我正在尝试根据 diagnosticType 和更高级别的 patientKey 和 encounterKey 聚合文档。这是文档的样子:

{
    "_id" : ObjectId("...."),
    "patientKey" : 1,
    "encounterKey" : 2,
    "diagnosisType" : "medical_history",
    "diagnosisCode" : "Z87.81",
}

{
    "_id" : ObjectId("...."),
    "patientKey" : 1,
    "encounterKey" : 2,
    "diagnosisType" : "problem_list",
    "diagnosisCode" : "2x.2",
}

{
    "_id" : ObjectId("...."),
    "patientKey" : 1,
    "encounterKey" : 3,
    "diagnosisType" : "medical_history",
    "diagnosisCode" : "D21.01",
}
{
    "_id" : ObjectId("...."),
    "patientKey" : 1,
    "encounterKey" : 3,
    "diagnosisType" : "medical_history",
    "diagnosisCode" : "X2.31",
}
{
    "_id" : ObjectId("...."),
    "patientKey" : 1,
    "encounterKey" : 3,
    "diagnosisType" : "problem_list",
    "diagnosisCode" : "p.2342",
}

这是我希望聚合的样子:

{
    "patientKey": 1,
    "encounters":[{
                   encounterKey: 2,
                   medical_history: [Z87.81],
                   problem_list: [2x.2]
                  },
                  {
                   encounterKey: 3,
                   medical_history: [D21.01, X2.31],
                   problem_list: [p.2342]
                  }]
 }

有什么想法可以解决这个问题吗？

我已经尝试了以下病史和问题列表，我最初的想法是先在问题列表上聚合，然后在病史上聚合，然后组合这些集合，然后根据遭遇进行另一次聚合，最后再进行一次聚合在患者身上。

但是合并异构文档是个问题。

db.collections.aggregate([
{$match:{diagnosisType:"Problem List"}},
{$group:{_id:"$encounterKey",
         "ProblemList": {$push:{$concat:"$diagnosisCode"}},
         "durableKey":{$first:"$durableKey"}}},
{$project:{_id:0,
            encounterKey:"$_id",
            ProblemList:1,
            durableKey:1}},
 {$out:"output"}
],{
allowDiskUse: true
}) 

最佳答案

您可以使用以下聚合:

db.col.aggregate([
    {
        $group: {
            _id: { patientKey: "$patientKey", encounterKey: "$encounterKey", diagnosisType: "$diagnosisType" },
            problem_list: { $push: "$diagnosisCode" }
        }
    },
    {
        $group: {
            _id: { patientKey: "$_id.patientKey", encounterKey: "$_id.encounterKey" },
            encounters: { $push: { k: "$_id.diagnosisType", v: "$problem_list" } }
        }
    },
    {
        $project: {
            encounters: { $arrayToObject: "$encounters" }
        }
    },
    {
        $group: {
            _id: "$_id.patientKey",
            encounters: {
                $push: {
                    encounterKey: "$_id.encounterKey",
                    problem_list: "$encounters.problem_list",
                    medical_history: "$encounters.medical_history"
                }
            }
        }
    },
    {
        $project: {
            patientKey: "$_id",
            _id: 0,
            encounters: 1
        }
    }
])

基本上你需要几个$group阶段来积累数据。另外你必须使用 $arrayToObject动态构建您的 JSON key 。

输出:

{
    "patientKey" : 1,
    "encounters" : [
        {
            "encounterKey" : 2,
            "problem_list" : [
                    "2x.2"
            ],
            "medical_history" : [
                    "Z87.81"
            ]
        },
        {
            "encounterKey" : 3,
            "problem_list" : [
                    "p.2342"
            ],
            "medical_history" : [
                    "D21.01",
                    "X2.31"
            ]
        }
    ]
}

关于mongodb - 如何基于唯一元素作为字段创建聚合？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/54043124/