的MapReduce /聚合：集團由值嵌套文檔中

Question

想象我有這樣一個集合：

{ 
    "_id": "10280", 
    "city": "NEW YORK", 
    "state": "NY", 
    "departments": [ 
      {"departmentType":"01", 
       "departmentHead":"Peter"}, 
      {"departmentType":"02", 
       "departmentHead":"John"} 
    ] 
}, 
{ 
    "_id": "10281", 
    "city": "LOS ANGELES", 
    "state": "CA", 
    "departments": [ 
      {"departmentType":"02", 
       "departmentHead":"Joan"}, 
      {"departmentType":"03", 
       "departmentHead":"Mary"} 
    ] 
}, 
{ 
    "_id": "10284", 
    "city": "MIAMI", 
    "state": "FL", 
    "department": [ 
    "departments": [ 
      {"departmentType":"01", 
       "departmentHead":"George"}, 
      {"departmentType":"02", 
       "departmentHead":"Harry"} 
    ] 
} 

我想獲得每departmentType計數，像這樣：

[{"departmentType":"01", "dCount":2}, 
{"departmentType":"02", "dCount":3}, 
{"departmentType":"03", "dCount":1} 
] 

爲此，我已經嘗試了幾乎所有的東西，但我在網上找到的所有示例都比較容易，其中group by是在文檔根級別的字段上完成的。相反，在這裏我試圖按部門類型進行分組，這似乎破壞了我發現的所有東西。

有關如何使用Mongoose的聚合實現或mapreduce來做到這一點的任何想法？

理想情況下，我想排除所有與< = 1的部門類型和按部門類型排序結果。

謝謝大家提前！

Answer 1

您需要$展開部門數組，這將爲數組中的每個條目創建一個文檔，以便您可以在管道中聚合它們。

不幸的是，您不能預先過濾departmentTypes < = 1，因爲$ size只會取一個確切的值，但您可以將其過濾出結果。這不是很好，但它的工作原理。此示例僅對具有2個部門的記錄進行預過濾，但僅用於演示，您可能希望放棄第一個$匹配，因爲我們過濾出< = 1，並在稍後對結果進行第二次$匹配;

db.runCommand({ 
    aggregate: "so", 
    pipeline: [ 
     { // filter out only records with 2 departments 
      $match: { 
       departments: { $size: 2 } 
      } 
     }, 
     // unwind - create a doc for each department in the array 
     { $unwind: "$departments" }, 
     { // aggregate sum of departments by type 
      $group: { 
       _id: "$departments.departmentType", 
       count: { $sum: 1 }, 
      } 
     }, 
     { // filter out departments with <=1 
      $match: { 
       count: { $gt: 1 }, 
      } 
     }, 
     { // rename fields as per example 
      $project: { 
       _id: 0, 
       departmentType: "$_id", 
       dCount: "$count", 
      } 
     } 
    ] 
}); 

請注意，我還假設你以前的JSON的樣品有一個錯字，和「部門」實際上並不存在。假設所有文檔與前兩個文檔具有相同的模式，此代碼將工作。

如果您對所獲得的實際字段名稱沒有打擾，請隨意放棄第一個$匹配項和最後一個$項目。