I am getting below exception if I do join in between two dataframes in spark (ver 1.5, scala 2.10).
Exception in thread "main" org.apache.spark.sql.AnalysisException: syntax error in attribute name: col1.;
at org.apache.spark.sql.catalyst.analysis.UnresolvedAttribute$.e$1(unresolved.scala:99)
at org.apache.spark.sql.catalyst.analysis.UnresolvedAttribute$.parseAttributeName(unresolved.scala:118)
at org.apache.spark.sql.catalyst.plans.logical.LogicalPlan.resolveQuoted(LogicalPlan.scala:182)
at org.apache.spark.sql.DataFrame.resolve(DataFrame.scala:158)
at org.apache.spark.sql.DataFrame.col(DataFrame.scala:653)
at com.nielsen.buy.integration.commons.Demo$.main(Demo.scala:62)
at com.nielsen.buy.integration.commons.Demo.main(Demo.scala)
Code works fine if column in dataframe does not contain any period . Please do help me out.
You can find the code that I am using.
import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.sql.SQLContext
import com.google.gson.Gson
import org.apache.spark.sql.types.StructType
import org.apache.spark.sql.types.StructField
import org.apache.spark.sql.types.StringType
import org.apache.spark.sql.Row
object Demo
{
lazy val sc: SparkContext = {
val conf = new SparkConf().setMaster("local")
.setAppName("demooo")
.set("spark.driver.allowMultipleContexts", "true")
new SparkContext(conf)
}
sc.setLogLevel("ERROR")
lazy val sqlcontext=new SQLContext(sc)
val data=List(Row("a","b"),Row("v","b"))
val dataRdd=sc.parallelize(data)
val schema = new StructType(Array(StructField("col.1",StringType,true),StructField("col2",StringType,true)))
val df1=sqlcontext.createDataFrame(dataRdd, schema)
val data2=List(Row("a","b"),Row("v","b"))
val dataRdd2=sc.parallelize(data2)
val schema2 = new StructType(Array(StructField("col3",StringType,true),StructField("col4",StringType,true)))
val df2=sqlcontext.createDataFrame(dataRdd2, schema2)
val val1="col.1"
val df3= df1.join(df2,df1.col(val1).equalTo(df2.col("col3")),"outer").show
}
In general, period is used to access members of a struct field.
The spark version you are using (1.5) is relatively old. Several such issues were fixed in later versions so if you upgrade it might just solve the issue.
That said, you can simply use withColumnRenamed to rename the column to something which does not have a period before the join.
So you basically do something like this:
val dfTmp = df1.withColumnRenamed(val1, "JOIN_COL")
val df3= dfTmp.join(df2,dfTmp.col("JOIN_COL").equalTo(df2.col("col3")),"outer").withColumnRenamed("JOIN_COL", val1)
df3.show
btw show returns a Unit so you probably meant df3 to be equal to the expression without it and do df3.show separately.
Related
what is the most efficient way to sort one column in data frame, convert it to list, and assign the first element to variable in scala. I tried the following
import org.apache.spark.SparkConf
import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.functions.{col, first, regexp_replace}
import org.apache.spark.sql.functions._
println(CONFIG.getString("spark.appName"))
val conf = new SparkConf()
.setAppName(CONFIG.getString("spark.appName"))
.setMaster(CONFIG.getString("spark.master"))
val spark: SparkSession = SparkSession.builder().config(conf).getOrCreate()
val df = spark.read.format("com.databricks.spark.csv").option("delimiter", ",").load("file.csv")
val dfb=df.sort(desc("_c0"))
val list=df.select(df("_c0")).distinct
but I'm still no able to save the first element as variable
Use select, orderBy, map & head
Assuming column _c0 is of type string, If it is different type you have to modify your column data type in _.getAs[<your column datatype>]
Check below code.
scala> import spark.implicits._
import spark.implicits._
scala> val first = df
.select($"_c0")
.orderBy($"_c0".desc)
.map(_.getAs[String](0))
.head
Or
scala> import spark.implicits._
import spark.implicits._
scala> val first = df
.select($"_c0")
.orderBy($"_c0".desc)
.head
.getAs[String](0)
I am new to Scala and I ran into the error while doing some practice.
I tried to convert RDD into DataFrame and following is my code.
package com.sclee.examples
import com.sun.org.apache.xalan.internal.xsltc.compiler.util.IntType
import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.sql.Row
import org.apache.spark.sql.types.{LongType, StringType, StructField, StructType};
object App {
def main(args: Array[String]): Unit = {
val conf = new SparkConf().setAppName("examples").setMaster("local")
val sc = new SparkContext(conf)
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext.implicits._
case class Person(name: String, age: Long)
val personRDD = sc.makeRDD(Seq(Person("A",10),Person("B",20)))
val df = personRDD.map({
case Row(val1: String, val2: Long) => Person(val1,val2)
}).toDS()
// val ds = personRDD.toDS()
}
}
I followed the instructions in Spark documentation and also referenced some blogs showing me how to convert rdd into dataframe but the I got the error below.
Error:(20, 27) Unable to find encoder for type stored in a Dataset. Primitive types (Int, String, etc) and Product types (case classes) are supported by importing sqlContext.implicits._ Support for serializing other types will be added in future releases.
val df = personRDD.map({
Although I tried to fix the problem by myself but failed. Any help will be appreciated.
