如何计算顶点相似于graphx邻居 [英] how to compute vertex similarity to neighbors in graphx
问题描述
假设有一个简单的图形,如:
VAL用户= sc.parallelize(阵列(
(1L,SEQ(M,2014年,40376,空,N,1,拉贾斯坦)),
(2L,SEQ(M,2009年,20231,NULL,N,1,拉贾斯坦)),
(3L,SEQ(F,2016年,40376,空,N,1,拉贾斯坦))
))
VAL边缘= sc.parallelize(阵列(
边缘(1L,2L,),
边缘(1L,3L,),
边缘(2L,3L,)))
VAL图=图(用户,边缘)
我要计算每个顶点多少类似于其对每个属性的邻居。
理想的输出(RDD一个数据框或)将持有这些结果:
1L:0.5,0.5,0.5,1.0,1.0,1.0,1.0
2L:0.5,0.0,0.0,1.0,1.0,1.0,1.0
3L:0.0,0.5,0.5,1.0,1.0,1.0,1.0
例如,对于1L所述第一值表示在2个邻居,仅1共享相同的值...
我与aggregateMessage玩只是为了计算有多少邻居也有类似的属性值,但无济于事迄今:
VAL结果= graph.aggregateMessages [(智力,SEQ [任何])](
//建立消息
SENDMSG = {
//地图功能
三重=>
//发送消息到目标顶点
triplet.sendToDst(1,triplet.srcAttr)
//发送消息到源顶点
triplet.sendToSrc(1,triplet.dstAttr)
} //试图指望有类似性质的邻居
{情况下((CNT1,发件人),(CNT2,接收器))=>
VAL为prop1 =如果(发送者(0)==接收机(0))1d的别的0D
VAL prop2 = IF(Math.abs(发送者(1).asInstanceOf [INT] - 接收器(1).asInstanceOf [INT])3;)1D其他0D
VAL prop3 =如果(发送者(2)==接收机(2))1d的别的0D
VAL prop4 =如果(发送者(3)==接收机(3))1d的别的0D
VAL prop5 =如果(发送者(4)==接收机(4))1d的别的0D
VAL prop6 =如果(发送者(5)==接收机(5))1d的别的0D
VAL prop7 =如果(发送者(6)==接收机(6))1d的别的0D
(CNT1 + CNT2,SEQ(为prop1,prop2,prop3,prop4,prop5,prop6,prop7))
}
)
这让我对每个顶点正确的邻里大小,但没有总结权值:
//> (1,(2,列表(0.0,0.0,0.0,1.0,1.0,1.0,1.0)))
// | (2,(2,列表(0.0,1.0,1.0,1.0,1.0,1.0,1.0)))
// | (3,(2,列表(1.0,0.0,0.0,1.0,1.0,1.0,1.0)))
因为在你的code,任何款项它并不值求和。而且你的逻辑是错误的。 mergeMsg
接收消息没有(的消息
,电流
)对。尝试是这样的:
进口breeze.linalg.DenseVector高清compareAttrs(XS:序号[任何]伊苏:序号[任何])=
DenseVector(xs.zip(YS).MAP {壳体(X,Y)=>如果(X == y)的1L别的0L} .toArray)VAL结果= graph.aggregateMessages [(长,DenseVector [龙])](
三重=> {
VAL comparedAttrs = compareAttrs(triplet.dstAttr,triplet.srcAttr)
triplet.sendToDst(1L,comparedAttrs)
triplet.sendToSrc(1L,comparedAttrs)
},
{情况下((CNT1,V1),(CNT2,V2))=> (CNT1 + CNT2,V1 + V2)}
)result.mapValues(KV =方式>(kv._2.map(_ toDouble)/ kv._1.toDouble))收集。
//阵列(
//(1,DenseVector(0.5,0.0,0.5,1.0,1.0,1.0,1.0)),
//(2,DenseVector(0.5,0.0,0.0,1.0,1.0,1.0,1.0)),
//(3,DenseVector(0.0,0.0,0.5,1.0,1.0,1.0,1.0)))
Suppose to have a simple graph like:
val users = sc.parallelize(Array(
(1L, Seq("M", 2014, 40376, null, "N", 1, "Rajastan")),
(2L, Seq("M", 2009, 20231, null, "N", 1, "Rajastan")),
(3L, Seq("F", 2016, 40376, null, "N", 1, "Rajastan"))
))
val edges = sc.parallelize(Array(
Edge(1L, 2L, ""),
Edge(1L, 3L, ""),
Edge(2L, 3L, "")))
val graph = Graph(users, edges)
I'd like to compute how much each vertex is similar to its neighbors on each attribute.
