Question

我正在尝试基于诊断类型并在更高级别patientKey和encounterKey上聚合文档。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",
}

这是我希望聚合显示的样子：

{
    "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]
                  }]
 }

有什么办法可以解决这个问题吗？

我尝试过以下操作以获取病历，然后再列出问题列表，最初的想法是先在“问题列表”上汇总，然后再在“病历”上汇总，然后合并集合，然后根据遇到的情况再进行汇总，最后再进行汇总在耐心的。

但是合并异类文档是一个问题。

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
})

Answer 1

您可以使用以下汇总：

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，您需要几个$group阶段来累积数据。另外，您必须使用$arrayToObject动态构建JSON密钥。

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"
            ]
        }
    ]
}

如何创建基于唯一元素作为字段的聚合？

1 个答案: