1

我的数据如下所示：

 {
       "_id":ObjectId("516fbf68067323ce2ea5b4b8"),
       "title":"GVPKFlFIXdLUaLM",
       "release_year":1913,
       "country_of_origin":"sWdXLXUfun",
       "length_in_minutes":147,
       "plot_summary":"bmwYkyyiSymHJYoXEPauPNjdKoFANDgcDImVelDGPuPJmLhyWOuNXjurNyGp",
       "director":"rNDFhhxGIo",
       "language":"oYeWskT",
       "popularity":5.2,
       "genre":"jDwdaMhuT",
       "actors":[
          {
             "id":2740,
             "name":"actor2740",
             "dob":1989,
             "alt_name":"PBpXPqJwmftpfcR",
             "pob":"DFoxETDuhAdDGNE"
          },
          {
             "id":3143,
             "name":"actor3143",
             "dob":1953,
             "alt_name":"AHnVvTviSKuvNZO",
             "pob":"KBUdvbnvNkXmddk"
          }
       ]
    }

起初我以为 Mongo 有一个错误。我尝试使用聚合函数来解决假设的业务问题。（编辑：我并不是说我解决了一个 mongo 问题或者我希望人们帮助我创建一个算法，只是为了确认 MongoDB 的潜在错误）

db.movies.aggregate([{$match:{popularity:{$gte:7.3}}},
     {$project:{actors:1,popularity:1}},
     {$unwind:"$actors"},
     {$group:{_id:"$actors.id",avgPop:{$avg:"$popularity"},
              docsByTag : { $sum : 1 }, popSum:{$sum:"$popularity"}}},
    {$match:{avgPop:{$gte:7.5}}}]);

我关注的结果（编辑 $sum：1 而不是 0）

{
            "_id" : 1383,
            "avgPop" : 8.772857142857141,
            "docsByTag" : 28,
            "popSum" : 245.63999999999996
        },

但是当我手动验证结果时。

db.movies.find({"actors.name":"actor1383"},{title:1,popularity:1,_id: 0})

{ "title" : "kZFfBwtAfVNobEq", "popularity" : 8.54 }
{ "title" : "kyOeSorYUWyJmjK", "popularity" : 8.11 }
{ "title" : "rvSdJCgEkkpYgFB", "popularity" : 8.36 }
{ "title" : "SwcgHTgZqqcYJja", "popularity" : 8.68 }
{ "title" : "XmcidmdwtDlNoKw", "popularity" : 7.33 }
{ "title" : "gwThvrWifoKCvyG", "popularity" : 7.94 }
{ "title" : "RdUsAFIxTnntTZR", "popularity" : 6.91 }
{ "title" : "RwhJlORFdvtDtpO", "popularity" : 5.13 }
{ "title" : "TuDfcWhNkQFeycl", "popularity" : 9.93 }
{ "title" : "xTVkwnyvftKQraC", "popularity" : 7.27 }
{ "title" : "HYMjUFlSXgnWVTx", "popularity" : 6.94 }
{ "title" : "ZPPyAUdGMeVQhbK", "popularity" : 8.48 }
{ "title" : "kEITAiMMrWTECGM", "popularity" : 9.42 }
{ "title" : "asNsLYKjvHlihXZ", "popularity" : 9.86 }
{ "title" : "ctEmciXPhbMtspt", "popularity" : 8.85 }
{ "title" : "DHjFtctccwDHtlf", "popularity" : 5.5 }
{ "title" : "ElUqbLqkoKrJPVl", "popularity" : 8.26 }
{ "title" : "XdTCieKsWtTbfZa", "popularity" : 5.72 }
{ "title" : "EeNqOPSuKiHuWRs", "popularity" : 5.91 }
{ "title" : "YgysqxcesvPryMY", "popularity" : 6.05 }
{ "title" : "eARvpGydsWilquc", "popularity" : 7.34 }
{ "title" : "NDpdkhSUfePDYjH", "popularity" : 7.28 }
{ "title" : "wUGKLBwijftQKgU", "popularity" : 8.97 }
{ "title" : "UHVGUmAcjBgAPBp", "popularity" : 7.44 }
{ "title" : "NKTKEKfbxFrudVi", "popularity" : 9.4 }
{ "title" : "AeByTKwsEQuQBYG", "popularity" : 8.97 }
{ "title" : "nZskARfGbhYRxdY", "popularity" : 9.16 }
{ "title" : "nBenZrikXFFrrnq", "popularity" : 7.58 }
{ "title" : "GdEFwoKgqjhHvjM", "popularity" : 6.3 }
{ "title" : "grpKTHgnYcDNyXH", "popularity" : 7.16 }
{ "title" : "hXhOqknvjIYJIaT", "popularity" : 5.24 }
{ "title" : "rggTJENnVeuqQVI", "popularity" : 9.95 }
{ "title" : "ABvGVFHkgOumMPO", "popularity" : 9.56 }
{ "title" : "baVkepHniIURUFH", "popularity" : 9.28 }
{ "title" : "PUYXlhPwbanMDmT", "popularity" : 9.6 }
{ "title" : "IJbqonvsVeorDMv", "popularity" : 7.82 }
{ "title" : "iAhyATKYpCVjtMw", "popularity" : 5.88 }
{ "title" : "uDECLFQGTOVnyvC", "popularity" : 6.25 }
{ "title" : "rTwfCYLfLwgPcbH", "popularity" : 8.38 }
{ "title" : "GRyKjecBHQhvYJk", "popularity" : 9.11 }
{ "title" : "GyEaSHoprUvGmZM", "popularity" : 9.92 }

