MongoDB按数组中的元素分组 [英] MongoDB group by elements in array

问题描述

我有一个看起来像这样的集合:

I have a collection that looks like this:

{
  "id": "id1",
  "tags": ['a', 'b']
},
{
  "id": "id2",
  "tags": ['b', 'c']
},
{
  "id": "id3",
  "tags": ['a', 'c']
}

如何进行按标签"中的每个元素分组的查询?数组，所以结果看起来像这样?:

How can I make a query that groups by every element in the "tags" array, so the result looks like this?:

{'a': 2},
{'b': 2},
{'c': 2}

(其中 2 是出现的次数，计数).感谢您的帮助！

(where 2 is the number of times it appears, the count). Thanks for your help!

推荐答案

你可以使用这个聚合查询:

You can use this aggregation query:

• 首先 $unwind 数组来解构像对象一样的访问.
• 然后 $group 通过 tags 和 $sum 1 得到总数.
• 最后使用 $replaceRoot 和 $arrayToObject 来获得所需的输出.

• First $unwind the array to deconstruct an access like objects.
• Then $group by tags and $sum 1 to get the total.
• And last use $replaceRoot with $arrayToObject to get the desired output.

db.collection.aggregate([
  {
    "$unwind": "$tags"
  },
  {
    "$group": {
      "_id": "$tags",
      "count": {
        "$sum": 1
      }
    }
  },
  {
    "$replaceRoot": {
      "newRoot": {
        "$arrayToObject": [
          [
            {
              "k": "$_id",
              "v": "$count"
            }
          ]
        ]
      }
    }
  }
])

示例此处

作为补充，如果您想获得排序值(a、b、c...)，您可以添加 $sort 阶段，如 这个例子

As an adittion, if you want to get sorted values (a, b, c...) you can add $sort stage like this example

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