聚合子文档，按 id 列表过滤它们，并按 id 分组值

Question

我有一个需要映射的 mongo ObjectId 列表，将一些响应（可能存在也可能不存在）分配给该 id，然后根据响应生成可用性平均值。例如，我有一个引用 Users 的 ObjectId 列表：

["5d978d372f263f41cc624727", "5d978d372f263f41cc624728", "5d978d372f263f41cc624729"]

还有一个 Events 的聚合列表，其中包含一个 employeeResponses 子文档，如果他们已经回复，则记录他们的 _id、response 和任何事件 notes：

[
  {
    ...
    employeeResponses: [
      {
        _id: "5d978d372f263f41cc624727",
        response: "I want to work.",
        notes: ""
      },
      {
        _id: "5d978d372f263f41cc624728",
        response: "Available to work.",
        notes: ""
      }
    ]
  },
  {
    ...
    employeeResponses: [
      {
        _id: "5d978d372f263f41cc624727",
        response: "I want to work.",
        notes: ""
      },
      {
        _id: "5d978d372f263f41cc624728",
        response: "Prefer not to work.",
        notes: ""
      }
    ]
  },
  {
    ...
    employeeResponses: [
      {
        _id: "5d978d372f263f41cc624727",
        response: "Not available to work.",
        notes: ""
      },
      {
        _id: "5d978d372f263f41cc624729",
        response: "Not available to work.",
        notes: ""
      },
    ]
  }
]

我需要做的是在responses 数组中将他们的响应（或没有响应）分配给他们的 id：

[
  {
    "_id": "5d978d372f263f41cc624727",
    "responses": ["I want to work.", "I want to work.", "Not available to work."]
  },
  {
    "_id": "5d978d372f263f41cc624728",
    "responses": ["Available to work.", "Prefer not to work.", "No response."] 
  }
  {
    "_id": "5d978d372f263f41cc624729",
    "responses": ["No response.", "No response.", "Not available to work."] 
  }
]

然后，根据他们的响应，生成成员可用性平均值（available：["I want to work.", "Available to work"]；unavailable：["Prefer not to work.", "Not available to work.", "No response."]）：

[
  {
    "_id": "5d978d372f263f41cc624727",
    "availability": 66
  },
  {
    "_id": "5d978d372f263f41cc624728",
     "availability": 33,
  }
  {
    "_id": "5d978d372f263f41cc624729",
     "availability": 0,  
  }
]

在@mickl 的帮助下，我可以aggregate the responses for 1 ObjectId，但我正在努力将其扩展为 ObjectIds 列表，而不映射 ids 并为每个用户聚合 - 这似乎非常低效且成本高昂。

Answer 1

你可以试试这个：

原始示例：https://mongoplayground.net/p/_pDaErcuvkP

调整https://mongoplayground.net/p/oURNxgNiprl中的查询：

    db.collection.aggregate([
  {
    $unwind: {
      path: "$employeeResponses",          
    }
  },
  {
    "$group": {
      _id: "$employeeResponses._id",
      "availability": {
        "$sum": {
          "$cond": [
            {
              "$in": [
                "$employeeResponses.response",
                [
                  "Available to work.",
                  "I want to work."
                ]
              ]
            },
            1,
            0
          ]
        }
      }
    }
  }
])