MongoDB aggregation without duplication - mongodb

Question

I have the following records:
{ "_id" : 1, "c" : 120, "b" : [ { "f1" : 10 }, { "f1" : 10 } ] }
{ "_id" :2, "c" : 5, "b" : [ { "f1" : 10 }, { "f1" : 10 } ] }
I need the output this way:
{ "_id" : 1, 'total':140}
{ "_id" :2, 'total':25 }
where total = sum of value in 'c' with sum of values in f1 for same record.
When i unwind the field 'b' it creates two documents with same id and hence data is duplicated and when i sum it up, i get:
db.test2.aggregate([
{'$unwind':'$b'},
{'$project':{'total':{'$add':['$c','$b.f1']}}},
{'$group':{'_id':'$_id', 'total':{'$sum':'$total'}}}
])
outputs:
{ "_id" : 1, 'total':260}
{ "_id" :2, 'total':30 }
(not what i wanted, as it has added 120 and 5 again to total due to duplication during unwinding)
So i tried:
db.test2.aggregate([
{'$unwind':'$b'},
{'$group':{'_id':'$_id', 'c':{'$push': '$c'},'f1':{'$sum':'$b.f1'}}},
{'$project':{'total':{'$add':[{'$arrayElemAt':['$c',0]},'$f1']}}}
])
outputs:
{ "_id" : 1, 'total':140}
{ "_id" :2, 'total':25 }
( what i wanted)
Is there any other way to achieve this?

Answer 1

You can try below query. Sum operator to first calculate sum in array followed by add to calculate total with other field.
db.test2.aggregate([{
$project: {
total: {"$add":["$c", {"$sum":"$b.f1"}]}
}
}]

Answer 2

An alternative:
db.test2.aggregate([{
$project: {
_id: 0,
c: "$c",
b: {
$reduce: {
input: "$b.f1",
initialValue: 0,
in: {
$add: ["$$value", "$$this"]
}
}
}
}
},
{
$project: {
_id: 0,
total: {
$sum: ["$c", "$b"]
}
}
}
])
That would create result:
{
"total" : 140
}
{
"total" : 25
}
If you need the field _id then replace the _id: 0 in both $project to _id: 1
That would create this result:
{
"_id" : 1,
"total" : 140
}
{
"_id" : 2,
"total" : 25
}