Neptune - 如何使用比例权重 gremlin 获取到所有节点的距离
Neptune - How to get distance to all nodes with proportional weights gremlin
我很难在 gremlin 中找出以下场景的查询。这是有向图（可能是循环的）。
我想获得前N个有利节点，从节点“Jane”开始，其中favor定义为：
favor(Jane->Lisa) = edge(Jane,Lisa) / total weight from outwards edges of Lisa
favor(Jane->Thomas) = favor(Jane->Thomas) + favor(Jane->Lisa) * favor(Lisa->Thomas)
favor(Jane->Jerryd) = favor(Jane->Thomas) * favor(Thomas->Jerryd) + favor(Jane->Lisa) * favor(Lisa->Jerryd)
favor(Jane->Jerryd) = [favor(Jane->Thomas) + favor(Jane->Lisa) * favor(Lisa->Thomas)] * favor(Thomas->Jerryd) + favor(Jane->Lisa) * favor(Lisa->Jerryd)
and so .. on
这是同一张图表，我的意思是手工计算，
这很容易通过编程进行传输，但我不确定用 gremlin 甚至 sparql 查询它有多准确。
这是创建此示例图的查询：
g
.addV('person').as('1').property(single, 'name', 'jane')
.addV('person').as('2').property(single, 'name', 'thomas')
.addV('person').as('3').property(single, 'name', 'lisa')
.addV('person').as('4').property(single, 'name', 'wyd')
.addV('person').as('5').property(single, 'name', 'jerryd')
.addE('favor').from('1').to('2').property('weight', 10)
.addE('favor').from('1').to('3').property('weight', 20)
.addE('favor').from('3').to('2').property('weight', 90)
.addE('favor').from('2').to('4').property('weight', 50)
.addE('favor').from('2').to('5').property('weight', 90)
.addE('favor').from('3').to('5').property('weight', 100)
我要找的是：
[Lisa, computedFavor]
[Thomas, computedFavor]
[Jerryd, computedFavor]
[Wyd, computedFavor]
我正在努力结合循环图来调整权重。到目前为止，这是我能够查询的地方：https://gremlify.com/f2r0zy03oxc/2
g.V().has('name','jane'). // our starting node
repeat(
union(
outE() // get only outwards edges
).
otherV().simplePath()). // produce simple path
emit().
times(10). // max depth of 10
path(). // attain path
by(valueMap())
处理来自 stephen mallette 的评论：
favor(Jane->Jerryd) =
favor(Jane->Thomas) * favor(Thomas->Jerryd)
+ favor(Jane->Lisa) * favor(Lisa->Jerryd)
// note we can expand on favor(Jane->Thomas) in above expression
//
// favor(Jane->Thomas) is favor(Jane->Thomas)@directEdge +
// favor(Jane->Lisa) * favor(Lisa->Thomas)
//
计算示例
Jane to Lisa => 20/(10+20) => 2/3
Lisa to Jerryd => 100/(100+90) => 10/19
Jane to Lisa to Jerryd => 2/3*(10/19)
Jane to Thomas (directly) => 10/(10+20) => 1/3
Jane to Lisa to Thomas => 2/3 * 90/(100+90) => 2/3 * 9/19
Jane to Thomas => 1/3 + (2/3 * 9/19)
Thomas to Jerryd => 90/(90+50) => 9/14
Jane to Thomas to Jerryd => [1/3 + (2/3 * 9/19)] * (9/14)
Jane to Jerryd:
= Jane to Lisa to Jerryd + Jane to Thomas to Jerryd
= 2/3 * (10/19) + [1/3 + (2/3 * 9/19)] * (9/14)
这里是一些伪代码：
def get_favors(graph, label="jane", starting_favor=1):
start = graph.findNode(label)
queue = [(start, starting_favor)]
favors = {}
seen = set()
while queue:
node, curr_favor = queue.popleft()
# get total weight (out edges) from this node
total_favor = 0
for (edgeW, outNode) in node.out_edges:
total_favor = total_favor + edgeW
for (edgeW, outNode) in node.out_edges:
# if there are no favors for this node
# take current favor and provide proportional favor
if outNode not in favors:
favors[outNode] = curr_favor * (edgeW / total_favor)
# it already has some favor, so we add to it
# we add proportional favor
else:
favors[outNode] += curr_favor * (edgeW / total_favor)
# if we have seen this edge, and node ignore
# otherwise, transverse
if (edgeW, outNode) not in seen:
seen.add((edgeW, outNode))
queue.append((outNode, favors[outNode]))
# sort favor by value and return top X
return favors
这是一个 Gremlin 查询，我相信它可以正确应用您的公式。我将首先粘贴完整的最终查询，然后简单介绍所涉及的步骤。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','jerryd')).
