MongoDB query group with 'sub' group - mongodb

Question

From a product stocks log I have created a MongoDB collection. The relevant fields are: sku, stock and date. Every time a products stock is updated there is a new entry with the total stock.
The skus are made up of two parts. A parent part, say 'A' and a variant or child part, say '1', '2', '3', etc.. So a sku might look like this: 'A2'.
I can query for a single products stock, grouped by day, with this query:
[{
$match: {
sku: 'A2'
}
},
{
$group: {
_id: {
year: {$year: '$date'},
day: {$dayOfYear: '$date'}
},
stock: {
$min: '$stock'
},
date: {
$first: '$date'
}
}
},
{
$sort: {
date: 1
}
}]
Note: I want the minimum stock for each day.
But I need to query for all variations (minimum) stocks added up. I can change the $match object to:
[{
$match: {
sku: /^A/
}
}
How do I create a 'sub' group in the $group stage?
EDIT:
The data looks like this:
{
sku: 'A1',
date: '2015-01-01',
stock: 15
}
{
sku: 'A1',
date: '2015-01-01',
stock: 14
}
{
sku: 'A2',
date: '2015-01-01',
stock: 20
}
Two stocks for 'A1' and one for 'A2' on a single day. My query (all skus grouped by day) would give me stock 14 as a result ($min of the 3 values). But I want the result to be 34. 20 (min for A2) plus 14 (min for A1)

Answer 1

If you add the sku to the _id field in the group phase it will aggregate on that as well, i.e. group per sku, year & day.
db.stocks.aggregate(
[
{
$group: {
_id: {
sku: '$sku',
year: {$year: '$date'},
day: {$dayOfYear: '$date'}
},
stock: {
$min: '$stock'
},
date: {
$first: '$date'
}
}
},
{
$sort: {
date: 1
}
}]
)