按计数排序并获取父节点属性

Question

假设我有这样的结构：

{
    "_id": "Henry IV",
    "acts": [
    {
        "_id": "ACT I",
        "scenes": [
        {
            "_id": "SCENE I. London. The palace.",
            "speeches": [
            {
                "_id": 1,
                "speaker": "KING HENRY IV",
                "lines": [
                {
                    "_id": "1.1.1",
                    "text": "So shaken as we are, so wan with care,"
                },
                {
                    "_id": "1.1.2",
                    "text": "Find we a time for frighted peace to pant,"
                },
                {
                    "_id": "1.1.3",
                    "text": "And breathe short-winded accents of new broils"
                }]
            },
            {
                "_id": 2,
                "speaker": "WESTMORELAND",
                "lines": [
                {
                    "_id": "1.1.34",
                    "text": "My liege, this haste was hot in question,"
                },
                {
                    "_id": "1.1.35",
                    "text": "And many limits of the charge set down"
                }]
            }]
        }]
    }]
}

如何获得按“行”计数排序的结果，以及如何获得父节点属性？理想的结果如下所示：

{'play': 'Henry IV', 'act': 'ACT I', 'scene': 'SCENE I. London. The palace.', 'speech', '1', 'speaker': 'KING HENRY IV', 'line_count': 3}

{'play': 'Henry IV', 'act': 'ACT I', 'scene': 'SCENE I. London. The palace.', 'speech', '2', 'speaker': 'WESTMORELAND', 'line_count': 2}

Answer 1

您需要使用操作符将嵌套数组的扁平，并使用$project运算符将其转换为所需的输出，如下所示：

db.collection.aggregate([
  {
    $unwind: "$acts"
  },
  {
    $unwind: "$acts.scenes"
  },
  {
    $unwind: "$acts.scenes.speeches"
  },
  {
    $project: {
      _id: 0,
      play: "$_id",
      act: "$acts._id",
      scene: "$acts.scenes._id",
      speech: "$acts.scenes.speeches._id",
      speaker: "$acts.scenes.speeches.speaker",
      line_count: {
        $size: "$acts.scenes.speeches.lines"
      }
    }
  },
  //Add here your sorting condition
  {
    $sort: {
      act: 1,
      speech: 1,
      line_count: 1
    }
  }
])

MongoPlayground

注意:如果嵌套数组为空，则跳过文档。

皮蒙戈

from bson.son import SON
pipeline = [
     {"$unwind": "$acts" },
     ...
     {"$sort": SON([("act", 1), ("speech", 1), ("line_count", 1)])}
]
result = list(db.collection.aggregate(pipeline))