GraphFrames：合并具有相似列值的边缘节点

Posted 2023-04-15

【中文标题】GraphFrames：合并具有相似列值的边缘节点

【英文标题】：GraphFrames: Merge edge nodes with similar column values

【发布时间】：2020-08-28 04:48:42

【问题描述】：

tl;dr：如何简化图，删除具有相同 name 值的边缘节点？

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)

生成如下图（其中数字表示顶点 ID）：

从等于1 的顶点ID 开始，我想简化图形。这样具有相似name 值的节点将合并为单个节点。结果图看起来有点像 像这样：

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)。顶点的value 无关紧要，只是为了表明每个顶点都是唯一的。

我试图实现一个解决方案，但它很老套，效率极低，我确信有更好的方法（我也不认为它有效）：

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()

是否有更简单的方法来删除节点（及其相应的边），以便边节点在其name 上是唯一的？

【问题讨论】：

从集合中选择一个 vx 的任何规则，即为什么是 2、3、4 -> 2； 6, 7, 8 -> 6 ?

没有规则，这是随机选择。重要的是节点name 是唯一的。 id 和 value 在很大程度上无关紧要。

【参考方案1】：

您为什么不简单地将name 列视为新的id？

import graphframes

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'])

#create a dataframe with only one column
new_vertices = vertices.select(vertices.name.alias('id')).distinct()

#replace the src ids with the name column
new_edges = edges.join(vertices, edges.src == vertices.id, 'left')
new_edges = new_edges.select(new_edges.dst, new_edges.name.alias('src'))

#replace the dst ids with the name column
new_edges = new_edges.join(vertices, new_edges.dst == vertices.id, 'left')
new_edges = new_edges.select(new_edges.src, new_edges.name.alias('dst'))

#drop duplicate edges
new_edges = new_edges.dropDuplicates(['src', 'dst'])

new_edges.show()
new_vertices.show()

f = graphframes.GraphFrame(new_vertices, new_edges)

输出：

+---+----+
|src| dst|
+---+----+
|foo| baz|
|foo| bar|
|baz|blah|
+---+----+

+----+
|  id|
+----+
|blah|
| bar|
| foo|
| baz|
+----+