scala - significado - Unir dos RDD comunes con/sin Spark SQL

spark scala sql

Question

Necesito unir dos RDDs ordinarios en una o más columnas. Lógicamente, esta operación es equivalente a la operación de combinación de base de datos de dos tablas. Me pregunto si esto es posible solo a través de Spark SQL o hay otras formas de hacerlo.

Como ejemplo concreto, considere RDD r1 con la clave primaria ITEM_ID :

(ITEM_ID, ITEM_NAME, ITEM_UNIT, COMPANY_ID)

y RDD r2 con clave principal COMPANY_ID :

(COMPANY_ID, COMPANY_NAME, COMPANY_CITY)

Quiero unirme a r1 y r2 .

¿Cómo puede hacerse esto?

Answer 1

Algo como esto debería funcionar.

scala> case class Item(id:String, name:String, unit:Int, companyId:String)

scala> case class Company(companyId:String, name:String, city:String)

scala> val i1 = Item("1", "first", 2, "c1")

scala> val i2 = i1.copy(id="2", name="second")

scala> val i3 = i1.copy(id="3", name="third", companyId="c2")

scala> val items = sc.parallelize(List(i1,i2,i3))
items: org.apache.spark.rdd.RDD[Item] = ParallelCollectionRDD[14] at parallelize at <console>:20

scala> val c1 = Company("c1", "company-1", "city-1")

scala> val c2 = Company("c2", "company-2", "city-2")

scala> val companies = sc.parallelize(List(c1,c2))

scala> val groupedItems = items.groupBy( x => x.companyId) 
groupedItems: org.apache.spark.rdd.RDD[(String, Iterable[Item])] = ShuffledRDD[16] at groupBy at <console>:22

scala> val groupedComp = companies.groupBy(x => x.companyId)
groupedComp: org.apache.spark.rdd.RDD[(String, Iterable[Company])] = ShuffledRDD[18] at groupBy at <console>:20

scala> groupedItems.join(groupedComp).take(10).foreach(println)

14/12/12 00:52:32 INFO DAGScheduler: Job 5 finished: take at <console>:35, took 0.021870 s
(c1,(CompactBuffer(Item(1,first,2,c1), Item(2,second,2,c1)),CompactBuffer(Company(c1,company-1,city-1))))
(c2,(CompactBuffer(Item(3,third,2,c2)),CompactBuffer(Company(c2,company-2,city-2))))

Answer 2

Soumya Simanta dio una buena respuesta. Sin embargo, los valores en el RDD unido son Iterable , por lo que los resultados pueden no ser muy similares a la unión de tabla ordinaria.

Alternativamente, puedes:

val mappedItems = items.map(item => (item.companyId, item))
val mappedComp = companies.map(comp => (comp.companyId, comp))
mappedItems.join(mappedComp).take(10).foreach(println)

La salida sería:

(c1,(Item(1,first,2,c1),Company(c1,company-1,city-1)))
(c1,(Item(2,second,2,c1),Company(c1,company-1,city-1)))
(c2,(Item(3,third,2,c2),Company(c2,company-2,city-2)))

Answer 3

(Usando Scala) Supongamos que tiene dos RDD:

• emp: (empid, ename, departamento)

• departamento: (dname, departamento)

Lo siguiente es otra forma:

//val emp = sc.parallelize(Seq((1,"jordan",10), (2,"ricky",20), (3,"matt",30), (4,"mince",35), (5,"rhonda",30)))
val emp = sc.parallelize(Seq(("jordan",10), ("ricky",20), ("matt",30), ("mince",35), ("rhonda",30)))

val dept = sc.parallelize(Seq(("hadoop",10), ("spark",20), ("hive",30), ("sqoop",40)))

//val shifted_fields_emp = emp.map(t => (t._3, t._1, t._2))
val shifted_fields_emp = emp.map(t => (t._2, t._1))

val shifted_fields_dept = dept.map(t => (t._2,t._1))

shifted_fields_emp.join(shifted_fields_dept)
// Create emp RDD
val emp = sc.parallelize(Seq((1,"jordan",10), (2,"ricky",20), (3,"matt",30), (4,"mince",35), (5,"rhonda",30)))

// Create dept RDD
val dept = sc.parallelize(Seq(("hadoop",10), ("spark",20), ("hive",30), ("sqoop",40)))

// Establishing that the third field is to be considered as the Key for the emp RDD
val manipulated_emp = emp.keyBy(t => t._3)

// Establishing that the second field need to be considered as the Key for dept RDD
val manipulated_dept = dept.keyBy(t => t._2)

// Inner Join
val join_data = manipulated_emp.join(manipulated_dept)
// Left Outer Join
val left_outer_join_data = manipulated_emp.leftOuterJoin(manipulated_dept)
// Right Outer Join
val right_outer_join_data = manipulated_emp.rightOuterJoin(manipulated_dept)
// Full Outer Join
val full_outer_join_data = manipulated_emp.fullOuterJoin(manipulated_dept)

// Formatting the Joined Data for better understandable (using map)
val cleaned_joined_data = join_data.map(t => (t._2._1._1, t._2._1._2, t._1, t._2._2._1))

Esto dará la salida como:

// Imprime la salida cleaned_joined_data en la consola

scala> cleaned_joined_data.collect()
res13: Array[(Int, String, Int, String)] = Array((3,matt,30,hive), (5,rhonda,30,hive), (2,ricky,20,spark), (1,jordan,10,hadoop))

Answer 4

Spark SQL puede realizar join en SPARK RDDs.

A continuación, el código realiza una combinación de SQL en RDD de empresa y artículos

object SparkSQLJoin {

case class Item(id:String, name:String, unit:Int, companyId:String)
case class Company(companyId:String, name:String, city:String)

def main(args: Array[String]) {

    val sparkConf = new SparkConf()
    val sc= new SparkContext(sparkConf)
    val sqlContext = new SQLContext(sc)

    import sqlContext.createSchemaRDD

    val i1 = Item("1", "first", 1, "c1")
    val i2 = Item("2", "second", 2, "c2")
    val i3 = Item("3", "third", 3, "c3")
    val c1 = Company("c1", "company-1", "city-1")
    val c2 = Company("c2", "company-2", "city-2")

    val companies = sc.parallelize(List(c1,c2))
    companies.registerAsTable("companies")

    val items = sc.parallelize(List(i1,i2,i3))
    items.registerAsTable("items")

    val result = sqlContext.sql("SELECT * FROM companies C JOIN items I ON C.companyId= I.companyId").collect

    result.foreach(println)

    }
}

La salida se muestra como

     [c1,company-1,city-1,1,first,1,c1]
     [c2,company-2,city-2,2,second,2,c2]