MongoDB: $mod operator in aggregation pipeline - mongodb

Question

I have a restaurants collection that contains 3772 documents and I am trying to calculate the total number of documents that contain a score in first element of the grades array that's a multiple of 7 using the aggregation framework.
Query:
db.restaurants.aggregate([
{$project: {remainder: {$mod: ["$grades.0.score", 7]},
restaurant_id: 1,
name: 1,
grades: 1
}
},
{$match: {remainder: {$eq: 0}}},
{$group: {_id: null, total: {$sum: 1}}}
])
However, I am getting an error message that's caused by the use of the $mod operator in the $project pipeline stage. The error message is the following:
$mod only supports numeric types, not Array and NumberDouble
However, both $grades.0.score and 7 are integers, right? What should I change to make this query work as intended?
Example document:
{
"_id" : ObjectId("57290430139a4a37132c9e93"),
"address" : {
"building" : "469",
"coord" : [
-73.961704,
40.662942
],
"street" : "Flatbush Avenue",
"zipcode" : "11225"
},
"borough" : "Brooklyn",
"cuisine" : "Hamburgers",
"grades" : [
{
"date" : ISODate("2014-12-30T00:00:00Z"),
"grade" : "A",
"score" : 8
},
{
"date" : ISODate("2014-07-01T00:00:00Z"),
"grade" : "B",
"score" : 23
},
{
"date" : ISODate("2013-04-30T00:00:00Z"),
"grade" : "A",
"score" : 12
},
],
"name" : "Wendy'S",
"restaurant_id" : "30112340"
}

Answer 1

instead of $grades.0.score
put $grades[0].score
in your query.
the above is wrong. see below the correct form. As you want to filter by grades whose first score is a multiple of 7, you aggregation should start like this.
db.restaurants.aggregate([{$match: {"grades.0.score": {$mod: [7, 0]}}},{$group: {_id: null, total: {$sum: 1}}}])
I changed the grade.0.score to 7 and ran the command to check it is working or not, it seems it is working as you wanted.
> db.restaurants.find().pretty();
{
"_id" : 0,
"address" : {
"building" : "469",
"coord" : [
-73.961704,
40.662942
],
"street" : "Flatbush Avenue",
"zipcode" : "11225"
},
"borough" : "Brooklyn",
"cuisine" : "Hamburgers",
"grades" : [
{
"date" : ISODate("2014-12-30T00:00:00Z"),
"grade" : "A",
"score" : 7
},
{
"date" : ISODate("2014-07-01T00:00:00Z"),
"grade" : "B",
"score" : 23
},
{
"date" : ISODate("2013-04-30T00:00:00Z"),
"grade" : "A",
"score" : 12
}
],
"name" : "Wendy'S",
"restaurant_id" : "30112340"
> db.restaurants.aggregate([{$match: {"grades.0.score": {$mod: [7, 0]}}},{$group:{_id:null,count:{$sum:1}}} ])
{ "_id" : null, "count" : 1 }

Answer 2

First: why doesn't it work? Try:
db.restaurants.aggregate([
{$project: {
score0: "$grades.0.score",
restaurant_id: 1,
name: 1
}
}
])
You'll see that score0 returns [0 elements] so it does output an array hence the error message.
Based on this other question Get first element in array and return using Aggregate? (Mongodb), here is a solution to your problem:
db.restaurants.aggregate([
{$unwind: "$grades"},
{$group:{"_id":"$_id","grade0":{$first:"$grades"}}},
{$project: {
remainder: {$mod: ["$grade0.score", 7]},
restaurant_id: 1,
name: 1,
grade0: 1,
}
},
{$match: {remainder: {$eq: 0}}},
{$group: {_id: null, total: {$sum: 1}}}
])