将GraphFrames ShortestPath地图转换为Pyspark中的DataFrame行

Question

我正在尝试找到从GraphFrames函数短路路径输出的最有效方法，并将每个顶点的距离映射到新的DataFrame中的单个行中。我能够通过将距离列拉到字典中，然后从那里转换为pandas dataframe，然后转换回火花数据框，但我知道必须有更好的方法。

from graphframes import *
v = sqlContext.createDataFrame([
  ("a", "Alice", 34),
  ("b", "Bob", 36),
  ("c", "Charlie", 30),
], ["id", "name", "age"])
# Create an Edge DataFrame with "src" and "dst" columns
e = sqlContext.createDataFrame([
  ("a", "b", "friend"),
  ("b", "c", "follow"),
  ("c", "b", "follow"),
], ["src", "dst", "relationship"])
# Create a GraphFrame
g = GraphFrame(v, e)
results = g.shortestPaths(landmarks=["a", "b","c"])
results.select("id","distances").show()
+---+--------------------+
| id|           distances|
+---+--------------------+
|  a|Map(a -> 0, b -> ...|
|  b| Map(b -> 0, c -> 1)|
|  c| Map(c -> 0, b -> 1)|
+---+--------------------+

我想要的是在将ID保存到这样的同时，将输出放置在上面并使距离弄平：

+---+---+---------+      
| id| v | distance|
+---+---+---------+
|  a| a | 0       |
|  a| b | 1       |
|  a| c | 2       |
|  b| b | 0       |
|  b| c | 1       |
|  c| c | 0       |
|  c| b | 1       |
+---+---+---------+ 

谢谢。

Answer 1

您可以爆炸：

>>> from pyspark.sql.functions import explode
>>> results.select("id", explode("distances"))