]
Does exactly what it should do.
Thanks. :)