Mongodb 同时聚合(计数)多个字段

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我有这样的文档:

{ "_id" : "someuniqueeventid", "event" : "event_type_1", "date" : ISODate("2014-01-14T00:00:00Z"), }

我想按event"分组并计算一周中每一天发生的每种事件类型的数量.基本上，我想得到类似的东西:

I want to group by "event" and count how many of each event type occured in each day of the week. Basically, I want to get something like:

{ "_id": "event_type_1", "1": "number of event_type_1 for Monday", "2": "number of event_type_1 for Tuesday", ... }, { "_id": "event_type_2", ... }

不幸的是，我被困在:

db.data.aggregate([ {$project: {date_of_week: {$dayOfWeek: "$date"}, event: "$event"}}, {$group: {_id: "$event", .... } ])

有什么想法吗?

推荐答案

聚合框架不会基于数据创建键，你也不应该这样做，因为数据"不是关键但实际上是数据，所以你应该坚持这个模式.

The aggregation framework won't create keys based on data, nor should you even be doing so as "data" is not a key but actually data, so you should stick to the pattern.

这意味着你基本上可以这样做:

That means you can basically just do this:

db.data.aggregate([ { "$group": { "_id": { "event_type": "$event", "day": { "$dayOfWeek": "$date" } }, "count": { "$sum": 1 } }} ])

这将计算每个事件每周发生的次数，尽管在输出中的多个文档中，但这很容易更改为每个事件的单个文档:

And that will count the occurrences per day of week per event, albeit in multiple documents in the output, but this is easy to change to a single document per event:

db.data.aggregate([ { "$group": { "_id": { "event_type": "$event", "day": { "$dayOfWeek": "$date" } }, "count": { "$sum": 1 } }}, { "$group": { "_id": "$_id.event_type", "days": { "$push": { "day": "$_id.day", "count": "$count" } } }} ])

这是一个数组形式，但它仍然保存着你想要的结果.

And that is in an array form, but it still holds the results you want.

如果你真的一心想要做你的确切形式，那么你想做这样的事情:

If you are really bent on doing your exact form then you want to do something like this:

db.data.aggregate([ { "$group": { "_id": "$event", "1": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 1 ] }, 1, 0 ] } }, "2": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 2 ] }, 1, 0 ] } }, "3": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 3 ] }, 1, 0 ] } }, "4": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 4 ] }, 1, 0 ] } }, "5": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 5 ] }, 1, 0 ] } }, "6": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 6 ] }, 1, 0 ] } }, "7": { "$sum": { "$cond": [ { "$eq": [{ "$dayOfWeek": "$date" }, 7 ] }, 1, 0 ] } } }} )

但这真的很长，所以恕我直言，我会坚持第一个或第二个解决方案，因为它们更短且更易于阅读.

But that is really long winded so IMHO I would stick with the first or maybe second solution as they are shorter and more easy to read.