I'm a table in cassandra tfm.foehis that have data.
When i did the first charge of data from spark to cassandra, I used this set of commands:
import org.apache.spark.sql.functions._
import com.datastax.spark.connector._
import org.apache.spark.sql.cassandra._
val wkdir="/home/adminbigdata/tablas/"
val fileIn= "originales/22_FOEHIS2.csv"
val fileOut= "22_FOEHIS_PRE2"
val fileCQL= "22_FOEHISCQL"
val data = sc.textFile(wkdir + fileIn).filter(!_.contains("----")).map(_.trim.replaceAll(" +", "")).map(_.dropRight(1)).map(_.drop(1)).map(_.replaceAll(",", "")).filter(array => array(6) != "MOBIDI").filter(array => array(17) != "").saveAsTextFile(wkdir + fileOut)
val firstDF = spark.read.format("csv").option("header", "true").option("inferSchema", "true").option("mode", "DROPMALFORMED").option("delimiter", "|").load(wkdir + fileOut)
val columns: Array[String] = firstDF.columns
val reorderedColumnNames: Array[String] = Array("hoclic","hodtac","hohrac","hotpac","honrac","hocdan","hocdrs","hocdsl","hocol","hocpny","hodesf","hodtcl","hodtcm","hodtea","hodtra","hodtrc","hodtto","hodtua","hohrcl","hohrcm","hohrea","hohrra","hohrrc","hohrua","holinh","holinr","honumr","hoobs","hooe","hotdsc","hotour","housca","houscl","houscm","housea","houser","housra","housrc")
val secondDF= firstDF.select(reorderedColumnNames.head, reorderedColumnNames.tail: _*)
secondDF.write.cassandraFormat("foehis", "tfm").save()
But when I load new data using the same script, I get errors. I don't know what's wrong?
This is the message:
java.lang.UnsupportedOperationException: 'SaveMode is set to ErrorIfExists and Table
tfm.foehis already exists and contains data.
Perhaps you meant to set the DataFrame write mode to Append?
Example: df.write.format.options.mode(SaveMode.Append).save()" '
The error message clearly says you that you need to use Append mode & shows what you can do with it. In your case it happens because destination table already exists, and writing mode is set to "error if exists". If you still want to write data, the code should be following:
import org.apache.spark.sql.SaveMode
secondDF.write.cassandraFormat("foehis", "tfm").mode(SaveMode.Append).save()
Related
I am interested in being able to retrieve the location value of a Hive table given a Spark object (SparkSession). One way to obtain this value is by parsing the output of the location via the following SQL query:
describe formatted <table name>
I was wondering if there is another way to obtain the location value without having to parse the output. An API would be great in case the output of the above command changes between Hive versions. If an external dependency is needed, which would it be? Is there some sample spark code that can obtain the location value?
Here is the correct answer:
import org.apache.spark.sql.catalyst.TableIdentifier
lazy val tblMetadata = spark.sessionState.catalog.getTableMetadata(new TableIdentifier(tableName,Some(schema)))
You can also use .toDF method on desc formatted table then filter from dataframe.
DataframeAPI:
scala> :paste
spark.sql("desc formatted data_db.part_table")
.toDF //convert to dataframe will have 3 columns col_name,data_type,comment
.filter('col_name === "Location") //filter on colname
.collect()(0)(1)
.toString
Result:
String = hdfs://nn:8020/location/part_table
(or)
RDD Api:
scala> :paste
spark.sql("desc formatted data_db.part_table")
.collect()
.filter(r => r(0).equals("Location")) //filter on r(0) value
.map(r => r(1)) //get only the location
.mkString //convert as string
.split("8020")(1) //change the split based on your namenode port..etc
Result:
String = /location/part_table
First approach
You can use input_file_name with dataframe.
it will give you absolute file-path for a part file.
spark.read.table("zen.intent_master").select(input_file_name).take(1)
And then extract table path from it.
Second approach
Its more of hack you can say.
package org.apache.spark.sql.hive
import java.net.URI
import org.apache.spark.sql.catalyst.catalog.{InMemoryCatalog, SessionCatalog}
import org.apache.spark.sql.catalyst.parser.ParserInterface
import org.apache.spark.sql.internal.{SessionState, SharedState}
import org.apache.spark.sql.SparkSession
class TableDetail {
def getTableLocation(table: String, spark: SparkSession): URI = {
val sessionState: SessionState = spark.sessionState
val sharedState: SharedState = spark.sharedState
val catalog: SessionCatalog = sessionState.catalog
val sqlParser: ParserInterface = sessionState.sqlParser
val client = sharedState.externalCatalog match {
case catalog: HiveExternalCatalog => catalog.client
case _: InMemoryCatalog => throw new IllegalArgumentException("In Memory catalog doesn't " +
"support hive client API")
}
val idtfr = sqlParser.parseTableIdentifier(table)
require(catalog.tableExists(idtfr), new IllegalArgumentException(idtfr + " done not exists"))
val rawTable = client.getTable(idtfr.database.getOrElse("default"), idtfr.table)
rawTable.location
}
}
Here is how to do it in PySpark:
(spark.sql("desc formatted mydb.myschema")
.filter("col_name=='Location'")
.collect()[0].data_type)
Use this as re-usable function in your scala project
def getHiveTablePath(tableName: String, spark: SparkSession):String =
{
import org.apache.spark.sql.functions._
val sql: String = String.format("desc formatted %s", tableName)
val result: DataFrame = spark.sql(sql).filter(col("col_name") === "Location")
result.show(false) // just for debug purpose
val info: String = result.collect().mkString(",")
val path: String = info.split(',')(1)
path
}
caller would be
println(getHiveTablePath("src", spark)) // you can prefix schema if you have
Result (I executed in local so file:/ below if its hdfs hdfs:// will come):
+--------+------------------------------------+-------+
|col_name|data_type |comment|
+--------+--------------------------------------------+
|Location|file:/Users/hive/spark-warehouse/src| |
+--------+------------------------------------+-------+
file:/Users/hive/spark-warehouse/src
USE ExternalCatalog
scala> spark
res15: org.apache.spark.sql.SparkSession = org.apache.spark.sql.SparkSession#4eba6e1f
scala> val metastore = spark.sharedState.externalCatalog
metastore: org.apache.spark.sql.catalyst.catalog.ExternalCatalog = org.apache.spark.sql.hive.HiveExternalCatalog#24b05292
scala> val location = metastore.getTable("meta_data", "mock").location
location: java.net.URI = hdfs://10.1.5.9:4007/usr/hive/warehouse/meta_data.db/mock
I have a file which is file1snappy.parquet. It is having a complex data structure like a map, array inside that.After processing that I got final result.while writing that results to csv I am getting some error saying
"Exception in thread "main" java.lang.UnsupportedOperationException: CSV data source does not support map<string,bigint> data type."
