I have a DataFrame called source, a table from mysql
val source = sqlContext.read.jdbc(jdbcUrl, "source", connectionProperties)
I have converted it to rdd by
val sourceRdd = source.rdd
but its RDD[Row] I need RDD[String] to do transformations like
source.map(rec => (rec.split(",")(0).toInt, rec)), .subtractByKey(), etc..
Thank you
You can use Row. mkString(sep: String): String
method in a map
call like this :
val sourceRdd = source.rdd.map(_.mkString(","))
You can change the ","
parameter by whatever you want.
Hope this help you, Best Regards.
What is your schema?
If it's just a String, you can use:
import spark.implicits._
val sourceDS = source.as[String]
val sourceRdd = sourceDS.rdd // will give RDD[String]
Note: use sqlContext instead of spark in Spark 1.6 - spark is a SparkSession, which is a new class in Spark 2.0 and is a new entry point to SQL functionality. It should be used instead of SQLContext in Spark 2.x
You can also create own case classes.
Also you can map rows - here source is of type DataFrame, we use partial function in map function:
val sourceRdd = source.rdd.map { case x : Row => x(0).asInstanceOf[String] }.map(s => s.split(","))
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