The ideal output (an RDD or DataFrame) would hold these results:
1L: 0.5, 0.5, 0.5, 1.0, 1.0, 1.0, 1.0
2L: 0.5, 0.0, 0.0, 1.0, 1.0, 1.0, 1.0
3L: 0.0, 0.5, 0.5, 1.0, 1.0, 1.0, 1.0
For instance, the first value for 1L means that on 2 neighbors, just 1 share the same value...
I am playing with aggregateMessage just to count how many neighbors have a similar attribute value but with no avail so far:
val result = graph.aggregateMessages[(Int, Seq[Any])](
// build the message
sendMsg = {
// map function
triplet =>
// send message to destination vertex
triplet.sendToDst(1, triplet.srcAttr)
// send message to source vertex
triplet.sendToSrc(1, triplet.dstAttr)
}, // trying to count neighbors with similar property
{ case ((cnt1, sender), (cnt2, receiver)) =>
val prop1 = if(sender(0) == receiver(0)) 1d else 0d
val prop2 = if(Math.abs(sender(1).asInstanceOf[Int] - receiver(1).asInstanceOf[Int])<3) 1d else 0d
val prop3 = if(sender(2) == receiver(2)) 1d else 0d
val prop4 = if(sender(3) == receiver(3)) 1d else 0d
val prop5 = if(sender(4) == receiver(4)) 1d else 0d
val prop6 = if(sender(5) == receiver(5)) 1d else 0d
val prop7 = if(sender(6) == receiver(6)) 1d else 0d
(cnt1 + cnt2, Seq(prop1, prop2, prop3, prop4, prop5, prop6, prop7))
}
)
this gives me the correct neighborhood size for each vertex but is not summing up the values right:
//> (1,(2,List(0.0, 0.0, 0.0, 1.0, 1.0, 1.0, 1.0)))
//| (2,(2,List(0.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0)))
//| (3,(2,List(1.0, 0.0, 0.0, 1.0, 1.0, 1.0, 1.0)))
It doesn't sum values because there is no sum in your code. Moreover your logic is wrong. mergeMsg
receives messages not (message
, current
) pairs. Try something like this:
import breeze.linalg.DenseVector
def compareAttrs(xs: Seq[Any], ys: Seq[Any]) =
DenseVector(xs.zip(ys).map{ case (x, y) => if (x == y) 1L else 0L}.toArray)
val result = graph.aggregateMessages[(Long, DenseVector[Long])](
triplet => {
val comparedAttrs = compareAttrs(triplet.dstAttr, triplet.srcAttr)
triplet.sendToDst(1L, comparedAttrs)
triplet.sendToSrc(1L, comparedAttrs)
},
{ case ((cnt1, v1), (cnt2, v2)) => (cnt1 + cnt2, v1 + v2) }
)
result.mapValues(kv => (kv._2.map(_.toDouble) / kv._1.toDouble)).collect
// Array(
// (1,DenseVector(0.5, 0.0, 0.5, 1.0, 1.0, 1.0, 1.0)),
// (2,DenseVector(0.5, 0.0, 0.0, 1.0, 1.0, 1.0, 1.0)),
// (3,DenseVector(0.0, 0.0, 0.5, 1.0, 1.0, 1.0, 1.0)))
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