MongoDB aggregate group by sum of distinct column - mongodb

Question

I have analytics collection with the below sample data.
{ "_id" : ObjectId("55f996a4e4b0cc9c0a392594"), "action" : "apiUploadFile", "assetId" : "55f996a4e4b0cc9c0a392593" },
{ "_id" : ObjectId("5603d384e4b0cf75af10be88"), "action" : "agAsset", "assetId" : "55f996a4e4b0cc9c0a392593"},
{ "_id" : ObjectId("5603d395e4b0cf75af10becc"), "action" : "aAD", "assetId" : "55f996a4e4b0cc9c0a392593" },
{ "_id" : ObjectId("5603d395e4b0cf75af10becd"), "action" : "mobCmd", "assetId" : "55f996a4e4b0cc9c0a392593", sessionId : "123"},
{ "_id" : ObjectId("5603d395e4b0cf75af10bece"), "action" : "mobCmd", "assetId" : "55f996a4e4b0cc9c0a392593", sessionId : "1234" },
{ "_id" : ObjectId("5603d395e4b0cf75af10becf"), "action" : "mobCmd", "assetId" : "55f996a4e4b0cc9c0a392593", sessionId : "1234" }
I need find sum of analytics group by 'assetId' and then for each 'action' type. I have come up with the below query
db.analytics.aggregate(
[
{
$match : {
'assetId' : { "$ne": null }
}
},
{$group :{
_id:
{
assId:'$assetId'
},
viewCount:{
$sum:{
$cond: [ { $eq: [ '$action', 'agAsset' ] }, 1, 0 ]
}
},
sessionCount:{
$sum:{
$cond: [ { $eq: [ '$action', 'mobCmd' ] }, 1, 0 ]
}
}
}
}]
)
This works great except for the fact that I can not find the 'sessionCount' using distinct 'sessionId'. For example here is the current output
{ "_id" : { "assId" : "55f996a4e4b0cc9c0a392593" }, "viewCount" : 1, "sessionCount" : 3 }
The expected output is
{ "_id" : { "assId" : "55f996a4e4b0cc9c0a392593" }, "viewCount" : 1, "sessionCount" : 2 }
I need find the sessionCount for action='mobCmd' and has distinct values for sessionId. How can use distinct inside $sum operation of the 'sessionCount' section?

Answer 1

You will need to group your documents on a compound _id field.
db.collection.aggregate([
{ "$match": { "assetId": { "$ne": null }}},
{ "$group": {
"_id": { "assId": "$assetId", "sessionId": "$sessionId" },
"viewCount": {
"$sum": {
"$cond": [
{ "$eq": [ "$action", "agAsset" ] },
1,
0
]
}
},
"sessionCount": {
"$sum": {
"$cond": [
{ "$eq": [ "$action", "mobCmd" ] },
1,
0
]
}
}
}}
])
Which yields:
{ "_id" : { "assId" : "55f996a4e4b0cc9c0a392593", "sessionId" : "1234" }, "viewCount" : 0, "sessionCount" : 2 }
{ "_id" : { "assId" : "55f996a4e4b0cc9c0a392593", "sessionId" : "123" }, "viewCount" : 0, "sessionCount" : 1 }
{ "_id" : { "assId" : "55f996a4e4b0cc9c0a392593" }, "viewCount" : 1, "sessionCount" : 0 }
Or use the $addToSet operator to return an array of unique sessionId and $unwind the array then regroup your documents.
db.collection.aggregate([
{ "$match": { "assetId": { "$ne": null }}},
{ "$group": {
"_id": "$assetId",
"sessionId": { "$addToSet": "$sessionId" },
"viewCount": {
"$sum": {
"$cond": [
{ "$eq": [ "$action", "agAsset" ] },
1,
0
]
}
}
}},
{ "$unwind": "$sessionId" },
{ "$group": {
"_id": "$_id",
"viewCount": { "$first": "$viewCount" },
"sessionCount": { "$sum": 1 }
}}
])
Which returns:
{ "_id" : "55f996a4e4b0cc9c0a392593", "viewCount" : 1, "sessionCount" : 2 }