MongoDB Exercise - Find the restaurants achieved highest average score

Write a MongoDB query to find the restaurants achieved highest average score.

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([
  {$unwind: "$grades"},
  {$group: {
    _id: "$restaurant_id",
avgScore: {$avg: "$grades.score"}
  }},
  {$sort: {avgScore: -1}},
  {$limit: 1},
  {$project: {_id: 1, avgScore: 1}}
])

Output:

[ { _id: '40393488', avgScore: 38.6 } ]

Explanation:

The given query in MongoDB calculates the average score of each restaurant and returns only the restaurant with the highest average score, along with its restaurant_id and avgScore fields.

The $unwind stage that split each document of the grades array into multiple documents.

The $group operator grouped by the restaurant_id field, and a new field called avgScore is created using the $avg aggregation operator to calculate the average score of each restaurant.

The $sort then sorts the resulting documents in descending order by the avgScore field.

The $limit returns only the first document, meaning the restaurant with the highest average score.

Note: This output is generated using MongoDB server version 3.6

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