Re: Re: what is the best way to do cartesian
Thanks a lot for your reply, but i have tried the built-in RDD.cartesian()
method before, it didn't make it faster.
qinwei
From: Alex BoisvertDate: 2014-04-26 00:32To: userSubject: Re: what is the best
way to do cartesianYou might want to try the built-in RDD.cartesian() method.
On Thu, Apr 24, 2014 at 9:05 PM, Qin Wei wei@dewmobile.net wrote:
Hi All,
I have a problem with the Item-Based Collaborative Filtering Recommendation
Algorithms in spark.
The basic flow is as below:
(Item1 , (User1 ,
Score1))
RDD1 == (Item2 , (User2 , Score2))
(Item1 , (User2 ,
Score3))
(Item2 , (User1 ,
Score4))
RDD1.groupByKey == RDD2
(Item1, ((User1, Score1),
(User2, Score3)))
(Item2, ((User1, Score4),
(User2, Score2)))
The similarity of Vector ((User1, Score1), (User2, Score3)) and
((User1, Score4), (User2, Score2)) is the similarity of Item1 and
Item2.
In my situation, RDD2 contains 20 million records, my spark programm is
extreamly slow, the source code is as below:
val conf = new
SparkConf().setMaster(spark://211.151.121.184:7077).setAppName(Score
Calcu Total).set(spark.executor.memory,
20g).setJars(Seq(/home/deployer/score-calcu-assembly-1.0.jar))
val sc = new SparkContext(conf)
val mongoRDD = sc.textFile(args(0).toString,
400)
val jsonRDD = mongoRDD.map(arg = new
JSONObject(arg))
val newRDD = jsonRDD.map(arg = {
var score =
haha(arg.get(a).asInstanceOf[JSONObject])
// set score to 0.5 for testing
arg.put(score, 0.5)
arg
})
val resourceScoresRDD = newRDD.map(arg =
(arg.get(rid).toString.toLong, (arg.get(zid).toString,
arg.get(score).asInstanceOf[Number].doubleValue))).groupByKey().cache()
val resourceScores =
resourceScoresRDD.collect()
val bcResourceScores =
sc.broadcast(resourceScores)
val simRDD =
resourceScoresRDD.mapPartitions({iter =
val m = bcResourceScores.value
for{ (r1, v1) - iter
(r2, v2) - m
if r1 r2
} yield (r1, r2, cosSimilarity(v1,
v2))}, true).filter(arg = arg._3 0.1)
println(simRDD.count)
And I saw this in Spark Web UI:
http://apache-spark-user-list.1001560.n3.nabble.com/file/n4807/QQ%E6%88%AA%E5%9B%BE20140424204018.png
http://apache-spark-user-list.1001560.n3.nabble.com/file/n4807/QQ%E6%88%AA%E5%9B%BE20140424204001.png
My standalone cluster has 3 worker node (16 core and 32G RAM),and the
workload of the machine in my cluster is heavy when the spark program is
running.
Is there any better way to do the algorithm?
Thanks!
--
View this message in context:
http://apache-spark-user-list.1001560.n3.nabble.com/what-is-the-best-way-to-do-cartesian-tp4807.html
Sent from the Apache Spark User List mailing list archive at Nabble.com.
Re: what is the best way to do cartesian
You might want to try the built-in RDD.cartesian() method.
On Thu, Apr 24, 2014 at 9:05 PM, Qin Wei wei@dewmobile.net wrote:
Hi All,
I have a problem with the Item-Based Collaborative Filtering Recommendation
Algorithms in spark.
The basic flow is as below:
(Item1, (User1 ,
Score1))
RDD1 ==(Item2, (User2 , Score2))
(Item1, (User2 ,
Score3))
(Item2, (User1 ,
Score4))
RDD1.groupByKey == RDD2
(Item1, ((User1,
Score1),
(User2, Score3)))
(Item2, ((User1,
Score4),
(User2, Score2)))
The similarity of Vector ((User1, Score1), (User2, Score3)) and
((User1, Score4), (User2, Score2)) is the similarity of Item1 and
Item2.
