I want to create parquet file in hdfs and then read it through hive as external table. I'm struck with stage failures in spark-shell while writing parquet files.
Spark Version: 1.5.2
Scala Version: 2.10.4
Java: 1.7
Input file:(employee.txt)
1201,satish,25
1202,krishna,28
1203,amith,39
1204,javed,23
1205,prudvi,23
In Spark-Shell:
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)
val employee = sc.textFile("employee.txt")
employee.first()
val schemaString = "id name age"
import org.apache.spark.sql.Row;
import org.apache.spark.sql.types.{StructType, StructField, StringType};
val schema = StructType(schemaString.split(" ").map(fieldName ⇒ StructField(fieldName, StringType, true)))
val rowRDD = employee.map(_.split(",")).map(e ⇒ Row(e(0).trim.toInt, e(1), e(2).trim.toInt))
val employeeDF = sqlContext.createDataFrame(rowRDD, schema)
val finalDF = employeeDF.toDF();
sqlContext.setConf("spark.sql.parquet.compression.codec", "snappy")
var WriteParquet= finalDF.write.parquet("/user/myname/schemaParquet")
When I type the last command I get,
ERROR
SPARK APPLICATION MANAGER
I even tried increasing the executor memory, its still failing.
Also Importantly , finalDF.show() is producing the same error.
So, I believe I have made a logical error here.
Thanks for supporting
The issue here is you are creating a schema with all the fields/columns type defaulted to StringType. But while passing the values in the schema, the value of Id and Age is being converted to Integer as per the code.Hence, throwing the Matcherror while running.
The data types of columns in the schema should match the data type of values being passed to it. Try the below code.
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)
val employee = sc.textFile("employee.txt")
employee.first()
//val schemaString = "id name age"
import org.apache.spark.sql.Row;
import org.apache.spark.sql.types._;
val schema = StructType(StructField("id", IntegerType, true) :: StructField("name", StringType, true) :: StructField("age", IntegerType, true) :: Nil)
val rowRDD = employee.map(_.split(" ")).map(e ⇒ Row(e(0).trim.toInt, e(1), e(2).trim.toInt))
val employeeDF = sqlContext.createDataFrame(rowRDD, schema)
val finalDF = employeeDF.toDF();
sqlContext.setConf("spark.sql.parquet.compression.codec", "snappy")
var WriteParquet= finalDF.write.parquet("/user/myname/schemaParquet")
This code should run fine.
Related
I am trying to convert following code to run on the spark 1.6 but, on which I am facing certain issues. while converting the sparksession to context
object TestData {
def makeIntegerDf(spark: SparkSession, numbers: Seq[Int]): DataFrame =
spark.createDataFrame(
spark.sparkContext.makeRDD(numbers.map(Row(_))),
StructType(List(StructField("column", IntegerType, nullable = false)))
)
}
How Do I convert it to make it run on spark 1.6
SparkSession is supported from spark 2.0 on-wards only. So if you want to use spark 1.6 then you would need to create SparkContext and sqlContext in driver class and pass them to the function.
so you can create
val conf = new SparkConf().setAppName("simple")
val sparkContext = new SparkContext(conf)
val sqlContext = new SQLContext(sparkContext)
and then call the function as
val callFunction = makeIntegerDf(sparkContext, sqlContext, numbers)
And your function should be as
def makeIntegerDf(sparkContext: SparkContext, sqlContext: SQLContext, numbers: Seq[Int]): DataFrame =
sqlContext.createDataFrame(
sparkContext.makeRDD(numbers.map(Row(_))),
StructType(List(StructField("column", IntegerType, nullable = false)))
)
The only main difference here is the use of spark which is a spark session as opposed to spark context.
So you would do something like this:
object TestData {
def makeIntegerDf(sc: SparkContext, sqlContext: SQLContext, numbers: Seq[Int]): DataFrame =
sqlContext.createDataFrame(
sc.makeRDD(numbers.map(Row(_))),
StructType(List(StructField("column", IntegerType, nullable = false)))
)
}
Of course you would need to create a spark context instead of spark session in order to provide it to the function.