The following code works:
import org.apache.spark.rdd.RDD
import org.apache.spark.sql.SparkSession
case class Person(name: String, age: Long)
object SparkTest {
def main(args: Array[String]): Unit = {
// use the SparkSession of Spark 2
val spark = SparkSession
.builder()
.appName("Spark SQL basic example")
.config("spark.some.config.option", "some-value")
.getOrCreate()
import spark.implicits._
// this your RDD - just a sample how to create an RDD
val personRDD: RDD[Person] = spark.sparkContext.parallelize(Seq(Person("A",10),Person("B",20)))
// the sparksession has a method to convert to an Dataset
val ds = spark.createDataset(personRDD)
println(ds.count())
}
}
I made the following changes:
use SparkSession instead of SparkContext and SqlContext
move Person class out of the App (I'm not sure why I had to do
this)
use createDataset for conversion
However, I guess it's pretty uncommon to do this conversion and you probably want to read your input directly into an Dataset using the read method
Using Spark 1.6.1 I want to call the number of times a UDF is called. I want to do this because I have a very expensive UDF (~1sec per call) and I suspect the UDF being called more often than the number of records in my dataframe, making my spark job slower than necessary.
Although I could not reproduce this situation, I came up with a simple example showing that the number of calls to the UDF seems to be different (here: less) than the number of rows, how can that be?
import org.apache.spark.sql.SQLContext
import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.sql.functions.udf
object Demo extends App {
val conf = new SparkConf().setMaster("local[4]").setAppName("Demo")
val sc = new SparkContext(conf)
sc.setLogLevel("WARN")
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
val callCounter = sc.accumulator(0)
val df= sc.parallelize(1 to 10000,numSlices = 100).toDF("value")
println(df.count) // gives 10000
val myudf = udf((d:Int) => {callCounter.add(1);d})
val res = df.withColumn("result",myudf($"value")).cache
println(res.select($"result").collect().size) // gives 10000
println(callCounter.value) // gives 9941
}
If using an accumulator is not the right way to call the counts of the UDF, how else could I do it?
Note: In my actual Spark-Job, get a call-count which is about 1.7 times higher than the actual number of records.
Spark applications should define a main() method instead of extending scala.App. Subclasses of scala.App may not work correctly.
import org.apache.spark.sql.SQLContext
import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.sql.functions.udf
object Demo extends App {
def main(args: Array[String]): Unit = {
val conf = new SparkConf().setAppName("Simple Application").setMaster("local[4]")
val sc = new SparkContext(conf)
// [...]
}
}
This should solve your problem.
I would like to use spark SQL in an Intellij IDEA SBT project.
Even though I have imported the library the code does not seem to import it.
Spark Core seems to be working however.
You can't create a DataFrame from a scala List[A]. You need first to create an RDD[A], and then transform that to a DataFrame. You also need an SQLContext:
val conf = new SparkConf()
.setMaster("local[*]")
.setAppName("test")
val sc = new SparkContext(conf)
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
val test = sc.parallelize(List(1,2,3,4)).toDF
For reference this is how the Spark 2.0 boilerplate with spark sql should look like:
import org.apache.spark.sql.SparkSession
object Test {
def main(args: Array[String]) {
val spark = SparkSession.builder()
.master("local")
.appName("some name")
.getOrCreate()
import spark.sqlContext.implicits._
}
}
I am a newbie with Apache spark as well with Scala programming language.
What I am trying to achieve is to extract the data from my local mongoDB database for then to save it in a parquet format using Apache Spark with the hadoop-connector
This is my code so far:
package com.examples
import org.apache.spark.{SparkContext, SparkConf}
import org.apache.spark.rdd.RDD
import org.apache.hadoop.conf.Configuration
import org.bson.BSONObject
import com.mongodb.hadoop.{MongoInputFormat, BSONFileInputFormat}
import org.apache.spark.sql
import org.apache.spark.sql.SQLContext
object DataMigrator {
def main(args: Array[String])
{
val conf = new SparkConf().setAppName("Migration App").setMaster("local")
val sc = new SparkContext(conf)
val sqlContext = new SQLContext(sc)
// Import statement to implicitly convert an RDD to a DataFrame
import sqlContext.implicits._
val mongoConfig = new Configuration()
mongoConfig.set("mongo.input.uri", "mongodb://localhost:27017/mongosails4.case")
val mongoRDD = sc.newAPIHadoopRDD(mongoConfig, classOf[MongoInputFormat], classOf[Object], classOf[BSONObject]);
val count = countsRDD.count()
// the count value is aprox 100,000
println("================ PRINTING =====================")
println(s"ROW COUNT IS $count")
println("================ PRINTING =====================")
}
}
The thing is that in order to save data to a parquet file format first its necessary to convert the mongoRDD variable to Spark DataFrame. I have tried something like this:
// convert RDD to DataFrame
val myDf = mongoRDD.toDF() // this lines throws an error
myDF.write.save("my/path/myData.parquet")
and the error I get is this:
Exception in thread "main" scala.MatchError: java.lang.Object (of class scala.reflect.internal.Types.$TypeRef$$anon$6)
do you guys have any other idea how could I convert the RDD to a DataFrame so that I can save data in parquet format?
Here's the structure of one Document in the mongoDB collection : https://gist.github.com/kingtrocko/83a94238304c2d654fe4
Create a Case class representing the data stored in your DBObject.
case class Data(x: Int, s: String)
Then, map the values of your rdd to instances of your case class.
val dataRDD = mongoRDD.values.map { obj => Data(obj.get("x"), obj.get("s")) }
Now with your RDD[Data], you can create a DataFrame with the sqlContext
val myDF = sqlContext.createDataFrame(dataRDD)
That should get you going. I can explain more later if needed.