Mongoid Group By或MongoDb group by in rails [英] Mongoid Group By or MongoDb group by in rails

问题描述

我有一个mongo表，其中包含如下统计数据....

I have a mongo table that has statistical data like the following....

• course_id
• 状态which is a string, played or completed
使用 Mongoid的时间戳记功能

• course_id
• status which is a string, played or completed
• and timestamp information using Mongoid's Timestamping feature

所以我的课如下...

so my class is as follows...

class Statistic
  include Mongoid::Document
  include Mongoid::Timestamps
  include Mongoid::Paranoia

  field :course_id, type: Integer
  field :status, type: String # currently this is either play or complete

我想获得一门课程每天总播放次数的计数.例如 12年8月1日有2场比赛，12年8月2日有6场比赛.等等，因此，我将使用带有"course_id"和"action"的created_at timestamp字段.问题是我没有在Mongoid中看到按方法分组.我相信mongodb现在有一个了，但是我不确定在rails 3中如何做到这一点.

I want to get a daily count of total # of plays for a course. So for example... 8/1/12 had 2 plays, 8/2/12 had 6 plays. Etc. I would therefore be using the created_at timestamp field, with course_id and action. The issue is I don't see a group by method in Mongoid. I believe mongodb has one now, but I'm unsure of how that would be done in rails 3.

我可以使用每个表浏览整个表，并以递增的方式一起破解一些地图或哈希，但是如果该课程具有100万个视图，则检索并迭代超过100万条记录可能会很混乱.有一种干净的方法可以做到这一点吗?

I could run through the table using each, and hack together some map or hash in rails with incrementation, but what if the course has 1 million views, retrieving and iterating over a million records could be messy. Is there a clean way to do this?

推荐答案

如注释中所述，您可以为此目的使用map/reduce.因此，您可以在模型中定义以下方法( http://mongoid.org/en/mongoid/docs /querying.html#map_reduce )

As mentioned in comments you can use map/reduce for this purpose. So you could define the following method in your model ( http://mongoid.org/en/mongoid/docs/querying.html#map_reduce )

def self.today
  map = %Q{
    function() {
      emit(this.course_id, {count: 1})
    }
  }

  reduce = %Q{
    function(key, values) {
      var result = {count: 0};
      values.forEach(function(value) {
        result.count += value.count;
      });
      return result;
    }
  }

  self.where(:created_at.gt => Date.today, status: "played").
    map_reduce(map, reduce).out(inline: true)
end

这将导致以下结果:

[{"_id"=>1.0, "value"=>{"count"=>2.0}}, {"_id"=>2.0, "value"=>{"count"=>1.0}}] 

其中_id是course_id，count是播放次数.

MongoDB中也有专用的 group 方法，但我不确定如何使用Mongoid 3中裸露的mongodb集合.我还没有机会深入研究代码.

There is also dedicated group method in MongoDB but I am not sure how to get to the bare mongodb collection in Mongoid 3. I did not have a chance to dive into code that much yet.

您可能想知道为什么我发出一个文档{count: 1}，因为它无关紧要，我本可以发出空文档或任何东西，然后对每个值始终向result.count加1.问题是，如果仅对特定键执行一次发射操作(在我的示例中course_id仅被播放一次)，则不会调用reduce，因此最好以相同的格式发射文档.

You may wonder why I emit a document {count: 1} as it does not matter that much and I could have just emitted empty document or anything and then always add 1 to the result.count for every value. The thing is that reduce is not called if only one emit has been done for particular key (in my example course_id has been played only once) so it is better to emit documents in the same format as result.

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