MongoDB Exercise - Determine which borough has the most restaurants scoring 90 or higher and receiving an "A"

Write a MongoDB query to find the borough with the highest number of restaurants that have a grade of "A" and a score greater than or equal to 90.

Structure of 'restaurants' collection :

{
  "address": {
     "building": "1007",
     "coord": [ -73.856077, 40.848447 ],
     "street": "Morris Park Ave",
     "zipcode": "10462"
  },
  "borough": "Bronx",
  "cuisine": "Bakery",
  "grades": [
     { "date": { "$date": 1393804800000 }, "grade": "A", "score": 2 },
     { "date": { "$date": 1378857600000 }, "grade": "A", "score": 6 },
     { "date": { "$date": 1358985600000 }, "grade": "A", "score": 10 },
     { "date": { "$date": 1322006400000 }, "grade": "A", "score": 9 },
     { "date": { "$date": 1299715200000 }, "grade": "B", "score": 14 }
  ],
  "name": "Morris Park Bake Shop",
  "restaurant_id": "30075445"
}

Query:

db.restaurants.aggregate([
  {
    $match: {
      "grades.grade": "A",
      "grades.score": { $gte: 90 }
    }
  },
  {
    $group: {
      _id: "$borough",
count: { $sum: 1 }
    }
  },
  {
    $sort: { count: -1 }
  },
  {
    $limit: 1
  }
]);

Output:

[ { _id: 'Manhattan', count: 3 } ]

Explanation:

The code provided in MongoDB that finds the borough with the highest number of restaurants that have a grade of "A" and a score greater than or equal to 90.

The first stage of the aggregation $match which filters the documents based on the conditions -

checks whether the restaurant has a grade of "A" and it has a score greater than or equal to 90.

The $group stage, groups the matching documents by borough and counts the number of documents in each group.

The stage $sort, which sorts the results by the count field in descending order.

Note: This output is generated using MongoDB server version 3.6

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