给出大于或等于 7.3 的 27 个元素的子集

{ "title" : "kZFfBwtAfVNobEq", "popularity" : 8.54 }
{ "title" : "kyOeSorYUWyJmjK", "popularity" : 8.11 }
{ "title" : "rvSdJCgEkkpYgFB", "popularity" : 8.36 }
{ "title" : "SwcgHTgZqqcYJja", "popularity" : 8.68 }
{ "title" : "XmcidmdwtDlNoKw", "popularity" : 7.33 }
{ "title" : "gwThvrWifoKCvyG", "popularity" : 7.94 }
{ "title" : "TuDfcWhNkQFeycl", "popularity" : 9.93 }
{ "title" : "ZPPyAUdGMeVQhbK", "popularity" : 8.48 }
{ "title" : "kEITAiMMrWTECGM", "popularity" : 9.42 }
{ "title" : "asNsLYKjvHlihXZ", "popularity" : 9.86 }
{ "title" : "ctEmciXPhbMtspt", "popularity" : 8.85 }
{ "title" : "ElUqbLqkoKrJPVl", "popularity" : 8.26 }
{ "title" : "eARvpGydsWilquc", "popularity" : 7.34 }
{ "title" : "wUGKLBwijftQKgU", "popularity" : 8.97 }
{ "title" : "UHVGUmAcjBgAPBp", "popularity" : 7.44 }
{ "title" : "NKTKEKfbxFrudVi", "popularity" : 9.4 }
{ "title" : "AeByTKwsEQuQBYG", "popularity" : 8.97 }
{ "title" : "nZskARfGbhYRxdY", "popularity" : 9.16 }
{ "title" : "nBenZrikXFFrrnq", "popularity" : 7.58 }
{ "title" : "rggTJENnVeuqQVI", "popularity" : 9.95 }
{ "title" : "ABvGVFHkgOumMPO", "popularity" : 9.56 }
{ "title" : "baVkepHniIURUFH", "popularity" : 9.28 }
{ "title" : "PUYXlhPwbanMDmT", "popularity" : 9.6 }
{ "title" : "IJbqonvsVeorDMv", "popularity" : 7.82 }
{ "title" : "rTwfCYLfLwgPcbH", "popularity" : 8.38 }
{ "title" : "GRyKjecBHQhvYJk", "popularity" : 9.11 }
{ "title" : "GyEaSHoprUvGmZM", "popularity" : 9.92 }

