查询MongoDB数组并使用最匹配的元素进行排序 [英] Query MongoDB array and sort with the most matched elements

问题描述

在以下情况下，我需要您的专业知识.

I need your expertise on the following situation.

我有这样的收藏夹:

"array" : {
    "item" : 1,
    "1" : [100, 130, 255],
}

"array" : {
    "item" : 2,
    "1" " [0, 70, 120],
}

"array" : {
    "item" : 3,
    "1" : [100, 90, 140],

}

我正在这样查询这个集合:

I am querying this collection as such:

 db.test.find(array.1 : {$in : [100, 80, 140]});

这将返回项目编号1和3，因为它将提供的数组中的任何值与集合中的值进行匹配. 但是，我想对该数组进行排序，以便为我提供更多相似数字的结果. 结果应分别为项目3和项目1.

This returns me item number 1 and 3 since it matches any values in the provided array with the ones in the collection. However I would like to sort this array to give me the results with more similar numbers. The result should be items 3 and 1 respectively.

但是，我可以获取结果并使用k最近邻算法对数组进行排序.但是，处理庞大的数据集使此操作非常不可取(或者是?) MongoDB中是否有提供此功能的功能? 我正在使用Java，是否有足够的算法来实现这一目标? 感谢您的帮助.

I can however grab the results and use a k-nearest neighbor algorithm to sort the array. However dealing with huge datasets makes this very undesirable (or is it?) Are there any functions within MongoDB to provide this? I am using Java, any algorithms to achieve this fast enough? Any help is appreciated.

谢谢.

推荐答案

您可以使用聚合框架执行此操作，尽管这并不容易.问题在于没有$in运算符作为聚合框架的一部分.因此，您必须以编程方式匹配数组中的每个项目，这变得非常混乱. edit :如果$in帮助您过滤掉大部分内容，则重新排列顺序，以使匹配项排在第一位.

You can do this with the aggregation framework, although it's not easy. The trouble lies with there not being an $in operator as part of the aggregation framework. So you have to programatically match each of the items in the array, which gets very messy. edit: reordered so that the match is first, in case that $in helps you filter a good portion out.

db.test.aggregate(
  {$match:{"array.1":{$in:[100, 140,80]}}}, // filter to the ones that match
  {$unwind:"$array.1"}, // unwinds the array so we can match the items individually
  {$group: { // groups the array back, but adds a count for the number of matches
    _id:"$_id", 
    matches:{
      $sum:{
        $cond:[
          {$eq:["$array.1", 100]}, 
          1, 
          {$cond:[
            {$eq:["$array.1", 140]}, 
            1, 
            {$cond:[
              {$eq:["$array.1", 80]}, 
              1, 
              0
              ]
            }
            ]
          }
          ]
        }
      }, 
    item:{$first:"$array.item"}, 
    "1":{$push:"$array.1"}
    }
  }, 
  {$sort:{matches:-1}}, // sorts by the number of matches descending
  {$project:{matches:1, array:{item:"$item", 1:"$1"}}} // rebuilds the original structure
);

输出:

{
"result" : [
    {
        "_id" : ObjectId("50614c02162d92b4fbfa4448"),
        "matches" : 2,
        "array" : {
            "item" : 3,
            "1" : [
                100,
                90,
                140
            ]
        }
    },
    {
        "_id" : ObjectId("50614bb2162d92b4fbfa4446"),
        "matches" : 1,
        "array" : {
            "item" : 1,
            "1" : [
                100,
                130,
                255
            ]
        }
    }
],
"ok" : 1
}

如果最后将matches字段排除在$project之外，则可以将其保留在结果之外.

You can leave the matches field out of the result if you leave it out of the $project at the end.

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