我想在 $bucket 函数中单独处理 $group 的结果。

我的数据:

{ "_id" : 1, "title" : "The Pillars of Society", "artist" : "Grosz", "year" : 1926, "type": 1}
{ "_id" : 2, "title" : "Melancholy III", "artist" : "Munch", "year" : 1902, "type": 1}
{ "_id" : 3, "title" : "Dancer", "artist" : "Miro", "year" : 1925, "type": 1}
{ "_id" : 4, "title" : "The Great Wave off Kanagawa", "artist" : "Hokusai", "type": 1}
{ "_id" : 5, "title" : "The Persistence of Memory", "artist" : "Dali", "year" : 1931, "type": 3}
{ "_id" : 6, "title" : "Composition VII", "artist" : "Kandinsky", "year" : 1913, "type": 3}
{ "_id" : 7, "title" : "The Scream", "artist" : "Munch", "year" : 1893, "type": 2}
{ "_id" : 8, "title" : "Blue Flower", "artist" : "O'Keefe", "year" : 1918, "type": 2}

我需要做的:

1. 按类型分组 -> 返回 3 组
2. 按年份范围分类每个组在 1 年内返回
3. 在 2 中获取每个桶的统计数据，在 1 中获取每个组

我想在一个查询中处理所有问题，而不是多次运行“匹配”以按类型过滤。

预期输出:

{
    "1": {
        1900: [{ "_id" : 2, "title" : "Melancholy III", "artist" : "Munch", "year" : 1902, "type": 1}],
        1925: [{ "_id" : 1, "title" : "The Pillars of Society", "artist" : "Grosz", "year" : 1926, "type": 1}, { "_id" : 3, "title" : "Dancer", "artist" : "Miro", "year" : 1925, "type": 1}],
        "other": [{ "_id" : 4, "title" : "The Great Wave off Kanagawa", "artist" : "Hokusai", "type": 1}]
    }
    "2": {
        1900: [{ "_id" : 8, "title" : "Blue Flower", "artist" : "O'Keefe", "year" : 1918, "type": 2}],
        1925: [],
        "other": [{ "_id" : 7, "title" : "The Scream", "artist" : "Munch", "year" : 1893, "type": 2}]
    },
    "3": {
        1900: [{ "_id" : 6, "title" : "Composition VII", "artist" : "Kandinsky", "year" : 1913, "type": 3}],
        1925: [{ "_id" : 5, "title" : "The Persistence of Memory", "artist" : "Dali", "year" : 1931, "type": 3}],
        "other": []
    }
}

我知道如何执行一个桶、一个组和一个匹配，但我不知道如何将它们组合在同一个聚合管道中以获得预期的结果而不重复匹配查询(什么是可能的，例如，使用$方面)。

最佳答案

可以使用此管道实现类似的输出:

db.artwork.aggregate([
    { $bucket: {
        groupBy: "$year",
        boundaries: [1900, 1925, 1950],
        default: "Other",
        output: { titles: { $push: "$$ROOT" }}
    }},
    { $unwind: "$titles" },
    { $group: {
        _id: { type: "$titles.type", year: "$_id" },
        titles: { $push: "$titles" }
    }},
    { $project: {
        _id: 0,
        type: "$_id.type",
        year: "$_id.year",
        titles: 1
    }},
    { $group: {
        _id: { type: "$type" },
        buckets: { $push: "$$ROOT" }
    }},
    { $project: {
        _id: 0,
        type: "$_id.type",
        buckets: { year: 1, titles: 1 }
    }}
])

给出:

{
    "type" : 1,
    "buckets" : [
        {
            "year" : 1900,
            "titles" : [ { "_id" : 2, "title" : "Melancholy III", "artist" : "Munch", "year" : 1902, "type" : 1 } ]
        },
        {
            "year" : 1925,
            "titles" : [
                { "_id" : 1, "title" : "The Pillars of Society", "artist" : "Grosz", "year" : 1926, "type" : 1 },
                { "_id" : 3, "title" : "Dancer", "artist" : "Miro", "year" : 1925, "type" : 1}
            ]
        },
        {
            "year" : "Other",
            "titles" : [ { "_id" : 4, "title" : "The Great Wave off Kanagawa", "artist" : "Hokusai", "type" : 1 } ]
        }
    ]
},

{
    "type" : 2,
    "buckets" : [
        {
            "year" : 1900,
            "titles" : [ { "_id" : 8, "title" : "Blue Flower", "artist" : "O'Keefe", "year" : 1918, "type" : 2 } ]
        },
        {
            "year" : "Other",
            "titles" : [ { "_id" : 7, "title" : "The Scream", "artist" : "Munch", "year" : 1893, "type" : 2 } ]
        }
    ]
},

{
    "type" : 3,
    "buckets" : [
        {
            "year" : 1900,
            "titles" : [ { "_id" : 6, "title" : "Composition VII", "artist" : "Kandinsky", "year" : 1913, "type" : 3 } ]
        },
        {
            "year" : 1925,
            "titles" : [ { "_id" : 5, "title" : "The Persistence of Memory", "artist" : "Dali", "year" : 1931, "type" : 3 } ]
        }
    ]
}

项目阶段主要是格式化数据，但这里重要的是首先获取所有将用于对项目进行分组的数据(每次失误的分类是输出中唯一我们不需要的还没有)，用 $unwind 分隔每个项目，并使用 $group 阶段嵌套数组。

您想要的输出可以使用 $cond 运算符实现，但它不是动态的，因为您必须显式地编写字段。有关此的更多信息，请参阅此帖子 MongoDB aggregate group array to key : sum value

关于mongodb - 如何 $bucket 来自以前的 $group 的多个集合，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42543427/