couchdb地图/缩小视图:仅计算最新商品 [英] couchdb map/reduce view: counting only the most recent items

问题描述

我有以下文件.关键字的时间戳位置.

I have the following documents. Time stamped positions of keywords.

{
  _id: willem-aap-1234,
  keyword:aap,
  position: 10,
  profile: { name: willem },
  created_at: 1234
},
{
  _id: willem-aap-2345,
  keyword:aap,
  profile: { name: willem },
  created_at: 2345
},
{
  _id: oliver-aap-1235,
  keyword:aap,
  profile: { name: oliver },
  created_at: 1235
},
{
  _id: oliver-aap-2346,
  keyword:aap,
  profile: { name: oliver },
  created_at: 2346
}

每个配置文件查找最新的关键字.名称可以通过以下方式完成:

Finding the most recent keywords per profile.name can be done by:

map: function(doc) {
if(doc.profile)
    emit(
        [doc.profile.name, doc.keyword, doc.created_at], 
        { keyword : doc.keyword, position : doc.position, created_at: doc.created_at }
    );
}

reduce: function(keys, values, rered) {
  var r = values[0];
  for (var i=1; i<values.length; i++)
    if (r.created_at < values[i].created_at)
      r = values[i];
  return r;
}

然后使用

reduce : true,
group_level : 2,
startkey : [aname],
endkey : [aname,{}]

这为我提供了名称为aname的配置文件的最新文档.

This gives me the most recent documents for the profile with name aname.

但是现在我想计算每个关键字的所有最新文档，并对位置求和.我只能尝试仅使用map/reduce来解决这个问题.

But now I want to count all most recent documents per keyword, and sum the positions. I cannot get my head around this trying to do it with map/reduce only.

我的用户案例是:

1. 根据个人资料，用户和关键字查找最新文档
2. 计算每个关键字的唯一profile.name的数量
3. 对每个关键字的最新文档的位置求和

使它起作用的唯一方法是使用以下列表函数:

The only way I can make it work is using the following list function:

function(head, req) {
  var row;
  var counts = {};
  while (row = getRow()) {
    var v = row.value;
    var k = v.keyword;

    if (v.position) {
      if (!counts[k])
        counts[k] = { 
          position : 0,
          count : 0
        }
      counts[k].position += v.position;
      counts[k].count++;
    }
  }

  return JSON.stringify(counts);
}

有人能想到一种更好的方法，仅使用map/reduce吗?

Can anyone think of a better way to do this, using map/reduce only?

谢谢

推荐答案

某些部分的含义仍然有些模糊(例如，什么是位置"?).

The meaning of some parts are still a bit cloudy (for example, what is a "position"?).

但是从纯粹的形式上看，您的列表似乎在keyword上创建了索引，而地图在[profile, keyword, timestamp]上创建了索引.

But from a pure formal point of view, it seems that your list creates an index on keyword while your map created an index on [profile, keyword, timestamp].

如果您确实需要不同的索引，则需要几个映射，每个索引一个.唯一的例外是，当您已经在[a,b,c]上具有地图时，可以更改组级别"并获取其他两个索引:[a,b]和[a].

If you really need different indexes then you need several maps, one per index. The only exception is when you already have a map on [a,b,c], you can change the "group level" and get two other indexes: [a,b] and [a].

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