.....11> sack().
.....12> sum()
==>0.768170426065163
查询从 Jane 开始，一直遍历，直到检查到 Jerry D 的所有路径。沿途为每个遍历器维护一个sack，其中包含为每个关系计算的权重值相乘。第 6 行的计算找到所有可能来自先前顶点的边权重值，第 7 行的 math 步骤用于将当前边上的权重除以该总和。在最后，第 12 行将每个计算结果相加。如果删除最后的 sum 步骤，您可以看到中间结果。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','jerryd')).
.....11> sack()
==>0.2142857142857143
==>0.3508771929824561
==>0.2030075187969925
要查看经过的路线，可以将 path 步骤添加到查询中。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','jerryd')).
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight'),
.....16> sack()).fold())
==>[[jane,10,thomas,90,jerryd],0.2142857142857143]
==>[[jane,20,lisa,100,jerryd],0.3508771929824561]
==>[[jane,20,lisa,90,thomas,90,jerryd],0.2030075187969925]
这种方法还考虑了添加任何直接连接，根据您的公式，如果我们使用 Thomas 作为目标，我们可以看到。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','thomas')).
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight'),
.....16> sack()).fold())
==>[[jane,10,thomas],0.3333333333333333]
==>[[jane,20,lisa,90,thomas],0.3157894736842105]
不需要这些额外的步骤，但是包含 path 在调试这样的查询时很有用。此外，这不是必需的，但也许只是出于一般兴趣，我要补充一点，您也可以从此处获得最终答案，但我包含的第一个查询就是您真正需要的。
g.withSack(1).V().
has('name','jane').
repeat(outE().
sack(mult).
by(project('w','f').
by('weight').
by(outV().outE().values('weight').sum()).
math('w / f')).
inV().
simplePath()).
until(has('name','thomas')).
local(
union(
path().
by('name').
by('weight'),
sack()).fold().tail(local)).
sum()
==>0.6491228070175439
如果有任何不清楚的地方或者我有mis-understood公式，请告诉我。
编辑添加
要查找 Jane 可联系到的所有人的结果，我必须稍微修改查询。最后的 unfold 只是为了让结果更容易阅读。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> emit().
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight').unfold(),
.....16> sack()).fold()).
.....17> group().
.....18> by(tail(local,2).limit(local,1)).
.....19> by(tail(local).sum()).
.....20> unfold()
==>jerryd=0.768170426065163
==>wyd=0.23182957393483708
==>lisa=0.6666666666666666
==>thomas=0.6491228070175439
第 17 行的最后 group 步骤使用 path 结果计算找到的每个唯一名称的总支持度。要查看路径，您可以 运行 删除 group 步骤的查询。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> emit().
