Sum the different grades by date in MongoDB [duplicate] - mongodb

Question

This question already has an answer here:
Mongodb count distinct with multiple group fields
(1 answer)
Closed 6 years ago.
I'm using the restaurants dataset from the MongoDB website. A document has arrays like the following:
{
"grades" : [
{
"date" : ISODate("2014-06-10T00:00:00.000Z"),
"grade" : "A"
},
{
"date" : ISODate("2013-06-05T00:00:00.000Z"),
"grade" : "B",
"score" : 7
},
{
"date" : ISODate("2012-04-13T00:00:00.000Z"),
"grade" : "A"
},
{
"date" : ISODate("2011-10-12T00:00:00.000Z"),
"grade" : "A"
}
]
}
I'm trying to get a list of all dates, with a count of how many of each grade there was on that day.
I've got this far:
db.restaurants.aggregate([{
$unwind : {
path: '$grades'
}
}, {
$group: {
_id: '$grades.date',
grades: {
$push: '$grades.grade'
}
}
}])
Which gives me each date and the grades on that date.
How do I now count the number of each unique grade?

Answer 1

Figured it out with thanks to this question.
The solution is actually much simpler than I was thinking:
db.restaurants.aggregate([{
$unwind : {
path: '$grades'
}
}, {
$group: {
_id: {
date: '$grades.date',
grade: '$grades.grade'
},
count: {
$sum: 1
}
}
}])
This gives a result like:
/* 1 */
{
"_id" : {
"date" : ISODate("2014-06-23T00:00:00.000Z"),
"grade" : "C"
},
"count" : 4
}
/* 2 */
{
"_id" : {
"date" : ISODate("2011-11-01T00:00:00.000Z"),
"grade" : "C"
},
"count" : 3
}
/* 3 */
{
"_id" : {
"date" : ISODate("2014-05-06T00:00:00.000Z"),
"grade" : "A"
},
"count" : 121
}
/* 4 */
{
"_id" : {
"date" : ISODate("2012-08-21T00:00:00.000Z"),
"grade" : "C"
},
"count" : 5
}
/* 5 */
{
"_id" : {
"date" : ISODate("2013-09-04T00:00:00.000Z"),
"grade" : "C"
},
"count" : 4
}