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 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 >