大厦层次使用星火

Question

试想一下，我有这样的树：

- One 
    - One one 
    - One two 
    - One two one 
    - One two two 
    - One two three 
     - One two three one 
    - One three 
    - One three one 
    - One three two 
    - One three three 
    - One four 
    - One five 

数据明智它很简单也只是一个父子关系：

+-------------------+---------------+ 
|  Child  | Parent  | 
+-------------------+---------------+ 
| One    |    | 
| One one   | One   | 
| One two   | One   | 
| One two one  | One two  | 
| One two two  | One two  | 
| One two three  | One two  | 
| One two three one | One two three | 
| One three   | One   | 
| One three one  | One three  | 
| One three two  | One three  | 
| One three three | One three  | 
| One four   | One   | 
| One five   | One   | 
+-------------------+---------------+ 

现在想什么，我做的是：

• 我有两个项目的列表，让我们说One three three和One two three one
• 我想建造树父母的休息根级别

在RDBMS中，我会简单地写使用CTE和UNION ALL，但是我无法找到使用是否有可能在星火递归查询数据集或DataFrame，可能是由于缺乏Scala/Python知识。任何帮助，将不胜感激。

输出应该如下：

- One 
    - One two 
    - One two three 
     - One two three one 
    - One three 
    - One three three 

Answer 1

您可以使用基于Graphx的解决方案来执行递归查询（父/子或层次查询）。这是许多数据库称为递归公用表表达式（CTE）提供的功能或通过SQL第

连接请参见本文的详细信息：https://www.qubole.com/blog/processing-hierarchical-data-using-spark-graphx-pregel-api/