I have created a schema with following code
val schema= new StructType().add("city", StringType, true).add("female", IntegerType, true).add("male", IntegerType, true)
Created a RDD from
val data = spark.sparkContext.textFile("cities.txt")
Converted to RDD of Row to apply schema
val cities = data.map(line => line.split(";")).map(row => Row.fromSeq(row.zip(schema.toSeq)))
val citiesRDD = spark.sqlContext.createDataFrame(cities, schema)
This gives me an error
java.lang.RuntimeException: Error while encoding: java.lang.RuntimeException: scala.Tuple2 is not a valid external type for schema of string
You don't need a schema to create a Row, you need the schema when you create the DataFrame. You also need to introduce some logic how to convert your splitted line (which produces 3 strings) into integers:
here a minimal solution without exception-handling:
val data = sc.parallelize(Seq("Bern;10;12")) // mock for real data
val schema = new StructType().add("city", StringType, true).add("female", IntegerType, true).add("male", IntegerType, true)
val cities = data.map(line => {
val Array(city,female,male) = line.split(";")
Row(
city,
female.toInt,
male.toInt
)
}
)
val citiesDF = sqlContext.createDataFrame(cities, schema)
I normally use case-classes to create a dataframe, because spark can infer the schema from the case class:
// "schema" for dataframe, define outside of main method
case class MyRow(city:Option[String],female:Option[Int],male:Option[Int])
val data = sc.parallelize(Seq("Bern;10;12")) // mock for real data
import sqlContext.implicits._
val citiesDF = data.map(line => {
val Array(city,female,male) = line.split(";")
MyRow(
Some(city),
Some(female.toInt),
Some(male.toInt)
)
}
).toDF()
I'm building a schema that is rather large so I am using the example of progamatical schema creation from the documentation.
val schemaString = "field1,...,field126"
val schema = StructType(schemaString.split(",").map(fieldName => StructField(fieldName.trim, StringType, true)))
This works fine but I need to have all fields as DoubleType for my ML function. I changed the StringType to DoubleType and I get an error.
val schemaString = "field1,...,field126"
val schema = StructType(schemaString.split(",").map(fieldName => StructField(fieldName.trim, DoubleType, true)))
Error:
Exception in thread "main" java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Double
at scala.runtime.BoxesRunTime.unboxToDouble(BoxesRunTime.java:119)
I know I can shift to creating the schema manually but with 126 fields the code gets bulky.
val schema = new StructType()
.add("ColumnA", IntegerType)
.add("ColumnB", StringType)
val df = sqlContext.read
.schema(schema)
.format("com.databricks.spark.csv")
.delimiter(",")
.load("/path/to/file.csv")
I think there is no need to pass your own schema , It will infer it automatically , if your csv file contains the name of the columns then it will take it too if you set the header as true.
This will work simply(not-tested) :
val df = sqlContext.read
.format("com.databricks.spark.csv")
.option("header", "true")
.option("inferSchema", "true")
.load("data/sample.csv")
It will give you a dataframe and if you have the column name to then just set header as true !
I have a text file on HDFS and I want to convert it to a Data Frame in Spark.
I am using the Spark Context to load the file and then try to generate individual columns from that file.
val myFile = sc.textFile("file.txt")
val myFile1 = myFile.map(x=>x.split(";"))
After doing this, I am trying the following operation.
myFile1.toDF()
I am getting an issues since the elements in myFile1 RDD are now array type.
How can I solve this issue?
Update - as of Spark 1.6, you can simply use the built-in csv data source:
spark: SparkSession = // create the Spark Session
val df = spark.read.csv("file.txt")
You can also use various options to control the CSV parsing, e.g.:
val df = spark.read.option("header", "false").csv("file.txt")
For Spark version < 1.6:
The easiest way is to use spark-csv - include it in your dependencies and follow the README, it allows setting a custom delimiter (;), can read CSV headers (if you have them), and it can infer the schema types (with the cost of an extra scan of the data).