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight').unfold(),
.....16> sack()).fold())
==>[jane,10,thomas,0.3333333333333333]
==>[jane,20,lisa,0.6666666666666666]
==>[jane,10,thomas,50,wyd,0.11904761904761904]
==>[jane,10,thomas,90,jerryd,0.2142857142857143]
==>[jane,20,lisa,90,thomas,0.3157894736842105]
==>[jane,20,lisa,100,jerryd,0.3508771929824561]
==>[jane,20,lisa,90,thomas,50,wyd,0.11278195488721804]
==>[jane,20,lisa,90,thomas,90,jerryd,0.2030075187969925]
这 is quite elegant and best for the environment involved with Neptune and Python. I offer a second for reference, in case others come across this question. From the moment I saw this question I could only ever picture it as being solved with a VertexProgram 以 OLAP 方式与 GraphComputer 相结合。结果，我很难用其他方式思考它。当然，使用 VertexProgram 需要像 Java 这样的 JVM 语言，并且不能直接与 Neptune 一起使用。我想我最接近的解决方法是使用 Java，从 Neptune 获取 subgraph()，然后 运行 在本地 TinkerGraph 中自定义 VertexProgram，这会非常快。
更一般地说，在没有 Python/Neptune 要求的情况下，根据图的性质和需要遍历的数据量，将算法转换为 VertexProgram 并不是一个坏方法。由于没有太多关于这个主题的内容，我想我会在这里提供它的核心代码。这是它的核心：
@Override
public void execute(final Vertex vertex, final Messenger<Double> messenger, final Memory memory) {
// on the first pass calculate the "total favor" for all vertices
// and pass the calculated current favor forward along incident edges
// only for the "start vertex"
if (memory.isInitialIteration()) {
copyHaltedTraversersFromMemory(vertex);
final boolean startVertex = vertex.value("name").equals(nameOfStartVertrex);
final double initialFavor = startVertex ? 1d : 0d;
vertex.property(VertexProperty.Cardinality.single, FAVOR, initialFavor);
vertex.property(VertexProperty.Cardinality.single, TOTAL_FAVOR,
IteratorUtils.stream(vertex.edges(Direction.OUT)).mapToDouble(e -> e.value("weight")).sum());
if (startVertex) {
final Iterator<Edge> incidents = vertex.edges(Direction.OUT);
memory.add(VOTE_TO_HALT, !incidents.hasNext());
while (incidents.hasNext()) {
final Edge incident = incidents.next();
messenger.sendMessage(MessageScope.Global.of(incident.inVertex()),
(double) incident.value("weight") / (double) vertex.value(TOTAL_FAVOR));
}
}
} else {
// on future passes, sum all the incoming "favor" and add it to
// the "favor" property of each vertex. then once again pass the
// current favor to incident edges. this will keep happening
// until the message passing stops.
final Iterator<Double> messages = messenger.receiveMessages();
final boolean hasMessages = messages.hasNext();
if (hasMessages) {
double adjacentFavor = IteratorUtils.reduce(messages, 0.0d, Double::sum);
vertex.property(VertexProperty.Cardinality.single, FAVOR, (double) vertex.value(FAVOR) + adjacentFavor);
final Iterator<Edge> incidents = vertex.edges(Direction.OUT);
memory.add(VOTE_TO_HALT, !incidents.hasNext());
while (incidents.hasNext()) {
final Edge incident = incidents.next();
messenger.sendMessage(MessageScope.Global.of(incident.inVertex()),
adjacentFavor * ((double) incident.value("weight") / (double) vertex.value(TOTAL_FAVOR)));
}
}
}
}
然后上面的执行为：
ComputerResult result = graph.compute().program(FavorVertexProgram.build().name("jane").create()).submit().get();
GraphTraversalSource rg = result.graph().traversal();
Traversal elements = rg.V().elementMap();
并且“元素”遍历产量：
{id=0, label=person, ^favor=1.0, name=jane, ^totalFavor=30.0}
{id=2, label=person, ^favor=0.6491228070175439, name=thomas, ^totalFavor=140.0}
{id=4, label=person, ^favor=0.6666666666666666, name=lisa, ^totalFavor=190.0}
{id=6, label=person, ^favor=0.23182957393483708, name=wyd, ^totalFavor=0.0}
