Question 116582 in Mongodb Question posted in
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MongoDB sort data by most number of records

I want to sort with the most occurrences of any value in a collection.

Eg.

{
  "id": "ID",
  "fruit": 'Apple'
}, 
{
  "id": "ID",
  "fruit": 'Banana'
}, 
{
  "id": "ID",
  "fruit": 'Apple'
}, 
{
  "id": "ID",
  "fruit": 'Orange'
}, 
{
  "id": "ID",
  "fruit": 'Apple'
}, 
{
  "id": "ID",
  "fruit": 'Banana'
}, 
{
  "id": "ID",
  "fruit": 'Apple'
}

I need fruit with the name Apple should on top because Apple occurs more time than other fruit in above collection

Tags:

2

Answers


  1. 0

    You’ll need to use the aggregation framework to do that.

    You can use this MongoDB playground to test it, you need to run the following aggregation pipeline:

    db.collection.aggregate([
  {
    $group: {
      _id: "$fruit",
      count: {
        $sum: 1
      },
      documents: {
        $push: "$$ROOT"
      }
    }
  },
  {
    $sort: {
      count: -1
    }
  },
  {
    $unwind: {
      path: "$documents"
    }
  },
  {
    $replaceWith: "$documents"
  }
])
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  2. 0

    You could use the "$sortByCount" aggregation stage to accomplish this.

    db.collection.aggregate([
  {
    "$sortByCount": "$fruit"
  }
])

    Example output:

    [
  {
    "_id": "Apple",
    "count": 4
  },
  {
    "_id": "Banana",
    "count": 2
  },
  {
    "_id": "Orange",
    "count": 1
  }
]

    Try it on mongoplayground.net.

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