Mongodb计算所有按条件匹配的对象中的所有数组元素 [英] Mongodb count all array elements in all objects matching by criteria

问题描述

我有一个这样的对象的活动日志集合:

I have a collection that is log of activity on objects like this:

{
    "_id" : ObjectId("55e3fd1d7cb5ac9a458b4567"),
    "object_id" : "1",
    "activity" : [ 
        {
            "action" : "test_action",
            "time" : ISODate("2015-08-31T00:00:00.000Z")
        },
        {
            "action" : "test_action",
            "time" : ISODate("2015-08-31T00:00:22.000Z")
        }
    ]
}

{
    "_id" : ObjectId("55e3fd127cb5ac77478b4567"),
    "object_id" : "2",
    "activity" : [ 
        {
            "action" : "test_action",
            "time" : ISODate("2015-08-31T00:00:00.000Z")
        }
    ]
}

{
    "_id" : ObjectId("55e3fd0f7cb5ac9f458b4567"),
    "object_id" : "1",
    "activity" : [ 
        {
            "action" : "test_action",
            "time" : ISODate("2015-08-30T00:00:00.000Z")
        }
    ]
}

如果我进行以下查询:

db.objects.find({
    "createddate": {$gte : ISODate("2015-08-30T00:00:00.000Z")},
    "activity.action" : "test_action"}
    }).count()

它返回包含test_action"的文档计数(本集中 3 个)，但我需要获取所有 test_actions 的计数(本集中 4 个).我该怎么做?

it returns count of documents containing "test_action" (3 in this set), but i need to get count of all test_actions (4 on this set). How do i do that?

推荐答案

最高效"的方法是跳过 $unwind 总而言之 $group 进行计数.本质上过滤"数组得到 $size 的结果到 $sum:

The most "performant" way to do this is to skip the $unwind altogther and simply $group to count. Essentially "filter" arrays get the $size of the results to $sum:

db.objects.aggregate([
    { "$match": {
        "createddate": {
            "$gte": ISODate("2015-08-30T00:00:00.000Z")
        },
        "activity.action": "test_action"
    }},
    { "$group": {
        "_id": null,
        "count": {
            "$sum": {
                "$size": {
                    "$setDifference": [
                        { "$map": {
                            "input": "$activity",
                            "as": "el",
                            "in": {
                                "$cond": [ 
                                    { "$eq": [ "$$el.action", "test_action" ] },
                                    "$$el",
                                    false
                                ]
                            }               
                        }},
                        [false]
                    ]
                }
            }
        }
    }}
])

MongoDB 的未来版本将具有 $filter，这使得这变得更加简单:

Future releases of MongoDB will have $filter, which makes this much more simple:

db.objects.aggregate([
    { "$match": {
        "createddate": {
            "$gte": ISODate("2015-08-30T00:00:00.000Z")
        },
        "activity.action": "test_action"
    }},
    { "$group": {
        "_id": null,
        "count": {
            "$sum": {
                "$size": {
                    "$filter": {
                        "input": "$activity",
                        "as": "el",
                        "cond": {
                            "$eq": [ "$$el.action", "test_action" ]
                        }
                    }
                }
            }
        }
    }}
])

使用 $unwind 会导致文档去规范化并有效地为每个数组条目创建一个副本.在可能的情况下，您应该避免这种情况，因为这通常会带来极高的成本.相比之下，过滤和计算每个文档的数组条目要快得多.与许多阶段相比，简单的 $match 和 $group 管道也是如此.

Using $unwind causes the documents to de-normalize and effectively creates a copy per array entry. Where possible you should avoid this due the the often extreme cost. Filtering and counting array entries per document is much faster by comparison. As is a simple $match and $group pipeline compared to many stages.

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