如何在对象内部排序并减去MongoDB中对象内部的值答案

【问题标题】：how to sort inside the object and subtract the values inside the objects in MongoDB如何在对象内部排序并减去MongoDB中对象内部的值
【发布时间】：2021-10-24 09:52:37
【问题描述】：

输入文件

  "data": {
    "2021-05-05": {
      "12": 2.1,
      "15": 0.2,
      "06": 2.6,
      "09": 2.2
    },
    "2021-05-06": {
      "12": 2.1,
      "15": 0.2,
      "06": 2,
      "09": 1.7
    }
}

中间步骤数据objects 将与键连接并按时间排序。

  "data": {
    "2021-05-05": {
      "2021-05-05T06:00:00:0Z": 2.6,
      "2021-05-05T09:00:00:0Z": 2.2,
      "2021-05-05T12:00:00:0Z": 2.1,
      "2021-05-05T15:00:00:0Z": 0.2

    },
    "2021-05-06": {
      "2021-05-06T12:00:00:0Z": 2.1,
    }
}

排序后。下一步将是。 Subtract 从value_last - value_first 相对于时间。

输出文档

Document 1


{
  "data": {
      "time_first" : "2021-05-05T06:00:00:0Z",
      "value_first": 2.6,
      "time_last": "2021-05-05T15:00:00:0Z"
      "value_last": "0.2",
      "difference_value_last-value_first" : -0.4

    }

Document 2 如果只包含一个值，则上个月的value_last成为下一个文档的value_first 。

    "data": {
      "time_first" : "2021-05-05T15:00:00:0Z",
      "value_first": 0.2,
      "time_last":  "2021-05-06T12:00:00:0Z",
      "value_last": : 2.1 ,
      "difference_value_last-value_first" : 1.9
    }

【问题讨论】：

标签： arrays mongodb mongoose mongodb-query aggregation-framework

【解决方案1】：

你可以使用聚合

$objectToArray 把对象变成一个数组，所以我们可以得到键：值对，它定义了名称“k”，有助于获取数据
$unwind 解构数组
$sort 按条件排序
$group 查找我们排序的第一个和最后一个元素
$concat 拼接字符串日期时间字符串

这里是代码

db.collection.aggregate([
  {  $project: { data: { "$objectToArray": "$data} } },
  {
    $project: {
      data: {
        $map: {
          input: "$data",
          in: {
            "dk": "$$this.k",
            "dv": { "$objectToArray": "$$this.v" }
          }
        }
      }
    }
  },
  { "$unwind": "$data" },
  { $unwind: "$data.dv" },
  {
    $sort: {
      "data.dk": -1,
      "data.dv.v": -1
    }
  },
  {
    $group: {
      _id: "$data.dk",
      firstData: { $first: "$data.dv"  },
      lastData: { $last: "$data.dv" }
    }
  },
  {
    $project: {
      "time_first": {
        $concat: [ "$_id", "T", "$firstData.k", ":00:00:0Z" ]
      },
      "time_last": {
        $concat: [ "$_id", "T", "$lastData.k", ":00:00:0Z" ]
      },
      "value_first": "$firstData.v",
      "value_last": "$lastData.v",
      "difference_value_last-value_first": {
        "$subtract": [ "$lastData.v", "$firstData.v" ]
      }
    }
  }
])

工作Mongo playground

【讨论】：