比聚合函数少一。

所以我认为可能是聚合被破坏并将其重写为 mapReduce

// make sure we're using the right db; this is the same as "use aggdb;" in shell
db = db.getSiblingDB("recommendations"); //Put your MongoLab database name here.



var mapFunc2 = function() {
                       for (var idx = 0; idx < this.actors.length; idx++) {
                           var key = this.actors[idx].id;
                           var value = {
                                         count: 1,
                                         pop: this.popularity
                                       };
                           emit(key, value);
                       }
                    };

var reduceFunc2 = function(keyActor, countObjVals) {


                     reducedVal = { actor: keyActor, count: 0, pop: 0, pop_list : [] };

                     for (var idx = 0; idx < countObjVals.length; idx++) {
                         reducedVal.count += countObjVals[idx].count;
                         reducedVal.pop += countObjVals[idx].pop;
                         reducedVal.pop_list = reducedVal.pop_list.concat(countObjVals[idx].pop);
                     }

                     return reducedVal;
                  };

var finalizeFunc2 = function (key, reducedVal) {

                       reducedVal.avg = reducedVal.pop/reducedVal.count;

                       return reducedVal;

                    };


result = db.movies.mapReduce( mapFunc2,
                     reduceFunc2,
                     {
                       out: { merge: "mre" },
                       query: { popularity:
                                  { $gte: 7.3 }
                              },
                       finalize: finalizeFunc2
                     }
                   )
cursor = db.map_reduce_example.find()                  

while(cursor.hasNext()){
    printjson(cursor.next());

}

结果又差一

{
    "_id" : 1383,
    "value" : {
        "actor" : 1383,
        "count" : 28,
        "pop" : 245.63999999999996,
        "avg" : 8.772857142857141
    }
}

所以我开始调试，当谈到保存数组中每部电影的流行度时，我看到了一些奇怪的事情。

{“_id”：1，“值”：{“演员”：1，“计数”：13，“流行”：114.97，“pop_list”：[7.47、8.52、9.95、17.4、7.4、19.43、8.46、17.21 , 9.24, 9.89 ], "平均" : 8.843846153846155 } }

在这里，奇怪的是计数是 13，但元素的数量是 10。这是因为

7.4 7.4
7.47    7.47
8.07    1
8.14    2
8.46    8.46
8.52    8.52
9.14    1
9.24    9.24
9.26    2
9.57    3
9.86    3
9.89    9.89
9.95    9.95

其中 1,2,3 对应于

1   17.21=9.14+8.07
2   17.4=8.14+9.26
3   19.43=9.57+9.86

{“_id”：2，“值”：{“演员”：2，“计数”：14，“流行”：120.91999999999999，“pop_list”：[35.239999999999995、7.58、35.56、9.35、25.839999999999996、7] " : 8.637142857142857 } } 但是，以上内容完全是神秘的，因为我所有的平均值只有 2 位小数精度。

在这一点上真的很困惑。我相信这篇文章可能对那些被困在同一类型的计数问题上的人有所帮助。

4

1 回答 1

1

聚合框架和 mapreduce 都犯“错误”的可能性很小，所以我会要求您验证您如何将它们的结果与您的期望进行比较。

在您的聚合中，您按"actors.id"字段分组。但是您手动验证的查询是：

db.movies.find({"actors.name":"actor1383"},{title:1,popularity:1,_id: 0})

是否有证据表明您的“actors.name”和“actors.id”字段匹配 100%？

进行浮点运算时，高于 2 位的精度是正常的，无需担心。这与要求 5 和 10 的平均值并得到 7.5 没有什么不同，即使 5 和 10 都没有小数点后的数字。

“差异”可能来自另一个地方。如果你有这样的文件：

{人气：7.6，​​演员：[{id：1383，... ...}，{id：1383，... ...}}

您现在将只有一个对此有贡献的顶级文档，但是当您展开演员数组时，您现在有两个文档由此产生，其中两个文档都有 actor.id 1383。您能否验证每个演员在每个顶级仅出现一次文档？如果不是，那将导致您看到的差异。

于 2013-05-08T04:35:43.313 回答