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 ?
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Thanks&Best regards!
San.Luo
