javascript - Mongodb 聚合以对数组中最常见的项目进行排序？

Question

集合中的一个项目包含一个字符串数组。我想查找数组中具有最匹配元素的项目并对其进行排序。

考虑一个集合:

[
    {
        "item_name":"Item_1",
        "tags":["A","B","C","D","E"]
    },
    {
        "item_name":"Item_2",
        "tags":["A","B","D","E","G"]
    },
    {
        "item_name":"Item_3",
        "tags":["B","C","E","H"]
    }

]

我想根据 ["B","D","G","F"] 这样的数组对集合进行排序这将返回

[

    {
        "item_name":"Item_2",
        "tags":["A","B","D","E","G"]
    },
    {
        "item_name":"Item_1",
        "tags":["A","B","C","D","E"]
    },
    {
        "item_name":"Item_3",
        "tags":["B","C","E","H"]
    }

]

预期顺序为 Item_2、Item_1，然后是 Item_3，因为

• Item_2 匹配 3 个项目(“B”、“D”和“G”)
• 然后是具有 2 个匹配项(“B”和“D”)的 Item_1
• 最后 Item_3 有 1 个匹配(“B”)

如果不在mongodb中，JavaScript方法也会受到赞赏

Answer 1

如果 tags 中的值是唯一的，您可以使用 tags 的交集大小和使用 $size 的查询数组和 $setIntersection

db.collection.aggregate([
  {
    $set: {
      matchedCount: {
        $size: {
          $setIntersection: ["$tags", ["B","D","G","F"]]
        }
      }
    }
  },
  {
    $sort: {
      matchedCount: -1
    }
  }
])