与Riak链接 - 搜索“共同朋友”

问题描述

任何一个与Riak一起工作的家庭成员都能解释如何在两个人之间寻找共同的朋友（例如，就像Facebook的共同朋友一样）？此外，计算要求理想情况下应该是最小的 - 即。服务器不需要经过很长的时间才能做计算。

我正在考虑将所有'人员'放在人员桶中，并且每个人员都会有一个朋友链接（这就像一对一 - 关系），那么对于每个人来说，都是反向关系 - 与另一个人链接（这就像一对一的关系）。我认为可以有效地做到这一点，计算/结果将立即可用，但只是无法弄清楚。

（好吧，我还没有真正开始使用Riak进行实际编码，所以对Riak的链接步行功能的更全面的解释将不胜感激。）

谢谢！

解决方案

我看到两种可能的解决方案：
$ b

1. MapReduce大致适用于这种情况：
   
   
   
   • 两个输入，A和B被比较
   
   
   • 链接阶段获取他们的朋友ID
   
   
   • 将它们映射到元组{{FriendID，1}）
   
   
   • 减少1s的总数
     
   • = 2
   
   
   • %%如果需要，可以将最后几步合并为一个 每个人都可以通过链接进行查询并与客户端的返回集相交 - 有点难看，但有效。
   
   

Can anyone familier with Riak explain how to do a link-walk to find common friends among two people (eg. just like Facebook's common friends)? Also, the computation requirement should ideally be minimal - ie. the server does not have to go through great length do do calculations.

I'm thinking of putting all 'persons' in a "people" bucket, and each 'person' will have a friends link (this is like a one-to-many relationship), then for each 'person' is reversed relationship-linked back to the other person (this is like a one-to-one relationship). I think it is possible to do this efficiently and calculation/result will be instantly available, but just can't figure it out.

(Alright, I have not exactly started real-coding with Riak, so a more thorough explanation on Riak's link walking feature would be appreciated.)

Thanks!

解决方案

I see two possible solutions:

1. MapReduce job with approximately this scenario:
   • Two inputs with people A and B being compared
   • Link phase fetching their friends IDs
   • Map them to tuples {FriendID, 1}
   • Reduce sum the 1s
   • Reduce filter where sum = 2
   • %% Last few steps can be combined into one if necessary
2. Just issue a link-walking query for each person and intersect returned sets on client - a bit ugly, but works.

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