如何获取 mongoDB 集合的排名位置？答案

【问题标题】：How to get ranking position of a mongoDB collection?如何获取 mongoDB 集合的排名位置？
【发布时间】：2017-08-08 02:38:33
【问题描述】：

我有一个如下所示的 mongoDB 集合：

{
    "_id": 1,
    "name": "John Doe",
    "company": "Acme",
    "email": "john.doe@acme.com",
    "matches": [171844, 169729, 173168, 174310, 168752, 174972, 172959, 169546]
}
{
    "_id": 2,
    "name": "Bruce Wayne",
    "company": "Wayne Enterprises",
    "email": "bruce@wayne.com",
    "matches": [171844, 232333, 233312, 123456]
}
{
    "_id": 3,
    "name": "Tony Stark",
    "company": "Stark Industries",
    "email": "tony@stark.com",
    "matches": [173844, 155729, 133168, 199310, 132752, 139972]
}
{
    "_id": 4,
    "name": "Clark Kent",
    "company": "Daily Planet",
    "email": "clark.kent@planet.com",
    "matches": [169729, 174310, 168752]
}
{
    "_id": 5,
    "name": "Lois Lane",
    "company": "Daily Planet",
    "email": "lois.lane@planet.com",
    "matches": [172959, 169546]
}

我需要获得一个过滤后的用户列表，但使用一个键来显示用户的“排名”位置，该位置基于它拥有的“匹配”记录的数量。应该有一个global ranking 位置和一个company ranking 位置。

期望的结果应该是这样的（例如过滤 company='Daily Planet'）：

{
    _id: 4,
    name: 'Clark Kent',
    company: 'Daily Planet',
    email: 'clark.kent@planet.com',
    points: 3,     // <=
    globalRank: 4, // <=
    companyRank: 1 // <=
},
{
    _id: 5,
    name: 'Lois Lane',
    company: 'Daily Planet',
    email: 'lois.lane@planet.com',
    points: 2,     // <=
    globalRank: 4, // <=
    companyRank: 2 // <=
}

请注意，Clark Kent 在全球排名中排名第 4，因为他有 3 场比赛（John Doe、Bruce Wayne 和 Tony Stark 的比赛次数比他多），并且在公司排名中排名第 1，因为他的比赛次数比任何人都多每日星球用户。

但是，即使经过几天的研究，我也找不到方法。（我什至不知道如何做全球排名或公司排名）。

关于如何解决此问题或如何以不同方式解决问题的任何想法？

【问题讨论】：

标签： mongodb ranking

【解决方案1】：

基本思想是首先按照points 对点进行排序，然后将$push 放入一个数组中。这确保元素按排序顺序插入。然后我们$unwind 使用includeArrayIndex 属性生成排序数组中与排名相对应的元素的索引。

使用上述逻辑的流水线如下（尝试逐步进行以更好地理解）：-

aggregate([
    {
        $project: {
            _id: 1,
            name: "$name",
            company: "$company",
            email: "$email",
            points: {
                $size: "$matches"
            }
        }
    }, {
        $sort: {
            points: -1
        }
    },

    {
        $group: {
            _id: {},
            arr: {
                $push: {
                    name: '$name',
                    company: '$company',
                    email: '$email',
                    points: '$points'
                }
            }
        }
    }, {
        $unwind: {
            path: '$arr',
            includeArrayIndex: 'globalRank',
        }
    }, {
        $sort: {
            'arr.company': 1,
            'arr.points': -1
        }
    }, {
        $group: {
            _id: '$arr.company',
            arr: {
                $push: {
                    name: '$arr.name',
                    company: '$arr.company',
                    email: '$arr.email',
                    points: '$arr.points',
                    globalRank: '$globalRank'
                }
            }
        }
    }, {
        $unwind: {
            path: '$arr',
            includeArrayIndex: 'companyRank',
        }
    }, {
        $project: {
            _id: 0,
            name: '$arr.name',
            company: '$arr.company',
            email: '$arr.email',
            points: '$arr.points',
            globalRank: '$arr.globalRank',
            companyRank: '$companyRank'
        }
    }

