Get all possible combinations from array in MongoDB aggregation 🚀 - mongodb

Question

How to do aggregation ($group) by the same values ​​from the array? Not all at once, but few or all, if any. I can do $group by one word, but I also need all possible variations...
Collection example:
{"keywords": ["gta", "distribution", "keys"]}
{"keywords": ["gta", "online", "moto", "races"]}
{"keywords": ["gta", "online", "samp"]}
Result example:
"gta" - 3 matches
"online" - 2 matches
"gta online" - 2 matches

Answer 1

You could use $reduce to extract all combinations of pairs from an array. I've started from this post and I've added the current item, $unwind the
initial array and count the items :
db.test.aggregate([
{
$project: {
pairs: {
$reduce: {
input: { $range: [0, { $size: "$keywords" }] },
initialValue: [],
in: {
$concatArrays: [
"$$value",
[[{ $arrayElemAt: ["$keywords", "$$this"] }]],
{
$let: {
vars: { i: "$$this" },
in: {
$map: {
input: { $range: [{ $add: [1, "$$i"] }, { $size: "$keywords" }] },
in: [{ $arrayElemAt: ["$keywords", "$$i"] }, { $arrayElemAt: ["$keywords", "$$this"] }]
}
}
}
}
]
}
}
}
}
}, {
$unwind: "$pairs"
}, {
$group: {
_id: "$pairs",
count: { $sum: 1 }
}
}
])
Output :
{ "_id" : [ "online", "samp" ], "count" : 1 }
{ "_id" : [ "gta", "samp" ], "count" : 1 }
{ "_id" : [ "online", "races" ], "count" : 1 }
{ "_id" : [ "moto", "races" ], "count" : 1 }
{ "_id" : [ "gta", "keys" ], "count" : 1 }
{ "_id" : [ "races" ], "count" : 1 }
{ "_id" : [ "gta", "distribution" ], "count" : 1 }
{ "_id" : [ "samp" ], "count" : 1 }
{ "_id" : [ "distribution", "keys" ], "count" : 1 }
{ "_id" : [ "gta" ], "count" : 3 }
{ "_id" : [ "online" ], "count" : 2 }
{ "_id" : [ "keys" ], "count" : 1 }
{ "_id" : [ "gta", "online" ], "count" : 2 }
{ "_id" : [ "moto" ], "count" : 1 }
{ "_id" : [ "online", "moto" ], "count" : 1 }
{ "_id" : [ "distribution" ], "count" : 1 }
{ "_id" : [ "gta", "moto" ], "count" : 1 }
{ "_id" : [ "gta", "races" ], "count" : 1 }
If you need more combinations, you may need to update the $reduce stage above