Alternatively, if you know the schema you can create a case-class that represents it and map your RDD elements into instances of this class before transforming into a DataFrame, e.g.:
case class Record(id: Int, name: String)
val myFile1 = myFile.map(x=>x.split(";")).map {
case Array(id, name) => Record(id.toInt, name)
}
myFile1.toDF() // DataFrame will have columns "id" and "name"
I have given different ways to create DataFrame from text file
val conf = new SparkConf().setAppName(appName).setMaster("local")
val sc = SparkContext(conf)
raw text file
val file = sc.textFile("C:\\vikas\\spark\\Interview\\text.txt")
val fileToDf = file.map(_.split(",")).map{case Array(a,b,c) =>
(a,b.toInt,c)}.toDF("name","age","city")
fileToDf.foreach(println(_))
spark session without schema
import org.apache.spark.sql.SparkSession
val sparkSess =
SparkSession.builder().appName("SparkSessionZipsExample")
.config(conf).getOrCreate()
val df = sparkSess.read.option("header",
"false").csv("C:\\vikas\\spark\\Interview\\text.txt")
df.show()
spark session with schema
import org.apache.spark.sql.types._
val schemaString = "name age city"
val fields = schemaString.split(" ").map(fieldName => StructField(fieldName,
StringType, nullable=true))
val schema = StructType(fields)
val dfWithSchema = sparkSess.read.option("header",
"false").schema(schema).csv("C:\\vikas\\spark\\Interview\\text.txt")
dfWithSchema.show()
using sql context
import org.apache.spark.sql.SQLContext
val fileRdd =
sc.textFile("C:\\vikas\\spark\\Interview\\text.txt").map(_.split(",")).map{x
=> org.apache.spark.sql.Row(x:_*)}
val sqlDf = sqlCtx.createDataFrame(fileRdd,schema)
sqlDf.show()
If you want to use the toDF method, you have to convert your RDD of Array[String] into a RDD of a case class. For example, you have to do:
case class Test(id:String,filed2:String)
val myFile = sc.textFile("file.txt")
val df= myFile.map( x => x.split(";") ).map( x=> Test(x(0),x(1)) ).toDF()
You will not able to convert it into data frame until you use implicit conversion.
val sqlContext = new SqlContext(new SparkContext())
import sqlContext.implicits._
After this only you can convert this to data frame
case class Test(id:String,filed2:String)
val myFile = sc.textFile("file.txt")
val df= myFile.map( x => x.split(";") ).map( x=> Test(x(0),x(1)) ).toDF()
val df = spark.read.textFile("abc.txt")
case class Abc (amount:Int, types: String, id:Int) //columns and data types
val df2 = df.map(rec=>Amount(rec(0).toInt, rec(1), rec(2).toInt))
rdd2.printSchema
root
|-- amount: integer (nullable = true)
|-- types: string (nullable = true)
|-- id: integer (nullable = true)
A txt File with PIPE (|) delimited file can be read as :
df = spark.read.option("sep", "|").option("header", "true").csv("s3://bucket_name/folder_path/file_name.txt")
I know I am quite late to answer this but I have come up with a different answer:
val rdd = sc.textFile("/home/training/mydata/file.txt")
val text = rdd.map(lines=lines.split(",")).map(arrays=>(ararys(0),arrays(1))).toDF("id","name").show
You can read a file to have an RDD and then assign schema to it. Two common ways to creating schema are either using a case class or a Schema object [my preferred one]. Follows the quick snippets of code that you may use.
Case Class approach
case class Test(id:String,name:String)
val myFile = sc.textFile("file.txt")
val df= myFile.map( x => x.split(";") ).map( x=> Test(x(0),x(1)) ).toDF()
Schema Approach
import org.apache.spark.sql.types._
val schemaString = "id name"
val fields = schemaString.split(" ").map(fieldName => StructField(fieldName, StringType, nullable=true))
val schema = StructType(fields)
val dfWithSchema = sparkSess.read.option("header","false").schema(schema).csv("file.txt")
dfWithSchema.show()
The second one is my preferred approach since case class has a limitation of max 22 fields and this will be a problem if your file has more than 22 fields!
In the previous version, we used to have a 'saveAsOrcFile()' method on RDD. This is now gone! How do I save data in DataFrame in ORC File format?
def main(args: Array[String]) {
println("Creating Orc File!")
val sparkConf = new SparkConf().setAppName("orcfile")
val sc = new SparkContext(sparkConf)
val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)
val people = sc.textFile("/apps/testdata/people.txt")
val schemaString = "name age"
val schema = StructType(schemaString.split(" ").map(fieldName => {if(fieldName == "name") StructField(fieldName, StringType, true) else StructField(fieldName, IntegerType, true)}))
val rowRDD = people.map(_.split(",")).map(p => Row(p(0), new Integer(p(1).trim)))
//# Infer table schema from RDD**
val peopleSchemaRDD = hiveContext.createDataFrame(rowRDD, schema)
//# Create a table from schema**
peopleSchemaRDD.registerTempTable("people")
val results = hiveContext.sql("SELECT * FROM people")
results.map(t => "Name: " + t.toString).collect().foreach(println)
// Now I want to save this Dataframe(peopleSchemaRDD) in ORC Format. How do I do that?
}
Since Spark 1.4 you can simply use DataFrameWriter and set format to orc:
peopleSchemaRDD.write.format("orc").save("people")
or
peopleSchemaRDD.write.orc("people")