{id=8, label=person, ^favor=0.768170426065163, name=jerryd, ^totalFavor=0.0}
我很难在 gremlin 中找出以下场景的查询。这是有向图（可能是循环的）。
我想获得前N个有利节点，从节点“Jane”开始，其中favor定义为：
favor(Jane->Lisa) = edge(Jane,Lisa) / total weight from outwards edges of Lisa
favor(Jane->Thomas) = favor(Jane->Thomas) + favor(Jane->Lisa) * favor(Lisa->Thomas)
favor(Jane->Jerryd) = favor(Jane->Thomas) * favor(Thomas->Jerryd) + favor(Jane->Lisa) * favor(Lisa->Jerryd)
favor(Jane->Jerryd) = [favor(Jane->Thomas) + favor(Jane->Lisa) * favor(Lisa->Thomas)] * favor(Thomas->Jerryd) + favor(Jane->Lisa) * favor(Lisa->Jerryd)
and so .. on
这是同一张图表，我的意思是手工计算，
这很容易通过编程进行传输，但我不确定用 gremlin 甚至 sparql 查询它有多准确。
这是创建此示例图的查询：
g
.addV('person').as('1').property(single, 'name', 'jane')
.addV('person').as('2').property(single, 'name', 'thomas')
.addV('person').as('3').property(single, 'name', 'lisa')
.addV('person').as('4').property(single, 'name', 'wyd')
.addV('person').as('5').property(single, 'name', 'jerryd')
.addE('favor').from('1').to('2').property('weight', 10)
.addE('favor').from('1').to('3').property('weight', 20)
.addE('favor').from('3').to('2').property('weight', 90)
.addE('favor').from('2').to('4').property('weight', 50)
.addE('favor').from('2').to('5').property('weight', 90)
.addE('favor').from('3').to('5').property('weight', 100)
我要找的是：
[Lisa, computedFavor]
[Thomas, computedFavor]
[Jerryd, computedFavor]
[Wyd, computedFavor]
我正在努力结合循环图来调整权重。到目前为止，这是我能够查询的地方：https://gremlify.com/f2r0zy03oxc/2
g.V().has('name','jane'). // our starting node
repeat(
union(
outE() // get only outwards edges
).
otherV().simplePath()). // produce simple path
emit().
times(10). // max depth of 10
path(). // attain path
by(valueMap())
处理来自 stephen mallette 的评论：
favor(Jane->Jerryd) =
favor(Jane->Thomas) * favor(Thomas->Jerryd)
+ favor(Jane->Lisa) * favor(Lisa->Jerryd)
// note we can expand on favor(Jane->Thomas) in above expression
//
// favor(Jane->Thomas) is favor(Jane->Thomas)@directEdge +
// favor(Jane->Lisa) * favor(Lisa->Thomas)
//
计算示例
Jane to Lisa => 20/(10+20) => 2/3
Lisa to Jerryd => 100/(100+90) => 10/19
Jane to Lisa to Jerryd => 2/3*(10/19)
Jane to Thomas (directly) => 10/(10+20) => 1/3
Jane to Lisa to Thomas => 2/3 * 90/(100+90) => 2/3 * 9/19
Jane to Thomas => 1/3 + (2/3 * 9/19)
Thomas to Jerryd => 90/(90+50) => 9/14
Jane to Thomas to Jerryd => [1/3 + (2/3 * 9/19)] * (9/14)
Jane to Jerryd:
= Jane to Lisa to Jerryd + Jane to Thomas to Jerryd
= 2/3 * (10/19) + [1/3 + (2/3 * 9/19)] * (9/14)
这里是一些伪代码：
def get_favors(graph, label="jane", starting_favor=1):
start = graph.findNode(label)
queue = [(start, starting_favor)]
favors = {}
seen = set()
while queue:
node, curr_favor = queue.popleft()
# get total weight (out edges) from this node
total_favor = 0
for (edgeW, outNode) in node.out_edges:
total_favor = total_favor + edgeW
for (edgeW, outNode) in node.out_edges:
# if there are no favors for this node
# take current favor and provide proportional favor
if outNode not in favors:
favors[outNode] = curr_favor * (edgeW / total_favor)
# it already has some favor, so we add to it
# we add proportional favor
else:
favors[outNode] += curr_favor * (edgeW / total_favor)
# if we have seen this edge, and node ignore
# otherwise, transverse
if (edgeW, outNode) not in seen:
seen.add((edgeW, outNode))
queue.append((outNode, favors[outNode]))
# sort favor by value and return top X
return favors
这是一个 Gremlin 查询，我相信它可以正确应用您的公式。我将首先粘贴完整的最终查询，然后简单介绍所涉及的步骤。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','jerryd')).
