Convert csv file to dataframe in Spark 1.5.2 without databricks - scala

Question

I am trying to convert a csv file to a dataframe in Spark 1.5.2 with Scala without the use of the library databricks, as it is a community project and this library is not available. My approach was the following:
var inputPath = "input.csv"
var text = sc.textFile(inputPath)
var rows = text.map(line => line.split(",").map(_.trim))
var header = rows.first()
var data = rows.filter(_(0) != header(0))
var df = sc.makeRDD(1 to data.count().toInt).map(i => (data.take(i).drop(i-1)(0)(0), data.take(i).drop(i-1)(0)(1), data.take(i).drop(i-1)(0)(2), data.take(i).drop(i-1)(0)(3), data.take(i).drop(i-1)(0)(4))).toDF(header(0), header(1), header(2), header(3), header(4))
This code, even though it is quite a mess, works without returning any error messages. The problem comes when trying to display the data inside dfin order to verify the correctness of this method and later try to do some queries in df. The error code I am getting after executing df.show() is SPARK-5063. My questions are:
1) Why is it not possible to print the content of df?
2) Is there any other more straightforward method to convert a csv to a dataframe in Spark 1.5.2 without using the library databricks?

Answer 1

For spark 1.5.x can be used code snippet below to convert input into DF
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
// this is used to implicitly convert an RDD to a DataFrame.
import sqlContext.implicits._
// Define the schema using a case class.
// Note: Case classes in Scala 2.10 can support only up to 22 fields. To work around this limit,
// you can use custom classes that implement the DataClass interface with 5 fields.
case class DataClass(id: Int, name: String, surname: String, bdate: String, address: String)
// Create an RDD of DataClass objects and register it as a table.
val peopleData = sc.textFile("input.csv").map(_.split(",")).map(p => DataClass(p(0).trim.toInt, p(1).trim, p(2).trim, p(3).trim, p(4).trim)).toDF()
peopleData.registerTempTable("dataTable")
val peopleDataFrame = sqlContext.sql("SELECT * from dataTable")
peopleDataFrame.show()
Spark 1.5

Answer 2

You can create like this:
SparkSession spark = SparkSession
.builder()
.appName("RDDtoDF_Updated")
.master("local[2]")
.config("spark.some.config.option", "some-value")
.getOrCreate();
StructType schema = DataTypes
.createStructType(new StructField[] {
DataTypes.createStructField("eid", DataTypes.IntegerType, false),
DataTypes.createStructField("eName", DataTypes.StringType, false),
DataTypes.createStructField("eAge", DataTypes.IntegerType, true),
DataTypes.createStructField("eDept", DataTypes.IntegerType, true),
DataTypes.createStructField("eSal", DataTypes.IntegerType, true),
DataTypes.createStructField("eGen", DataTypes.StringType,true)});
String filepath = "F:/Hadoop/Data/EMPData.txt";
JavaRDD<Row> empRDD = spark.read()
.textFile(filepath)
.javaRDD()
.map(line -> line.split("\\,"))
.map(r -> RowFactory.create(Integer.parseInt(r[0]), r[1].trim(),Integer.parseInt(r[2]),
Integer.parseInt(r[3]),Integer.parseInt(r[4]),r[5].trim() ));
Dataset<Row> empDF = spark.createDataFrame(empRDD, schema);
empDF.groupBy("edept").max("esal").show();

Answer 3

Using Spark with Scala.
import org.apache.spark.sql.Row
import org.apache.spark.sql.types._
var hiveCtx = new HiveContext(sc)
var inputPath = "input.csv"
var text = sc.textFile(inputPath)
var rows = text.map(line => line.split(",").map(_.trim)).map(a => Row.fromSeq(a))
var header = rows.first()
val schema = StructType(header.map(fieldName => StructField(fieldName.asInstanceOf[String],StringType,true)))
val df = hiveCtx.createDataframe(rows,schema)
This should work.
But for creating dataframe, would recommend you to use Spark-CSV.