从 MongoDB 聚合中的数组中获取所有可能的组合

Question

如何对数组中的相同值进行聚合（$group）？不是一次全部，但很少或全部，如果有的话。我可以用一个词做到$group，但我还需要所有可能的变化...

集合示例：

{"keywords": ["gta", "distribution", "keys"]}
{"keywords": ["gta", "online", "moto", "races"]}
{"keywords": ["gta", "online", "samp"]}

结果示例：

“gta”- 3 场比赛
“在线”- 2 场比赛
“gta 在线”- 2 场比赛

Answer 1

您可以使用 $reduce to extract all combinations of pairs from an array. I've started from this post，我已经添加了当前项目，$unwind 初始数组并计算项目数：

db.test.aggregate([
    {
        $project: {
            pairs: {
                $reduce: {
                    input: { $range: [0, { $size: "$keywords" }] },
                    initialValue: [],
                    in: {
                        $concatArrays: [
                            "$$value",
                            [[{ $arrayElemAt: ["$keywords", "$$this"] }]],
                            {
                                $let: {
                                    vars: { i: "$$this" },
                                    in: {
                                        $map: {
                                            input: { $range: [{ $add: [1, "$$i"] }, { $size: "$keywords" }] },
                                            in: [{ $arrayElemAt: ["$keywords", "$$i"] }, { $arrayElemAt: ["$keywords", "$$this"] }]
                                        }
                                    }
                                }
                            }
                        ]
                    }
                }
            }
        }
    }, {
        $unwind: "$pairs"
    }, {
        $group: {
            _id: "$pairs",
            count: { $sum: 1 }
        }
    }
])

输出：

{ "_id" : [ "online", "samp" ], "count" : 1 }
{ "_id" : [ "gta", "samp" ], "count" : 1 }
{ "_id" : [ "online", "races" ], "count" : 1 }
{ "_id" : [ "moto", "races" ], "count" : 1 }
{ "_id" : [ "gta", "keys" ], "count" : 1 }
{ "_id" : [ "races" ], "count" : 1 }
{ "_id" : [ "gta", "distribution" ], "count" : 1 }
{ "_id" : [ "samp" ], "count" : 1 }
{ "_id" : [ "distribution", "keys" ], "count" : 1 }
{ "_id" : [ "gta" ], "count" : 3 }
{ "_id" : [ "online" ], "count" : 2 }
{ "_id" : [ "keys" ], "count" : 1 }
{ "_id" : [ "gta", "online" ], "count" : 2 }
{ "_id" : [ "moto" ], "count" : 1 }
{ "_id" : [ "online", "moto" ], "count" : 1 }
{ "_id" : [ "distribution" ], "count" : 1 }
{ "_id" : [ "gta", "moto" ], "count" : 1 }
{ "_id" : [ "gta", "races" ], "count" : 1 }

如果需要更多组合，可能需要更新上面的$reduce阶段

从 MongoDB 聚合中的数组中获取所有可能的组合

Get all possible combinations from array in MongoDB aggregation

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