Thank you guys, I got my code worked like below: val record75df =
sc.parallelize(listForRule75, numPartitions).map(x=> x.replace("|",
",")).map(_.split(",")).map(x =>
Mycaseclass4(x(0).toInt,x(1).toInt,x(2).toFloat,x(3).toInt)).toDF()
val userids = 1 to 10000
val uiddf = sc.parallelize(userids, numPartitions).toDF("userid")
record75df.registerTempTable("b")
uiddf.registerTempTable("a")
val rule75df = sqlContext.sql("select a.*,b.* from a join b")
rule75df.show
--------------------------------
Thanks&Best regards!
San.Luo
----- 原始邮件 -----
发件人：Ryan <ryan.hd....@gmail.com>
收件人：ayan guha <guha.a...@gmail.com>
抄送人：Riccardo Ferrari <ferra...@gmail.com>, luohui20...@sina.com, user
<user@spark.apache.org>
主题：Re: Re: Is there an operation to create multi record for every element in a
RDD?
日期：2017年08月09日 17点32分
rdd has a cartesian method
On Wed, Aug 9, 2017 at 5:12 PM, ayan guha <guha.a...@gmail.com> wrote:
If you use join without any condition in becomes cross join. In sql, it looks
like
Select a.*,b.* from a join b
On Wed, 9 Aug 2017 at 7:08 pm, <luohui20...@sina.com> wrote:
Riccardo and Ryan Thank you for your ideas.It seems that crossjoin is a new
dataset api after spark2.x. my spark version is 1.6.3. Is there a relative
api to do crossjoin? thank you.
--------------------------------
Thanks&Best regards!
San.Luo
----- 原始邮件 -----
发件人：Riccardo Ferrari <ferra...@gmail.com>
收件人：Ryan <ryan.hd....@gmail.com>
抄送人：luohui20...@sina.com, user <user@spark.apache.org>
主题：Re: Is there an operation to create multi record for every element in a RDD?
日期：2017年08月09日 16点54分
Depends on your Spark version, have you considered the Dataset api?
You can do something like:
val df1 = rdd1.toDF("userid")
val listRDD = sc.parallelize(listForRule77)
val listDF = listRDD.toDF("data")
df1.crossJoin(listDF).orderBy("userid").show(60,
truncate=false)+------+----------------------+|userid|data
|+------+----------------------+|1 |1,1,100.00|1483891200,||1
|1,1,100.00|1483804800,|...|1 |1,1,100.00|1488902400,|
|1 |1,1,100.00|1489075200,||1 |1,1,100.00|1488470400,|...
On Wed, Aug 9, 2017 at 10:44 AM, Ryan <ryan.hd....@gmail.com> wrote:
It's just sort of inner join operation... If the second dataset isn't very
large it's ok(btw, you can use flatMap directly instead of map followed by
flatmap/flattern), otherwise you can register the second one as a rdd/dataset,
and join them on user id.
On Wed, Aug 9, 2017 at 4:29 PM, <luohui20...@sina.com> wrote:
hello guys: I have a simple rdd like :val userIDs = 1 to 10000val rdd1 =
sc.parallelize(userIDs , 16) //this rdd has 10000 user id And I have a
List[String] like below:scala> listForRule77
res76: List[String] = List(1,1,100.00|1483286400, 1,1,100.00|1483372800,
1,1,100.00|1483459200, 1,1,100.00|1483545600, 1,1,100.00|1483632000,
1,1,100.00|1483718400, 1,1,100.00|1483804800, 1,1,100.00|1483891200,
1,1,100.00|1483977600, 3,1,200.00|1485878400, 1,1,100.00|1485964800,
1,1,100.00|1486051200, 1,1,100.00|1488384000, 1,1,100.00|1488470400,
1,1,100.00|1488556800, 1,1,100.00|1488643200, 1,1,100.00|1488729600,
1,1,100.00|1488816000, 1,1,100.00|1488902400, 1,1,100.00|1488988800,
1,1,100.00|1489075200, 1,1,100.00|1489161600, 1,1,100.00|1489248000,
1,1,100.00|1489334400, 1,1,100.00|1489420800, 1,1,100.00|1489507200,
1,1,100.00|1489593600, 1,1,100.00|1489680000, 1,1,100.00|1489766400)
scala> listForRule77.length
res77: Int = 29
I need to create a rdd containing 290000 records. for every userid in
rdd1 , I need to create 29 records according to listForRule77, each record
start with the userid, for example 1(the userid),1,1,100.00|1483286400.
My idea is like below:1.write a udfto add the userid to the beginning of every
string element of listForRule77.2.use val rdd2 = rdd1.map{x=>
List_udf(x))}.flatmap(), the result rdd2 maybe what I need.
My question: Are there any problems in my idea? Is there a better way to
do this ?
--------------------------------
Thanks&Best regards!
San.Luo
--
Best Regards,
Ayan Guha
