MongoDB 聚合不同文档中两个数组中的项目计数？

这是我的 MongoDB 集合架构：

company: String
model: String
cons: [String] // array of tags that were marked as "cons"
pros: [String] // array of tags that were marked as "pros"

我需要聚合它，以便得到以下输出：

[{
  "_id": {
    "company": "Lenovo",
    "model": "T400"
  },
  "tags": {
    tag: "SomeTag"
    pros: 124 // number of times, "SomeTag" tag was found in "pros" array in `Lenovo T400`
    cons: 345 // number of times, "SomeTag" tag was found in "cons" array in `Lenovo T400`
  }
}...]

我尝试执行以下操作：

var aggParams = {};
aggParams.push({ $unwind: '$cons' });
aggParams.push({ $unwind: '$pros' });
aggParams.push({$group: {
  _id: {
    company: '$company',
    model: '$model',
    consTag: '$cons'
  },
  consTagCount: { $sum: 1 }
}});
aggParams.push({$group: {
  _id: {
    company: '$_id.company',
    model: '$_id.model',
    prosTag: '$pros'
  },
  prosTagCount: { $sum: 1 }
}});
aggParams.push({$group: {
  _id: {
    company:'$_id.company',
    model: '$_id.model'
  },
  tags: { $push: { tag: { $or: ['$_id.consTag', '$_id.prosTag'] }, cons: '$consTagCount', pros: '$prosTagCount'} }
}});

但我得到了以下结果：

{
  "_id": {
    "company": "Lenovo",
    "model": "T400"
  },
  "tags": [
    {
      "tag": false,
      "pros": 7
    }
  ]
}

执行此操作的正确方法是什么aggregation?

是的，考虑到有多个数组，这有点困难，如果您同时尝试两个数组，您最终会得到“笛卡尔条件”，其中一个数组乘以另一个数组的内容。

因此，只需在开头组合数组内容，这可能表明您应该如何首先存储数据：

Model.aggregate(
    [
        { "$project": {
            "company": 1,
            "model": 1,
            "data": {
                "$setUnion": [
                    { "$map": {
                        "input": "$pros",
                        "as": "pro",
                        "in": {
                            "type": { "$literal": "pro" },
                            "value": "$$pro"
                        }
                    }},
                    { "$map": {
                        "input": "$cons",
                        "as": "con",
                        "in": {
                            "type": { "$literal": "con" },
                            "value": "$$con"
                        }
                    }}
                ]
            }
        }},
        { "$unwind": "$data" }
        { "$group": {
            "_id": { 
                "company": "$company",
                "model": "$model",
                "tag": "$data.value"
            },
            "pros": { 
                "$sum": { 
                    "$cond": [
                        { "$eq": [ "$data.type", "pro" ] },
                        1,
                        0
                    ]
                }
            },
            "cons": { 
                "$sum": { 
                    "$cond": [
                        { "$eq": [ "$data.type", "con" ] },
                        1,
                        0
                    ]
                }
            }
        }
    ], 
    function(err,result) {

    }
)

索非亚第一$project http://docs.mongodb.org/master/reference/operator/aggregation/project/上演$map http://docs.mongodb.org/master/reference/operator/aggregation/map/运算符将“type”值添加到每个数组的每个项目。这并不是说它真的很重要，因为无论如何所有项目都应该处理“唯一”，$setUnion http://docs.mongodb.org/master/reference/operator/aggregation/setUnion/运算符将每个数组“连接”为单个数组。

如前所述，您可能应该首先以这种方式存储。

然后处理$unwind http://docs.mongodb.org/master/reference/operator/aggregation/unwind/其次是$group http://docs.mongodb.org/master/reference/operator/aggregation/group/，其中每个“优点”和“缺点”然后通过评估$cond http://docs.mongodb.org/master/reference/operator/aggregation/cond/为它匹配的“类型”，要么返回1 or 0比赛分别在哪里true/false to the $sum http://docs.mongodb.org/master/reference/operator/aggregation/sum/聚合累加器。

这为您提供了一个“逻辑匹配”，可以根据指定的分组键对聚合操作中的每个相应“类型”进行计数。