.....11> sack().
.....12> sum()
==>0.768170426065163
查询从 Jane 开始，一直遍历，直到检查到 Jerry D 的所有路径。沿途为每个遍历器维护一个sack，其中包含为每个关系计算的权重值相乘。第 6 行的计算找到所有可能来自先前顶点的边权重值，第 7 行的 math 步骤用于将当前边上的权重除以该总和。在最后，第 12 行将每个计算结果相加。如果删除最后的 sum 步骤，您可以看到中间结果。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','jerryd')).
.....11> sack()
==>0.2142857142857143
==>0.3508771929824561
==>0.2030075187969925
要查看经过的路线，可以将 path 步骤添加到查询中。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','jerryd')).
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight'),
.....16> sack()).fold())
==>[[jane,10,thomas,90,jerryd],0.2142857142857143]
==>[[jane,20,lisa,100,jerryd],0.3508771929824561]
==>[[jane,20,lisa,90,thomas,90,jerryd],0.2030075187969925]
这种方法还考虑了添加任何直接连接，根据您的公式，如果我们使用 Thomas 作为目标，我们可以看到。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> until(has('name','thomas')).
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight'),
.....16> sack()).fold())
==>[[jane,10,thomas],0.3333333333333333]
==>[[jane,20,lisa,90,thomas],0.3157894736842105]
不需要这些额外的步骤，但是包含 path 在调试这样的查询时很有用。此外，这不是必需的，但也许只是出于一般兴趣，我要补充一点，您也可以从此处获得最终答案，但我包含的第一个查询就是您真正需要的。
g.withSack(1).V().
has('name','jane').
repeat(outE().
sack(mult).
by(project('w','f').
by('weight').
by(outV().outE().values('weight').sum()).
math('w / f')).
inV().
simplePath()).
until(has('name','thomas')).
local(
union(
path().
by('name').
by('weight'),
sack()).fold().tail(local)).
sum()
==>0.6491228070175439
如果有任何不清楚的地方或者我有mis-understood公式，请告诉我。
编辑添加
要查找 Jane 可联系到的所有人的结果，我必须稍微修改查询。最后的 unfold 只是为了让结果更容易阅读。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> emit().
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight').unfold(),
.....16> sack()).fold()).
.....17> group().
.....18> by(tail(local,2).limit(local,1)).
.....19> by(tail(local).sum()).
.....20> unfold()
==>jerryd=0.768170426065163
==>wyd=0.23182957393483708
==>lisa=0.6666666666666666
==>thomas=0.6491228070175439
第 17 行的最后 group 步骤使用 path 结果计算找到的每个唯一名称的总支持度。要查看路径，您可以 运行 删除 group 步骤的查询。
gremlin> g.withSack(1).V().
......1> has('name','jane').
......2> repeat(outE().
......3> sack(mult).
......4> by(project('w','f').
......5> by('weight').
......6> by(outV().outE().values('weight').sum()).
......7> math('w / f')).
......8> inV().
......9> simplePath()).
.....10> emit().
.....11> local(
.....12> union(
.....13> path().
.....14> by('name').
