MongoDB计数按数组元素分组的数组中的匹配字符串答案

【问题标题】：MongoDB Count on matching strings in array grouped by array elementMongoDB计数按数组元素分组的数组中的匹配字符串
【发布时间】：2021-05-14 14:28:46
【问题描述】：

我有一个存储调查答案的 MongoDB 集合。答案通常是带有“优秀”、“好”或“差”等响应的单选按钮。我正在尝试生成一个查询，该查询为每个问题返回给定响应的总数。响应当前存储在字符串数组中。数组中的位置 0 是问题 1 的答案，依此类推。

我目前有一个聚合查询，它以以下截断格式返回数据：

[{ 
    "name" : "Medical Visit Survey", 
    "question" : [
        "Ease of making an appointment?", 
        "About how long was your wait time before being seen by the provider?", 
        "Professionalism of person who took your call?"
    ], 
    "type" : [ "radiobutton", "radiobutton", "radiobutton" ], 
    "answers" : [ "Excellent",  "Less than 20 minutes", "Excellent" ]
},
{ 
    "name" : "Medical Visit Survey", 
    "question" : [
        "Ease of making an appointment?", 
        "About how long was your wait time before being seen by the provider?", 
        "Professionalism of person who took your call?"
    ], 
    "type" : [ "radiobutton",  "radiobutton", "radiobutton" ], 
    "answers" : ["Excellent",  "Less than 20 minutes",  "Very Good" ]
}]

产生类似于以下输出的最佳方法是什么：

[{
   "name" : "Medical Visit Survey",
   "question" : "Ease of making an appointment?",
   "type" : "radiobutton",
   "answers": { 
                 "Excellent": 2,
                 "Good": 3,
                 "Poor": 1
              }
  
},
{
   "name" : "Medical Visit Survey",
   "question" : "About how long was your wait time before being seen by the provider?",
   "type" : "radiobutton",
   "answers": { 
                 "Less than 20 minutes": 2,
                 "More than 20 minutes": 3,
                 "More than 60 minutes": 1
              }
  
}
]

我尝试过类似以下的查询：

[
  {$unwind: "$answers" },
  { $group: { _id: "$answers", count: { $sum: 1 } } }
]

输出根据给出的答案计算响应，但不考虑问题编号（数组中的元素位置）。

我有一个可能有用的 mongo 游乐场链接：https://mongoplayground.net/p/4_uM7khrMEM

任何帮助将不胜感激。

【问题讨论】：

你的例子没有加起来，你怎么能得到"Excellent": 4？只有三个？
我应该澄清数据是一个截断的例子。忽略总数。这只是一个格式化的例子。为了清楚起见，我将编辑问题。

标签： mongodb mongoose aggregation-framework nosql-aggregation

【解决方案1】：

我不确定有没有最好的方法，但我会建议一个聚合查询，

$unwind 解构 question 数组并在每个问题元素的 index 字段中包含数组索引
$arrayElemAt 选择提供的index 字段中的特定answer，对于type 字段也相同
$group by question 和 answer，选择必填字段并计算总数
$group 仅由 question 并以键值格式构造 answers 数组
$arrayToObject 将answers 数组转换为对象

[
  {
    $unwind: {
      path: "$question",
      includeArrayIndex: "index"
    }
  },
  {
    $group: {
      _id: {
        question: "$question",
        answer: { $arrayElemAt: ["$answers", "$index"] }
      },
      name: { $first: "$name" },
      type: { $first: { $arrayElemAt: ["$type", "$index"] } },
      count: { $sum: 1 }
    }
  },
  {
    $group: {
      _id: "$_id.question",
      answers: {
        $push: { k: "$_id.answer", v: "$count" }
      },
      name: { $first: "$name" },
      type: { $first: "$type" }
    }
  },
  { $addFields: { answers: { $arrayToObject: "$answers" } } }
]

Playground

【讨论】：