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 10000
> val 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 udf
> to 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
