I have a DataFrame called source, a table from mysql
val source = sqlContext.read.jdbc(jdbcUrl, "source", connectionProperties)
I have converted it to rdd by
val sourceRdd = source.rdd
but its RDD[Row] I need RDD[String]
to do transformations like
source.map(rec => (rec.split(",")(0).toInt, rec)), .subtractByKey(), etc..
Thank you
You can use Row. mkString(sep: String): String method in a map call like this :
val sourceRdd = source.rdd.map(_.mkString(","))
You can change the "," parameter by whatever you want.
Hope this help you, Best Regards.
What is your schema?
If it's just a String, you can use:
import spark.implicits._
val sourceDS = source.as[String]
val sourceRdd = sourceDS.rdd // will give RDD[String]
Note: use sqlContext instead of spark in Spark 1.6 - spark is a SparkSession, which is a new class in Spark 2.0 and is a new entry point to SQL functionality. It should be used instead of SQLContext in Spark 2.x
You can also create own case classes.
Also you can map rows - here source is of type DataFrame, we use partial function in map function:
val sourceRdd = source.rdd.map { case x : Row => x(0).asInstanceOf[String] }.map(s => s.split(","))
Related
I have a library in Scala for Spark which contains many functions.
One example is the following function to unite two dataframes that have different columns:
def appendDF(df2: DataFrame): DataFrame = {
val cols1 = df.columns.toSeq
val cols2 = df2.columns.toSeq
def expr(sourceCols: Seq[String], targetCols: Seq[String]): Seq[Column] = {
targetCols.map({
case x if sourceCols.contains(x) => col(x)
case y => lit(null).as(y)
})
}
// both df's need to pass through `expr` to guarantee the same order, as needed for correct unions.
df.select(expr(cols1, cols1): _*).union(df2.select(expr(cols2, cols1): _*))
}
I would like to use this function (and many more) to Dataset[CleanRow] and not DataFrames. CleanRow is a simple class here that defines the names and types of the columns.
My educated guess is to convert the Dataset into Dataframe using .toDF() method. However, I would like to know whether there are better ways to do it.
From my understanding, there shouldn't be many differences between Dataset and Dataframe since Dataset are just Dataframe[Row]. Plus, I think that from Spark 2.x the APIs for DF and DS have been unified, so I was thinking that I could pass either of them interchangeably, but that's not the case.
If changing signature is possible:
import spark.implicits._
import org.apache.spark.sql.DataFrame
import org.apache.spark.sql.Dataset
def f[T](d: Dataset[T]): Dataset[T] = {d}
// You are able to pass a dataframe:
f(Seq(0,1).toDF()).show
// res1: org.apache.spark.sql.Dataset[org.apache.spark.sql.Row] = [value: int]
// You are also able to pass a dataset:
f(spark.createDataset(Seq(0,1)))
// res2: org.apache.spark.sql.Dataset[Int] = [value: int]
ErrorHi I am trying to a new column to a Spark. I am trying in a data set where I want to add the percentage made by in all games.
The data set looks like this:
Name, Platform, Year, Genre, Publisher, NA_Sales, EU_Sales, JP_Sales, Other_Sales
val vgdataLines = sc.textFile("hdfs:///user/ashhall1616/bdc_data/t1/vgsales-small.csv")
val vgdata = vgdataLines.map(_.split(";"))
def toPercentage(x: Double): Double = {x * 100} val countPubl = vgdata.map(r => (r(4),1)).reduceByKey(_+_)
val addpercen = countPubl.withColumn("count", toPercentage($"count"/countPubl.count(_._2)))
I used withColumn() to add new column 'count' and expected output to be like:
(Ubisoft,3,15.0)
Can anyone tell whats wrong here?
You cannot use "withColumn" with an RDD.
You could do as follow
val addpercen = countPubl.map({case(key, value) => (key, value, toPercentage(value))})
use map to add a calculated value as new column and convert to a DataFrame if you want
import spark.implicits._
val myDf = addpercen.toDF("key","value","myNewColumn")
myDf.show()
Hope it helps.
You can not use withColumn with an RDD hence convert it to DataFrame as below and then use it
val countPubl : DataFrame = vgdata.map(r => (r(4),1)).reduceByKey(_+_).toDF()
If you still looking to use RDD then just converto it back to RDD once you add the with column as
val javaRdd : JavaRDD[Row] = countPubl.withColumn("...",col("...")).toJavaRDD
In our application, we are connecting spark with HBase, using the following code:
val hBaseRDD: RDD[(ImmutableBytesWritable, Result)] =
sparkSession.sparkContext.newAPIHadoopRDD(
conf,
classOf[TableInputFormat],
classOf[ImmutableBytesWritable],
classOf[Result]
)
val resultRDD: RDD[Result] = hBaseRDD.map(tuple => tuple._2)
But this provides us with an RDD of type Result.
We need RDD of type 'Row' to create DataFrame out of this RDD.
How can we do the same?
Thanks
In EMR Spark, I have a HadoopRDD
org.apache.spark.rdd.RDD[(org.apache.hadoop.io.Text, org.apache.hadoop.dynamodb.DynamoDBItemWritable)] = HadoopRDD[0] at hadoopRDD
I want to convert this to DataFrame org.apache.spark.sql.DataFrame.
Does anyone know how to do this?
First convert it to simple types. Let's say your DynamoDBItemWritable has just one string column:
val simple: RDD[(String, String)] = rdd.map {
case (text, dbwritable) => (text.toString, dbwritable.getString(0))
}
Then you can use toDF to get a DataFrame:
import sqlContext.implicits._
val df: DataFrame = simple.toDF()
The Spark documentation shows how to create a DataFrame from an RDD, using Scala case classes to infer a schema. I am trying to reproduce this concept using sqlContext.createDataFrame(RDD, CaseClass), but my DataFrame ends up empty. Here's my Scala code:
// sc is the SparkContext, while sqlContext is the SQLContext.
// Define the case class and raw data
case class Dog(name: String)
val data = Array(
Dog("Rex"),
Dog("Fido")
)
// Create an RDD from the raw data
val dogRDD = sc.parallelize(data)
// Print the RDD for debugging (this works, shows 2 dogs)
dogRDD.collect().foreach(println)
// Create a DataFrame from the RDD
val dogDF = sqlContext.createDataFrame(dogRDD, classOf[Dog])
// Print the DataFrame for debugging (this fails, shows 0 dogs)
dogDF.show()
The output I'm seeing is:
Dog(Rex)
Dog(Fido)
++
||
++
||
||
++
What am I missing?
Thanks!
All you need is just
val dogDF = sqlContext.createDataFrame(dogRDD)
Second parameter is part of Java API and expects you class follows java beans convention (getters/setters). Your case class doesn't follow this convention, so no property is detected, that leads to empty DataFrame with no columns.
You can create a DataFrame directly from a Seq of case class instances using toDF as follows:
val dogDf = Seq(Dog("Rex"), Dog("Fido")).toDF
Case Class Approach won't Work in cluster mode. It'll give ClassNotFoundException to the case class you defined.
Convert it a RDD[Row] and define the schema of your RDD with StructField and then createDataFrame like
val rdd = data.map { attrs => Row(attrs(0),attrs(1)) }
val rddStruct = new StructType(Array(StructField("id", StringType, nullable = true),StructField("pos", StringType, nullable = true)))
sqlContext.createDataFrame(rdd,rddStruct)
toDF() wont work either