在MongoDB中合并两个集合

我一直在尝试在MongoDB中使用MapReduce来完成我认为简单的过程。我不知道这是否是正确的方法，也不知道我是否应该使用MapReduce。我在谷歌上搜索了我想到的关键词，并试图找到我认为最成功的文档，但一无所获。也许我想得太多了？

我有两个集合：details和gpas

details是由一大堆（300多万）文档组成的。studentid元素可以重复两次，每个year一次，如下所示：

{ "_id" : ObjectId("4d49b7yah5b6d8372v640100"), "classes" : [1,17,19,21], "studentid" : "12345a", "year" : 1}
{ "_id" : ObjectId("4d76b7oij7s2d8372v640100"), "classes" : [2,12,19,22], "studentid" : "98765a", "year" : 1}
{ "_id" : ObjectId("4d49b7oij7s2d8372v640100"), "classes" : [32,91,101,217], "studentid" : "12345a", "year" : 2}
{ "_id" : ObjectId("4d76b7rty7s2d8372v640100"), "classes" : [1,11,18,22], "studentid" : "24680a", "year" : 1}
{ "_id" : ObjectId("4d49b7oij7s2d8856v640100"), "classes" : [32,99,110,215], "studentid" : "98765a", "year" : 2}
...

gpas具有与details具有相同studentid的元素。每个studentid只有一个条目，如下所示：

{ "_id" : ObjectId("4d49b7yah5b6d8372v640111"), "studentid" : "12345a", "overall" : 97, "subscore": 1}
{ "_id" : ObjectId("4f76b7oij7s2d8372v640213"), "studentid" : "98765a", "overall" : 85, "subscore": 5}
{ "_id" : ObjectId("4j49b7oij7s2d8372v640871"), "studentid" : "24680a", "overall" : 76, "subscore": 2}
...

最后，我想有一个集合，每个学生一行，格式如下：

{ "_id" : ObjectId("4d49b7yah5b6d8372v640111"), "studentid" : "12345a", "classes_1": [1,17,19,21], "classes_2": [32,91,101,217], "overall" : 97, "subscore": 1}
{ "_id" : ObjectId("4f76b7oij7s2d8372v640213"), "studentid" : "98765a", "classes_1": [2,12,19,22], "classes_2": [32,99,110,215], "overall" : 85, "subscore": 5}
{ "_id" : ObjectId("4j49b7oij7s2d8372v640871"), "studentid" : "24680a", "classes_1": [1,11,18,22], "classes_2": [], "overall" : 76, "subscore": 2}
...

我要做的是这样运行MapReduce：

var mapDetails = function() {
    emit(this.studentid, {studentid: this.studentid, classes: this.classes, year: this.year, overall: 0, subscore: 0});
};
var mapGpas = function() {
    emit(this.studentid, {studentid: this.studentid, classes: [], year: 0, overall: this.overall, subscore: this.subscore});
};
var reduce = function(key, values) {
    var outs = { studentid: "0", classes_1: [], classes_2: [], overall: 0, subscore: 0};
    values.forEach(function(value) {
        if (value.year == 0) {
            outs.overall = value.overall;
            outs.subscore = value.subscore;
        }
        else {
            if (value.year == 1) {
                outs.classes_1 = value.classes;
            }
            if (value.year == 2) {
                outs.classes_2 = value.classes;
            }
            outs.studentid = value.studentid;
        }
    });
    return outs;
};
res = db.details.mapReduce(mapDetails, reduce, {out: {reduce: 'joined'}})
res = db.gpas.mapReduce(mapGpas, reduce, {out: {reduce: 'joined'}})

但当我运行它时，这是我得到的集合：

{ "_id" : "12345a", "value" : { "studentid" : "12345a", "classes_1" : [ ], "classes_2" : [ ], "overall" : 97, "subscore" : 1 } }
{ "_id" : "98765a", "value" : { "studentid" : "98765a", "classes_1" : [ ], "classes_2" : [ ], "overall" : 85, "subscore" : 5 } }
{ "_id" : "24680a", "value" : { "studentid" : "24680a", "classes_1" : [ ], "classes_2" : [ ], "overall" : 76, "subscore" : 2 } }