.....15> by('weight').unfold(),
.....16> sack()).fold())
==>[jane,10,thomas,0.3333333333333333]
==>[jane,20,lisa,0.6666666666666666]
==>[jane,10,thomas,50,wyd,0.11904761904761904]
==>[jane,10,thomas,90,jerryd,0.2142857142857143]
==>[jane,20,lisa,90,thomas,0.3157894736842105]
==>[jane,20,lisa,100,jerryd,0.3508771929824561]
==>[jane,20,lisa,90,thomas,50,wyd,0.11278195488721804]
==>[jane,20,lisa,90,thomas,90,jerryd,0.2030075187969925]
这 GraphComputer 相结合。结果，我很难用其他方式思考它。当然，使用 VertexProgram 需要像 Java 这样的 JVM 语言，并且不能直接与 Neptune 一起使用。我想我最接近的解决方法是使用 Java，从 Neptune 获取 subgraph()，然后 运行 在本地 TinkerGraph 中自定义 VertexProgram，这会非常快。
更一般地说，在没有 Python/Neptune 要求的情况下，根据图的性质和需要遍历的数据量，将算法转换为 VertexProgram 并不是一个坏方法。由于没有太多关于这个主题的内容，我想我会在这里提供它的核心代码。这是它的核心：
@Override
public void execute(final Vertex vertex, final Messenger<Double> messenger, final Memory memory) {
// on the first pass calculate the "total favor" for all vertices
// and pass the calculated current favor forward along incident edges
// only for the "start vertex"
if (memory.isInitialIteration()) {
copyHaltedTraversersFromMemory(vertex);
final boolean startVertex = vertex.value("name").equals(nameOfStartVertrex);
final double initialFavor = startVertex ? 1d : 0d;
vertex.property(VertexProperty.Cardinality.single, FAVOR, initialFavor);
vertex.property(VertexProperty.Cardinality.single, TOTAL_FAVOR,
IteratorUtils.stream(vertex.edges(Direction.OUT)).mapToDouble(e -> e.value("weight")).sum());
if (startVertex) {
final Iterator<Edge> incidents = vertex.edges(Direction.OUT);
memory.add(VOTE_TO_HALT, !incidents.hasNext());
while (incidents.hasNext()) {
final Edge incident = incidents.next();
messenger.sendMessage(MessageScope.Global.of(incident.inVertex()),
(double) incident.value("weight") / (double) vertex.value(TOTAL_FAVOR));
}
}
} else {
// on future passes, sum all the incoming "favor" and add it to
// the "favor" property of each vertex. then once again pass the
// current favor to incident edges. this will keep happening
// until the message passing stops.
final Iterator<Double> messages = messenger.receiveMessages();
final boolean hasMessages = messages.hasNext();
if (hasMessages) {
double adjacentFavor = IteratorUtils.reduce(messages, 0.0d, Double::sum);
vertex.property(VertexProperty.Cardinality.single, FAVOR, (double) vertex.value(FAVOR) + adjacentFavor);
final Iterator<Edge> incidents = vertex.edges(Direction.OUT);
memory.add(VOTE_TO_HALT, !incidents.hasNext());
while (incidents.hasNext()) {
final Edge incident = incidents.next();
messenger.sendMessage(MessageScope.Global.of(incident.inVertex()),
adjacentFavor * ((double) incident.value("weight") / (double) vertex.value(TOTAL_FAVOR)));
}
}
}
}
然后上面的执行为：
ComputerResult result = graph.compute().program(FavorVertexProgram.build().name("jane").create()).submit().get();
GraphTraversalSource rg = result.graph().traversal();
Traversal elements = rg.V().elementMap();
并且“元素”遍历产量：
{id=0, label=person, ^favor=1.0, name=jane, ^totalFavor=30.0}
{id=2, label=person, ^favor=0.6491228070175439, name=thomas, ^totalFavor=140.0}
{id=4, label=person, ^favor=0.6666666666666666, name=lisa, ^totalFavor=190.0}
{id=6, label=person, ^favor=0.23182957393483708, name=wyd, ^totalFavor=0.0}
{id=8, label=person, ^favor=0.768170426065163, name=jerryd, ^totalFavor=0.0}