聚合组多个字段

Question

给定以下数据集：

{ "_id" : 1, "city" : "Yuma", "cat": "roads", "Q1" : 0, "Q2" : 25, "Q3" : 0, "Q4" : 0 }
{ "_id" : 2, "city" : "Reno", "cat": "roads", "Q1" : 30, "Q2" : 0, "Q3" : 0, "Q4" : 60 }
{ "_id" : 3, "city" : "Yuma", "cat": "parks", "Q1" : 0, "Q2" : 0, "Q3" : 45, "Q4" : 0 }
{ "_id" : 4, "city" : "Reno", "cat": "parks", "Q1" : 35, "Q2" : 0, "Q3" : 0, "Q4" : 0 }
{ "_id" : 5, "city" : "Yuma", "cat": "roads", "Q1" : 0, "Q2" : 15, "Q3" : 0, "Q4" : 20 }

我正在努力实现以下结果。如果 return 总计大于零，并且还将每个城市、cat 和 Qx 总计压缩到一条记录中，那就太好了。

{
    "city" : "Yuma",
    "cat" : "roads",
    "Q2total" : 40
}, 
{
    "city" : "Reno",
    "cat" : "roads",
    "Q1total" : 30
},
{
    "city" : "Reno",
    "cat" : "roads",
    "Q4total" : 60
},
{
    "city" : "Yuma",
    "cat" : "parks",
    "Q3total" : 45
},
{
    "city" : "Reno",
    "cat" : "parks",
    "Q1total" : 35
},
{
    "city" : "Yuma",
    "cat" : "roads",
    "Q4total" : 20
}

可能吗？

Answer 1

我们可以问，目的是什么？您的文档已经有一个很好的一致的对象结构，这是推荐的。让对象具有不同的键不是一个好主意。数据是 "data"，实际上不应该是键的名称。

考虑到这一点，聚合框架实际上遵循了这个意义，并且不允许从文档中包含的数据生成任意键名。但是您可以将输出作为数据点得到类似的结果：

db.junk.aggregate([
    // Aggregate first to reduce the pipeline documents somewhat
    { "$group": {
        "_id": {
            "city": "$city",
            "cat": "$cat"
        },
        "Q1": { "$sum": "$Q1" },
        "Q2": { "$sum": "$Q2" },
        "Q3": { "$sum": "$Q3" },
        "Q4": { "$sum": "$Q4" }
    }},

    // Convert the "quarter" elements to array entries with the same keys
    { "$project": {
        "totals": {
            "$map": {
                "input": { "$literal": [ "Q1", "Q2", "Q3", "Q4" ] },
                "as": "el",
                "in": { "$cond": [
                    { "$eq": [ "$$el", "Q1" ] },
                    { "quarter": "$$el", "total": "$Q1" },
                    { "$cond": [
                        { "$eq": [ "$$el", "Q2" ] },
                        { "quarter": "$$el", "total": "$Q2" },
                        { "$cond": [
                           { "$eq": [ "$$el", "Q3" ] },
                           { "quarter": "$$el", "total": "$Q3" },
                           { "quarter": "$$el", "total": "$Q4" }
                        ]}
                    ]}
                ]}
            }
        }
    }},

    // Unwind the array produced
    { "$unwind": "$totals" },

    // Filter any "0" resutls
    { "$match": { "totals.total": { "$ne": 0 } } },

    // Maybe project a prettier "flatter" output
    { "$project": {
        "_id": 0,
        "city": "$_id.city",
        "cat": "$_id.cat",
        "quarter": "$totals.quarter",
        "total": "$totals.total"
    }}
])

这给你这样的结果：

{ "city" : "Reno", "cat" : "parks", "quarter" : "Q1", "total" : 35 }
{ "city" : "Yuma", "cat" : "parks", "quarter" : "Q3", "total" : 45 }
{ "city" : "Reno", "cat" : "roads", "quarter" : "Q1", "total" : 30 }
{ "city" : "Reno", "cat" : "roads", "quarter" : "Q4", "total" : 60 }
{ "city" : "Yuma", "cat" : "roads", "quarter" : "Q2", "total" : 40 }
{ "city" : "Yuma", "cat" : "roads", "quarter" : "Q4", "total" : 20 }

您可以交替使用 mapReduce，它允许 "some" 键名称的灵活性。问题是你的聚合仍然是 "quarter"，所以你需要它作为主键的一部分，一旦发出就不能改变。

此外，在输出到集合后，您不能 "filter" 任何“0”的聚合结果，除非您可以接受"transform" 对输出集合的查询的第二个 mapReduce 操作。

值得注意的是，如果您使用 $project 和 $map 查看 "second" 管道阶段正在完成的工作，您会发现文档结构基本上被更改为有时您可以像原来那样交替构建您的文档，如下所示：

{
    "city" : "Reno", 
    "cat" : "parks"
    "totals" : [ 
        { "quarter" : "Q1", "total" : 35 }, 
        { "quarter" : "Q2", "total" : 0 }, 
        { "quarter" : "Q3", "total" : 0 }, 
        { "quarter" : "Q4", "total" : 0 }
    ]
},
{ 
    "city" : "Yuma", 
    "cat" : "parks"
    "totals" : [ 
        { "quarter" : "Q1", "total" : 0 }, 
        { "quarter" : "Q2", "total" : 0 }, 
        { "quarter" : "Q3", "total" : 45 }, 
        { "quarter" : "Q4", "total" : 0 } 
    ]
}

然后聚合操作就变得简单了，你的文档得到了如上所示的相同结果：

db.collection.aggregate([
    { "$unwind": "$totals" },
    { "$group": {
        "_id": {
            "city": "$city",
            "cat": "$cat",
            "quarter": "$totals.quarter"
        },
        "ttotal": { "$sum": "$totals.total" }
    }},
    { "$match": { "ttotal": { "$ne": 0 } },
    { "$project": {
        "_id": 0,
        "city": "$_id.city",
        "cat": "$_id.cat",
        "quarter": "$_id.quarter",
        "total": "$ttotal"
    }}
])

因此，考虑以这种方式构建文档并避免文档转换所需的任何开销可能更有意义。

我认为您会发现一致的键名可以形成更好的编程对象模型，您应该从键值而不是键名读取数据点。如果您确实需要，那么只需从对象中读取数据并在 post 处理中转换每个已聚合结果的键即可。

聚合组多个字段

Aggregate group multiple fields

mongodb

aggregation-framework