]);

查询的输出是

/* 1 */
{
    "companyRank" : NumberLong(0),
    "name" : "Bruce Wayne",
    "company" : "Wayne Enterprises",
    "email" : "bruce@wayne.com",
    "points" : 4,
    "globalRank" : NumberLong(2)
}

/* 2 */
{
    "companyRank" : NumberLong(0),
    "name" : "Tony Stark",
    "company" : "Stark Industries",
    "email" : "tony@stark.com",
    "points" : 6,
    "globalRank" : NumberLong(1)
}

/* 3 */
{
    "companyRank" : NumberLong(0),
    "name" : "Clark Kent",
    "company" : "Daily Planet",
    "email" : "clark.kent@planet.com",
    "points" : 3,
    "globalRank" : NumberLong(3)
}

/* 4 */
{
    "companyRank" : NumberLong(1),
    "name" : "Lois Lane",
    "company" : "Daily Planet",
    "email" : "lois.lane@planet.com",
    "points" : 2,
    "globalRank" : NumberLong(4)
}

/* 5 */
{
    "companyRank" : NumberLong(0),
    "name" : "John Doe",
    "company" : "Acme",
    "email" : "john.doe@acme.com",
    "points" : 8,
    "globalRank" : NumberLong(0)
}

此处的排名为 0。

【讨论】：

非常感谢！ “includeArrayIndex”是我要找的！！
@hyades，内存安全吗？如果有数千个文档怎么办？

【解决方案2】：

你期待这个结果吗？ . result_array 将保存最终结果。

var my_array = db.testCol.aggregate([{ $project: { _id:1, name:1, company:1, email:1, "points" : {$size: "$matches"}, "globalRank":{$literal: 0}, companyRank:{$literal: 0} } },
{$sort: {points : -1 } },
]).toArray()

var result_array = [];
var companyCount = {};
for (i = 0; i < my_array.length; i++) {
    var company_name = my_array[i].company
    if (companyCount[company_name] == null ){
        companyCount[company_name] = 1;
    }
    else{
        companyCount[company_name] = companyCount[company_name] + 1
    }
    result_array.push({ "_id" : my_array[i]._id, "name": my_array[i].name, "company" : my_array[i].company, "email" : my_array[i].email, "points" : my_array[i].points, "globalRank":i+1 , "companyRank" : companyCount[company_name]})
}

result_array

输出是：

[
        {
                "_id" : 1,
                "name" : "John Doe",
                "company" : "Acme",
                "email" : "john.doe@acme.com",
                "points" : 8,
                "globalRank" : 1,
                "companyRank" : 1
        },
        {
                "_id" : 3,
                "name" : "Tony Stark",
                "company" : "Stark Industries",
                "email" : "tony@stark.com",
                "points" : 6,
                "globalRank" : 2,
                "companyRank" : 1
        },
        {
                "_id" : 2,
                "name" : "Bruce Wayne",
                "company" : "Wayne Enterprises",
                "email" : "bruce@wayne.com",
                "points" : 4,
                "globalRank" : 3,
                "companyRank" : 1
        },
        {
                "_id" : 4,
                "name" : "Clark Kent",
                "company" : "Daily Planet",
                "email" : "clark.kent@planet.com",
                "points" : 3,
                "globalRank" : 4,
                "companyRank" : 1
        },
        {
                "_id" : 5,
                "name" : "Lois Lane",
                "company" : "Daily Planet",
                "email" : "lois.lane@planet.com",
                "points" : 2,
                "globalRank" : 5,
                "companyRank" : 2
        }
]

