我的数据如下:
{
"_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解决假设的业务问题。 (编辑:我不是说我解决了mongo问题,或者我希望别人帮我创建一个算法,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:{$gte:7.5}}}]);
结果我专注于(编辑$ sum:1而不是0)
{
"_id" : 1383,
"avgPop" : 8.772857142857141,
"docsByTag" : 28,
"popSum" : 245.63999999999996
},
但是当我手工验证结果时。
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27个元素的子集大于或等于7.3
{ "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 }
比集合函数少一个。
所以我认为聚合被破坏并重写为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());
}
结果再次被一个
关闭{
"_id" : 1383,
"value" : {
"actor" : 1383,
"count" : 28,
"pop" : 245.63999999999996,
"avg" : 8.772857142857141
}
}
所以我开始调试,在保存数组中每部电影的流行度时,我发现了一些奇怪的事情。
{“_ id”:1,“value”:{“actor”:1,“count”:13,“pop”:114.97,“pop_list”:[7.47,8.52,9.9,17.4,7.4,19.43, 8.46,17.21,9.24,9.89],“avg”:8.843846153846155}}
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
1 17.21=9.14+8.07
2 17.4=8.14+9.26
3 19.43=9.57+9.86
{“_ id”:2,“value”:{“actor”:2,“count”:14,“pop”:120.91999999999999,“pop_list”:[35.239999999999995,7.58,35.56,9.35,25.839999999999996,7.35] ,“avg”:8.637142857142857}} 但是,上面的内容完全是神秘的,因为我的平均值只有2位小数位。
此时真的很困惑。我确信这篇文章可能会对那些陷入困境的其他人有所帮助。
答案 0 :(得分:1)
聚合框架和mapreduce都不太可能造成错误&#34;所以我会请你验证你如何将他们的结果与你的期望进行比较。
在您的聚合中,您将对"actors.id"
字段进行分组。但您手动验证的查询是:
db.movies.find({"actors.name":"actor1383"},{title:1,popularity:1,_id: 0})
是否有证据证明你的&#34; actors.name&#34;和&#34; actors.id&#34;字段匹配100%?
当您进行浮点运算时,高于2位的精度是正常的,无需担心。它要求平均值为5和10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。
还有另一个地方的差异&#34;可能来自。如果你有这样的文件:
{受欢迎程度:7.6, 演员:[ {id:1383, ... ... }, {id:1383, ... ... } }
现在你只有一个顶级文档,但是当你展开actors数组时,你现在有两个文件,由此产生的两个都有actor.id 1383.你能验证每个actor只出现一次每个顶级文件?如果不是这样会导致你看到的差异。