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Create dataframe from list instead of RDD #363

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Create dataframe from list instead of RDD #363

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16 changes: 11 additions & 5 deletions spark-sql-application/src/main/scala/org/apache/spark/sql/FlintJobExecutor.scala
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Original file line number Diff line number Diff line change
Expand Up @@ -7,6 +7,8 @@ package org.apache.spark.sql

import java.util.Locale

import scala.collection.JavaConverters._

import com.amazonaws.services.glue.model.{AccessDeniedException, AWSGlueException}
import com.amazonaws.services.s3.model.AmazonS3Exception
import com.fasterxml.jackson.databind.ObjectMapper
Expand Down Expand Up @@ -177,12 +179,16 @@ trait FlintJobExecutor {
startTime: Long,
timeProvider: TimeProvider,
cleaner: Cleaner): DataFrame = {
// Create the schema dataframe
val schemaRows = result.schema.fields.map { field =>
Row(field.name, field.dataType.typeName)
}
// collect schema rows as List
val schemaRows = result.schema.fields
.map { field =>
Row(field.name, field.dataType.typeName)
}
.toList
.asJava

val resultSchema = spark.createDataFrame(
spark.sparkContext.parallelize(schemaRows),
schemaRows,
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will this process all data on driver node?

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yes, only schema.

StructType(
Seq(
StructField("column_name", StringType, nullable = false),
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