Reputation: 851
I have a streaming dataset, reading from kafka and trying to write to CSV
case class Event(map: Map[String,String])
def decodeEvent(arrByte: Array[Byte]): Event = ...//some implementation
val eventDataset: Dataset[Event] = spark
.readStream
.format("kafka")
.load()
.select("value")
.as[Array[Byte]]
.map(decodeEvent)
Event
holds Map[String,String]
inside and to write to CSV I'll need some schema.
Let's say all the fields are of type String
and so I tried the example from spark repo
val columns = List("year","month","date","topic","field1","field2")
val schema = new StructType() //Prepare schema programmatically
columns.foreach { field => schema.add(field, "string") }
val rowRdd = eventDataset.rdd.map { event => Row.fromSeq(
columns.map(c => event.getOrElse(c, "")
)}
val df = spark.sqlContext.createDataFrame(rowRdd, schema)
This gives error at runtime on line "eventDataset.rdd":
Caused by: org.apache.spark.sql.AnalysisException: Queries with streaming sources must be executed with writeStream.start();;
Below doesn't work because '.map' has a List[String] not Tuple
eventDataset.map(event => columns.map(c => event.getOrElse(c,""))
.toDF(columns:_*)
Is there a way to achieve this with programmatic schema and structured streaming datasets?
Upvotes: 3
Views: 1173
Reputation: 35249
I'd use much simpler approach:
import org.apache.spark.sql.functions._
eventDataset.select(columns.map(
c => coalesce($"map".getItem(c), lit("")).alias(c)
): _*).writeStream.format("csv").start(path)
but if you want something closer to the current solution skip RDD conversion
import org.apache.spark.sql.catalyst.encoders.RowEncoder
eventDataset.rdd.map(event =>
Row.fromSeq(columns.map(c => event.getOrElse(c,"")))
)(RowEncoder(schema)).writeStream.format("csv").start(path)
Upvotes: 2