热释光；dr：如何简化一个图，删除具有相同属性的边节点 name 价值观？
我有一个定义如下的图表：

import graphframes
from pyspark.sql import SparkSession

spark = SparkSession.builder.getOrCreate()
vertices = spark.createDataFrame([
    ('1', 'foo', '1'),
    ('2', 'bar', '2'),
    ('3', 'bar', '3'),
    ('4', 'bar', '5'),
    ('5', 'baz', '9'),
    ('6', 'blah', '1'),
    ('7', 'blah', '2'),
    ('8', 'blah', '3')
], ['id', 'name', 'value'])

edges = spark.createDataFrame([
    ('1', '2'),
    ('1', '3'),
    ('1', '4'),
    ('1', '5'),
    ('5', '6'),
    ('5', '7'),
    ('5', '8')
], ['src', 'dst'])

f = graphframes.GraphFrame(vertices, edges)

它会生成一个如下所示的图形（其中数字表示顶点id）：

从顶点id开始等于 1 ，我想把图表简化一下。使节点具有相似的 name 值合并到单个节点中。生成的图形如下所示：

注意我们只有一个 foo （id 1），一个 bar （id 2），一个 baz （id 5）和一个 blah （id 6）。这个 value 顶点的位置是不相关的，只是为了证明每个顶点是唯一的。
我试图实现一个解决方案，但它是黑客，非常低效，我肯定有一个更好的方法（我也不认为它的工作）：

f = graphframes.GraphFrame(vertices, edges)

# Get the out degrees for our nodes. Nodes that do not appear in

# this dataframe have zero out degrees.

outs = f.outDegrees

# Merge this with our nodes.

vertices = f.vertices
vertices = f.vertices.join(outs, outs.id == vertices.id, 'left').select(vertices.id, 'name', 'value', 'outDegree')
vertices.show()

# Create a new graph with our out degree nodes.

f = graphframes.GraphFrame(vertices, edges)

# Find paths to all edge vertices from our vertex ID = 1

# Can we make this one operation instead of two??? What if we have more than two hops?

one_hop = f.find('(a)-[e]->(b)').filter('b.outDegree is null').filter('a.id == "1"')
one_hop.show()

two_hop = f.find('(a)-[e1]->(b); (b)-[e2]->(c)').filter('c.outDegree is null').filter('a.id == "1"')
two_hop.show()

# Super ugly, but union the vertices from the `one_hop` and `two_hop` above, and unique

# on the name.

vertices = one_hop.select('a.*').union(one_hop.select('b.*'))
vertices = vertices.union(two_hop.select('a.*').union(two_hop.select('b.*').union(two_hop.select('c.*'))))
vertices = vertices.dropDuplicates(['name'])
vertices.show()

# Do the same for the edges

edges = two_hop.select('e1.*').union(two_hop.select('e2.*')).union(one_hop.select('e.*')).distinct()

# We need to ensure that we have the respective nodes from our edges. We do this  by

# Ensuring the referenced vertex ID is in our `vertices` in both the `src` and the `dst`

# columns - This does NOT seem to work as I'd expect!

edges = edges.join(vertices, vertices.id == edges.src, "left").select("src", "dst")
edges = edges.join(vertices, vertices.id == edges.dst, "left").select("src", "dst")
edges.show()

有没有更简单的方法来删除节点（及其对应的边）以便边节点在其上唯一 name ?