【讨论】：

【解决方案3】：

从 Mongo 5 开始，这是新的 $setWindowFields 聚合运算符的完美用例：

// { name: "John Doe",    firm: "Acme",              matches: [171844, 169729, 173168, 174310, 168752, 174972, 172959, 169546] }
// { name: "Bruce Wayne", firm: "Wayne Enterprises", matches: [171844, 232333, 233312, 123456] }
// { name: "Tony Stark",  firm: "Stark Industries",  matches: [173844, 155729, 133168, 199310, 132752, 139972] }
// { name: "Clark Kent",  firm: "Daily Planet",      matches: [169729, 174310, 168752] }
// { name: "Lois Lane",   firm: "Daily Planet",      matches: [172959, 169546] }
db.collection.aggregate([

  { $set: { pts: { $size: "$matches" } } },

  { $setWindowFields: {
    sortBy: { pts: -1 },
    output: { globRnk: { $rank: {} } }
  }},

  { $setWindowFields: {
    partitionBy: "$firm",
    sortBy: { pts: -1 },
    output: { firmRnk: { $rank: {} } }
  }}
])
// { name: "John Doe",   firm: "Acme",             pts: 8, globRnk: 1, firmRnk: 1, matches: [171844, 169729, 173168, 174310, 168752, 174972, 172959, 169546] }
// { name: "Tony Stark", firm: "Stark Industries", pts: 6, globRnk: 2, firmRnk: 1, matches: [173844, 155729, 133168, 199310, 132752, 139972] }
// { name: "Bruce Wayne",firm: "Wayne Enterprises",pts: 4, globRnk: 3, firmRnk: 1, matches: [171844, 232333, 233312, 123456] }
// { name: "Clark Kent", firm: "Daily Planet",     pts: 3, globRnk: 4, firmRnk: 1, matches: [169729, 174310, 168752] }
// { name: "Lois Lane",  firm: "Daily Planet",     pts: 2, globRnk: 5, firmRnk: 2, matches: [172959, 169546 ] }

第一个$setWindowFields 阶段添加全球排名：

按points的降序对文档进行排序：sortBy: { points: -1 }
并在每个文档中添加globRnk 字段 (output: { globRnk: { $rank: {} } })
- 这是基于排序字段points：globRnk: { $rank: {} }的所有文档中文档的排名。

第二个$setWindowFields 阶段与第一个阶段非常相似，除了现在在firm 定义的每个分区内计算的排名：partitionBy: "$firm"。

【讨论】：

【解决方案4】：

你使用 $match 条件。所以，你也试试这个..

db.rank.aggregate([{
    $match: {
        "company": "Daily Planet"
    }
}, {
    $project: {
        _id: 1,
        name: "$name",
        company: "$company",
        email: "$email",
        points: {
            $size: "$matches"
        }
    }
}, {
    $sort: {
        points: -1
    }
}, {
    $group: {
        _id: {},
        list: {
            $push: {
                name: '$name',
                company: '$company',
                email: '$email',
                points: '$points'
            }
        }
    }
}, {
    $unwind: {
        path: '$list',
        includeArrayIndex: 'globalRank',
    }
}, {
    $sort: {
        'list.company': 1,
        'list.points': -1
    }
}, {
    $group: {
        _id: '$list.company',
        list: {
            $push: {
                name: '$list.name',
                company: '$list.company',
                email: '$list.email',
                points: '$list.points',
                globalRank: '$globalRank'
            }
        }
    }
}, {
    $unwind: {
        path: '$list',
        includeArrayIndex: 'companyRank',
    }
}, {
    $project: {
        _id: 0,
        name: '$list.name',
        company: '$list.company',
        email: '$list.email',
        points: '$list.points',
        globalRank: '$list.globalRank',
        companyRank: '$companyRank'
    }
}]).pretty()

这样输出，

{
    "companyRank" : NumberLong(0),
    "name" : "Clark Kent",
    "company" : "Daily Planet",
    "email" : "clark.kent@planet.com",
    "points" : 3,
    "globalRank" : NumberLong(0)
}
{
    "companyRank" : NumberLong(1),
    "name" : "Lois Lane",
    "company" : "Daily Planet",
    "email" : "lois.lane@planet.com",
    "points" : 2,
    "globalRank" : NumberLong(1)
}

【讨论】：