我缺少类数组。

此外，顺便说一句，我如何访问生成的MapReduce value元素中的元素？MapReduce是否总是输出到value或其他名称？

这与MongoDB用户Google Groups上的一个问题类似
https://groups.google.com/group/mongodb-user/browse_thread/thread/60a8b683e2626ada?pli=1

答案参考了一个在线教程，该教程与您的示例类似：http://tebros.com/2011/07/using-mongodb-mapreduce-to-join-2-collections/

有关MongoDB中MapReduce的更多信息，请参阅文档：http://www.mongodb.org/display/DOCS/MapReduce

此外，在MongoDB Cookbook题为"使用版本化文档查找最大值和最小值"的文章的"Extras"部分，有一个关于MapReduce操作如何工作的有用的分步演练：http://cookbook.mongodb.org/patterns/finding_max_and_min/

如果你已经阅读了一些参考文件，请原谅我。我把它们包括在内是为了其他用户的利益，他们可能正在阅读这篇文章，并且是在MongoDB 中使用MapReduce的新手

Map函数中"emit"语句的输出与Reduce函数的输出匹配是很重要的。如果Map函数只输出一个文档，则Reduce函数可能根本无法运行，然后您的输出集合将具有不匹配的文档。

我稍微修改了您的map语句，以便使用两个独立的"类"数组以所需输出的格式发送文档
我还重新编写了reduce语句，将新类添加到classes_1和classes\u2数组中，前提是它们还不存在。

var mapDetails = function(){
    var output = {studentid: this.studentid, classes_1: [], classes_2: [], year: this.year, overall: 0, subscore: 0}
    if (this.year == 1) {
        output.classes_1 = this.classes;
    }
    if (this.year == 2) {
        output.classes_2 = this.classes;
    }
    emit(this.studentid, output);
};
var mapGpas = function() {
    emit(this.studentid, {studentid: this.studentid, classes_1: [], classes_2: [], year: 0, overall: this.overall, subscore: this.subscore});
};
var r = function(key, values) {
    var outs = { studentid: "0", classes_1: [], classes_2: [], overall: 0, subscore: 0};
    values.forEach(function(v){
        outs.studentid = v.studentid;
        v.classes_1.forEach(function(class){if(outs.classes_1.indexOf(class)==-1){outs.classes_1.push(class)}})
        v.classes_2.forEach(function(class){if(outs.classes_2.indexOf(class)==-1){outs.classes_2.push(class)}})
        if (v.year == 0) {
            outs.overall = v.overall;
            outs.subscore = v.subscore;
        }
    });
    return outs;
};
res = db.details.mapReduce(mapDetails, r, {out: {reduce: 'joined'}})
res = db.gpas.mapReduce(mapGpas, r, {out: {reduce: 'joined'}})

运行两个MapReduce操作会得到以下集合，该集合与您想要的格式相匹配：

> db.joined.find()
{ "_id" : "12345a", "value" : { "studentid" : "12345a", "classes_1" : [ 1, 17, 19, 21 ], "classes_2" : [ 32, 91, 101, 217 ], "overall" : 97, "subscore" : 1 } }
{ "_id" : "24680a", "value" : { "studentid" : "24680a", "classes_1" : [ 1, 11, 18, 22 ], "classes_2" : [ ], "overall" : 76, "subscore" : 2 } }
{ "_id" : "98765a", "value" : { "studentid" : "98765a", "classes_1" : [ 2, 12, 19, 22 ], "classes_2" : [ 32, 99, 110, 215 ], "overall" : 85, "subscore" : 5 } }
>

MapReduce始终以｛_id:"id"，value:"value"｝的形式输出文档在标题为"点符号（触及对象）"的文档中，有更多关于使用子文档的信息：http://www.mongodb.org/display/DOCS/Dot+符号+%28到达+进入+个对象%29

如果您希望MapReduce的输出以不同的格式显示，则必须在应用程序中以编程方式执行此操作。

希望这将提高您对MapReduce的理解，并使您离生成所需的输出集合又近了一步。祝你好运

您不能对此使用m/r，因为它被设计为仅应用于一个集合。从多个集合中读取会破坏分片兼容性，因此是不允许的。您可以使用新的聚合框架（2.1+）或在应用程序中执行您想要的操作。

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