Mongodb 基于嵌套文档集的聚合

假设我有接下来的5个文档：

{ "_id" : "1", "student" : "Oscar", "courses" : [ "A", "B" ] }
{ "_id" : "2", "student" : "Alan", "courses" : [ "A", "B", "C" ] }
{ "_id" : "3", "student" : "Kate", "courses" : [ "A", "B", "D" ] }
{ "_id" : "4", "student" : "John", "courses" : [ "A", "B", "C" ] }
{ "_id" : "5", "student" : "Bema", "courses" : [ "A", "B" ] }

我想对集合进行操作，这样它将按学生所修课程的集合（组合）返回一组学生（带他们的_id），并计算每个集合中有多少学生。

在上述示例中，我有3套（组合）课程和学生人数，如下所示：

---

1-

[“A”，“B”]

这是一个使用分组的聚合操作

db.students.aggregate([{
    $group: {
        // Uniquely identify the document.
        // The $ syntax queries on this field
        _id: "$courses",

        // Add 1 for each field found (effectively a counter)
        count: {$sum: 1}
    }
}]);

编辑：

如果课程可以按任何顺序排列，您可以按照编辑的问题中的建议再次执行

$unwind

、

$sort

和

$group

。也可以通过

mapReduce

来实现这一点，但我不确定哪一个更快

db.students.mapReduce(
    function () {
        // Use the sorted courses as the key
        emit(this.courses.sort(), this._id);
    },
    function (key, values) {
        return {"students": values, count: values.length};
    },
    {out: {inline: 1}}
)

---

谢谢是否可以将学生的

\u id

列表作为每个组的嵌套属性输出？您可以添加另一个组，如

学生：{$push:$\u id}

非常感谢！太好了！：）它似乎是根据元素在

课程中的排列方式进行分组的。例如，如果有一个记录
{“\u id”：“6”，“学生”：“Alan”，“课程”：[“C”，“B”，“a”]}
，则将其视为不同的组合。现在，我想我需要先进行排序，然后再进行分组…@oscar真棒谢谢分享你的发现。我也会看一看并更新我的答案。
{ "_id" : "1", "student" : "Oscar", "courses" : [ "A", "B" ] }
{ "_id" : "2", "student" : "Alan", "courses" : [ "A", "B", "C" ] }
{ "_id" : "3", "student" : "Kate", "courses" : [ "A", "B", "D" ] }
{ "_id" : "4", "student" : "John", "courses" : [ "A", "B", "C" ] }
{ "_id" : "5", "student" : "Bema", "courses" : [ "A", "B" ] }
{ "_id" : "6", "student" : "Alex", "courses" : [ "C", "A", "B" ] }

{ "_id" : [ "C", "A", "B" ], "count" : 1, "students" : [ "6" ] }
{ "_id" : [ "A", "B", "D" ], "count" : 1, "students" : [ "3" ] }
{ "_id" : [ "A", "B", "C" ], "count" : 2, "students" : [ "2", "4" ] }
{ "_id" : [ "A", "B" ], "count" : 2, "students" : [ "1", "5" ] }

db.students.aggregate([
    {$unwind: "$courses" },
    {$sort : {"courses": 1}}, 
    {$group: {_id: "$_id", courses: {$push: "$courses"}}}, 
    {$group: {_id: "$courses", count: {$sum:1}, students: {$push: "$_id"}}}
    ])

{ "_id" : [ "A", "B", "D" ], "count" : 1, "students" : [ "3" ] }
{ "_id" : [ "A", "B" ], "count" : 2, "students" : [ "5", "1" ] }
{ "_id" : [ "A", "B", "C" ], "count" : 3, "students" : [ "6", "4", "2" ] }

db.students.aggregate([{
    $group: {
        // Uniquely identify the document.
        // The $ syntax queries on this field
        _id: "$courses",

        // Add 1 for each field found (effectively a counter)
        count: {$sum: 1}
    }
}]);

db.students.mapReduce(
    function () {
        // Use the sorted courses as the key
        emit(this.courses.sort(), this._id);
    },
    function (key, values) {
        return {"students": values, count: values.length};
    },
    {out: {inline: 1}}
)