In my situation, RDD2 contains 20 million records, my spark programm is
extreamly slow, the source code is as below:
val conf = new
SparkConf().setMaster(spark://211.151.121.184:7077).setAppName(Score
Calcu Total).set(spark.executor.memory,
20g).setJars(Seq(/home/deployer/score-calcu-assembly-1.0.jar))
val sc = new SparkContext(conf)
val mongoRDD =
sc.textFile(args(0).toString,
400)
val jsonRDD = mongoRDD.map(arg = new
JSONObject(arg))
val newRDD = jsonRDD.map(arg = {
var score =
haha(arg.get(a).asInstanceOf[JSONObject])
// set score to 0.5 for testing
arg.put(score, 0.5)
arg
})
val resourceScoresRDD = newRDD.map(arg =
(arg.get(rid).toString.toLong, (arg.get(zid).toString,
arg.get(score).asInstanceOf[Number].doubleValue))).groupByKey().cache()
val resourceScores =
resourceScoresRDD.collect()
val bcResourceScores =
sc.broadcast(resourceScores)
val simRDD =
resourceScoresRDD.mapPartitions({iter =
val m = bcResourceScores.value
for{ (r1, v1) - iter
(r2, v2) - m
if r1 r2
} yield (r1, r2, cosSimilarity(v1,
v2))}, true).filter(arg = arg._3 0.1)
println(simRDD.count)
And I saw this in Spark Web UI:
http://apache-spark-user-list.1001560.n3.nabble.com/file/n4807/QQ%E6%88%AA%E5%9B%BE20140424204018.png
http://apache-spark-user-list.1001560.n3.nabble.com/file/n4807/QQ%E6%88%AA%E5%9B%BE20140424204001.png
My standalone cluster has 3 worker node (16 core and 32G RAM),and the
workload of the machine in my cluster is heavy when the spark program is
running.
Is there any better way to do the algorithm?
Thanks!
--
View this message in context:
http://apache-spark-user-list.1001560.n3.nabble.com/what-is-the-best-way-to-do-cartesian-tp4807.html
Sent from the Apache Spark User List mailing list archive at Nabble.com.
Re: what is the best way to do cartesian
Depending on the size of the rdd you could also do a collect broadcast and
then compute the product in a map function over the other rdd. If this is
the same rdd you might also want to cache it. This pattern worked quite
good for me
Le 25 avr. 2014 18:33, Alex Boisvert alex.boisv...@gmail.com a écrit :
You might want to try the built-in RDD.cartesian() method.
On Thu, Apr 24, 2014 at 9:05 PM, Qin Wei wei@dewmobile.net wrote:
Hi All,
I have a problem with the Item-Based Collaborative Filtering
Recommendation
Algorithms in spark.
The basic flow is as below:
(Item1, (User1 ,
Score1))
RDD1 ==(Item2, (User2 ,
Score2))
(Item1, (User2 ,
Score3))
(Item2, (User1 ,
Score4))
RDD1.groupByKey == RDD2
(Item1, ((User1,
Score1),
(User2, Score3)))
(Item2, ((User1,
Score4),
(User2, Score2)))
The similarity of Vector ((User1, Score1), (User2, Score3)) and
((User1, Score4), (User2, Score2)) is the similarity of Item1 and
Item2.
In my situation, RDD2 contains 20 million records, my spark programm is
extreamly slow, the source code is as below:
val conf = new
SparkConf().setMaster(spark://211.151.121.184:7077).setAppName(Score
Calcu Total).set(spark.executor.memory,
20g).setJars(Seq(/home/deployer/score-calcu-assembly-1.0.jar))
val sc = new SparkContext(conf)
val mongoRDD =
sc.textFile(args(0).toString,
400)
val jsonRDD = mongoRDD.map(arg = new
JSONObject(arg))
val newRDD = jsonRDD.map(arg = {
var score =
haha(arg.get(a).asInstanceOf[JSONObject])
// set score to 0.5 for testing
arg.put(score, 0.5)
arg
})
val resourceScoresRDD = newRDD.map(arg =
(arg.get(rid).toString.toLong, (arg.get(zid).toString,
arg.get(score).asInstanceOf[Number].doubleValue))).groupByKey().cache()
val resourceScores =
resourceScoresRDD.collect()
val bcResourceScores =
sc.broadcast(resourceScores)
val simRDD =
resourceScoresRDD.mapPartitions({iter =
val m = bcResourceScores.value
for{ (r1, v1) - iter
(r2, v2) - m
if r1 r2
} yield (r1, r2, cosSimilarity(v1,
v2))}, true).filter(arg = arg._3 0.1)
println(simRDD.count)
And I saw this in Spark Web UI:
http://apache-spark-user-list.1001560.n3.nabble.com/file/n4807/QQ%E6%88%AA%E5%9B%BE20140424204018.png
http://apache-spark-user-list.1001560.n3.nabble.com/file/n4807/QQ%E6%88%AA%E5%9B%BE20140424204001.png
My standalone cluster has 3 worker node (16 core and 32G RAM),and the
workload of the machine in my cluster is heavy when the spark program is
running.
Is there any better way to do the algorithm?
Thanks!
--
View this message in context:
http://apache-spark-user-list.1001560.n3.nabble.com/what-is-the-best-way-to-do-cartesian-tp4807.html
Sent from the Apache Spark User List mailing list archive at Nabble.com.
