Grouping different amounts together in MongoDB - mongodb

Question

If I have a set of objects each with the same description, but with different amounts.
{
{
"_id": "101",
"description": "DD from my employer1",
"amount": 1000.33
},
{
"_id": "102",
"description": "DD from my employer1",
"amount": 1000.34
},
{
"_id": "103",
"description": "DD from my employer1",
"amount": 1000.35
},
{
"_id": "104",
"description": "DD from employer1",
"amount": 5000.00
},
{
"_id": "105",
"description": "DD from my employer2",
"amount": 2000.33
},
{
"_id": "106",
"description": "DD from my employer2",
"amount": 2000.33
},
{
"_id": "107",
"description": "DD from my employer2",
"amount": 2000.33
}
}
Below, I am able to group them using the description:
{
{
"$group": {
"_id": {
"description": "$description"
},
"count": {
"$sum": 1
},
"_id": {
"$addToSet": "$_id"
}
}
},
{
"$match": {
"count": {
"$gte": 3
}
}
}
}
Is there a way to include all the amounts in the group (_ids: 101, 102, and 103 plus 105,106,107) even if they have a small difference, but exclude the bonus amount, which in the sample above is _id 104?
I don't believe it could be done in a group stage, but is there something that could be done at a later stage that could group _ids 101, 102 and 103 together and exclude _id 104. Basically, I want MongoDB to ignore the small differences in 101, 102, 103 and group them together since the are paychecks coming from the same employer.
I have been working with $stdDevPop, but can't get a solid formula down.
I am looking for a simple array output of just the _ids.
{
"result": [
"101",
"102",
"103",
"105",
"106",
"107"
]
}

Answer 1

You can do this by doing some math on the "amount" to round it down to the nearest 1000 and use that as the grouping _id:
db.collection.aggregate([
{ "$group": {
"_id": {
"$subtract": [
{ "$trunc": "$amount" },
{ "$mod": [
{ "$trunc": "$amount" },
1000
]}
]
},
"results": { "$push": "$_id" }
}},
{ "$redact": {
"$cond": {
"if": { "$gt": [ { "$size": "$results" }, 1 ] },
"then": "$$KEEP",
"else": "$$PRUNE"
}
}},
{ "$unwind": "$results" },
{ "$group": {
"_id": null,
"results": { "$push": "$results" }
}}
])
If your MongoDB is older than 3.2 then you would just need to use a long form with $mod of what $trunc is doing. And if your MongoDB is older than 2.6 then rather than $redact you would $match. So in the longer form this is:
db.collection.aggregate([
{ "$group": {
"_id": {
"$subtract": [
{ "$subtract": [
"$amount",
{ "$mod": [ "$amount", 1 ] }
]},
{ "$mod": [
{ "$subtract": [
"$amount",
{ "$mod": [ "$amount", 1 ] }
]},
1000
]}
]
},
"results": { "$push": "$_id" },
"count": { "$sum": 1 }
}},
{ "$match": { "count": { "$gt": 1 } } },
{ "$unwind": "$results" },
{ "$group": {
"_id": null,
"results": { "$push": "$results" }
}}
])
Either way the output is just the _id values whose amounts grouped to the boundaries with a count more than once.
{ "_id" : null, "results" : [ "105", "106", "107", "101", "102", "103" ] }
You could either add a $sort in there or live with sorting the result array in client code.

Answer 2

db.yourDBNameHere.aggregate( [
{ $match: { "amount" : { $lt : 5000 } } },
{ $project: { _id: 1 } },
])
that will grab the ID only of every transaction less than 5000$.