Mongodb aggregation to pass both a matched array and an unmatched array - node.js

Question

I've got a MongoDB / Nodes aggregation that looks a little like this (there are other values in there, but this is the basic idea).
[
{
'$unwind': {
'path': '$Vehicles'
}
},
{
'$match': {
'Vehicles.Manufacturer': 'FORD'
}
},
{
'$facet': {
'makes': [
{
'$group': {
'_id': '$Vehicles.Manufacturer',
'count': {
'$sum': 1
}
}
}
]
}
},
{
'$project': {
'makes': {
'$sortArray': {
'input': '$makes',
'sortBy': 1
}
}
}
}
]
This works fine. But I would also like to pass an unmatched list through. IE an an array of vehicles whose Manufacturer = FORD and an other list of all Manufacturer.
Can't get it to work. Any ideas please?
Thanks in advance.
Edit:-
The current output looks like this:
[{
"makes": [
{
"_id": "FORD",
"count": 285
}
]
}]
and ideally it would look something like this:
[{
"makes": [
{
"_id": "FORD",
"count": 285
}
],
"unfiltered_makes": [
{
"_id": "ABARTH",
"count": 1
},
{
"_id": "AUDI",
"count": 7
},
{
"_id": "BMW",
"count": 2
},
{
"_id": "CITROEN",
"count": 4
},
{
"_id": "DS",
"count": 1
},
{
"_id": "FIAT",
"count": 1
}.... etc
]
}]
The data looks a bit like this:
"Vehicles": [
{
"Id": 1404908,
"Manufacturer": "MG",
"Model": "3",
"Price": 11995 .... etc
},{
"Id": 1404909,
"Manufacturer": "FORD",
"ManufacturerId": 34,
"Model": "Focus",
"Price": 12000 .... etc
} ... etc
]

Answer 1

In this case you can do something like:
db.collection.aggregate([
{$unwind: "$Vehicles"},
{$group: {
_id: "$Vehicles.Manufacturer",
count: {$sum: 1}}
},
{$facet: {
makes: [{$match: {_id: "FORD"}}],
unfiltered_makes: [{$group: {_id: 0, data: {$push: "$$ROOT"}}}]
}
},
{$project: {makes: 1, unfiltered_makes: "$unfiltered_makes.data"}}
])
See how it works on the playground example
Another option is:
db.collection.aggregate([
{$unwind: "$Vehicles"},
{$group: {
_id: "$Vehicles.Manufacturer",
count: {$sum: 1}}
},
{$group: {
_id: 0,
unfiltered_makes: {$push: "$$ROOT"},
makes: {$push: {$cond: [{$eq: ["$_id", "FORD"]}, "$$ROOT", "$$REMOVE"]}}
}
}
])
See how it works on the playground example

Answer 2

Here's another way to do it using "$function" to generate a histogram of "Manufacturer" and format the returned array. The javascript function only traverses the "Vehicles" array once, so this may be fairly efficient, although I did not do algorithm timing comparisons on a large collection.
N.B.: I'm a javascript noob and there may be a better way to do this.
db.collection.aggregate([
{
"$set": {
"unfiltered_makes": {
"$function": {
// generate histogram of manufacturers and format output
"body": "function(makes) {const m = new Object();makes.forEach((elem) => {m[elem.Manufacturer] = m[elem.Manufacturer] + 1 || 1});return Object.entries(m).map(([make, count]) => {return {'_id':make, 'count':count}})}",
"args": ["$Vehicles"],
"lang": "js"
}
}
}
},
{
"$project": {
"_id": 0,
"unfiltered_makes": 1,
"makes": {
"$filter": {
"input": "$unfiltered_makes",
"as": "make",
"cond": {
"$eq": [
"$$make._id",
// your search "Manufacturer" goes here
"FORD"
]
}
}
}
}
}
])
Try it on mongoplayground.net.