Code which I have used:
val conf=new SparkConf().setAppName("student-example").setMaster("local")
val sc = new SparkContext(conf)
val sqlcontext = new org.apache.spark.sql.SQLContext(sc)
val datadf = sqlcontext.read.parquet("C:\\file1.snappy.parquet")
def sumaggr=udf((aggr: Map[String, collection.mutable.WrappedArray[Long]]) => if (aggr.keySet.contains("aggr")) aggr("aggr").sum else 0)
datadf.select(col("neid"),sumaggr(col("marks")).as("sum")).filter(col("sum") =!= 0).show(false)
datadf.write.format("com.databricks.spark.csv").option("header", "true").save("C:\\myfile.csv")
I tried converting datadf.toString() but still I am facing same issue.
How can write that result to CSV.
spark version 2.1.1
Spark CSV source supports only atomic types. You cannot store any columns that are non-atomic
I think best is to create a JSON for the column that has map<string,bigint> as a datatype and save it in csv as below.
import spark.implicits._
import org.apache.spark.sql.functions._
datadf.withColumn("column_name_with_map_type", to_json(struct($"column_name_with_map_type"))).write.csv("outputpath")
Hope this helps!
You are trying to save the output of
val datadf = sqlcontext.read.parquet("C:\\file1.snappy.parquet")
which I guess is a mistake as the udf function and all the aggregation done would go in vain if you do so
So I think you want to save the output of
datadf.select(col("neid"),sumaggr(col("marks")).as("sum")).filter(col("sum") =!= 0).show(false)
So you need to save it in a new dataframe variable and use that variable to save.
val finalDF = datadf.select(col("neid"),sumaggr(col("marks")).as("sum")).filter(col("sum") =!= 0)
finalDF.write.format("com.databricks.spark.csv").option("header", "true").save("C:\\myfile.csv")
And you should be fine.
I am trying to querying a Hive table from a map operation in Spark, but when it run a query the execution getting frozen.
This is my test code
val sc = new SparkContext(conf)
val datasetPath = "npiCodesMin.csv"
val sparkSession = SparkSession.builder().enableHiveSupport().getOrCreate()
val df = sparkSession.read.option("header", true).option("sep", ",").csv(datasetPath)
df.createOrReplaceTempView("npicodesTmp")
sparkSession.sql("DROP TABLE IF EXISTS npicodes");
sparkSession.sql("CREATE TABLE npicodes AS SELECT * FROM npicodesTmp");
val res = sparkSession.sql("SELECT * FROM npicodes WHERE NPI = '1588667638'") //This works
println(res.first())
val NPIs = sc.parallelize(List("1679576722", "1588667638", "1306849450", "1932102084"))//Some existing NPIs
val rows = NPIs.mapPartitions{ partition =>
val sparkSession = SparkSession.builder().enableHiveSupport().getOrCreate()
partition.map{code =>
val res = sparkSession.sql("SELECT * FROM npicodes WHERE NPI = '"+code+"'")//The program stops here
res.first()
}
}
rows.collect().foreach(println)
It loads the data from a CSV, creates a new Hive table and fills it with the CSV data.
Then, if I query the table from the master it works perfectly, but if I try to do that in a map operation the execution getting frozen.
It do not generate any error, it continue running without do anything.
The Spark UI shows this situation
Actually, I am not sure if I can query a table in a distributed way, I cannot find it in the documentation.
Any suggestion?
Thanks.
I am working on a spark program that inserts dataframe into Hive Table as below.
import org.apache.spark.sql.SaveMode
import org.apache.spark.sql._
val hiveCont = val hiveCont = new org.apache.spark.sql.hive.HiveContext(sc)
val partfile = sc.textFile("partfile")
val partdata = partfile.map(p => p.split(","))
case class partc(id:Int, name:String, salary:Int, dept:String, location:String)
val partRDD = partdata.map(p => partc(p(0).toInt, p(1), p(2).toInt, p(3), p(4)))
val partDF = partRDD.toDF()
partDF.registerTempTable("party")
hiveCont.sql("insert into parttab select id, name, salary, dept from party")
I know that Spark V2 has come out and we can use SparkSession object in it.
Can we use SparkSession object to directly insert the dataframe into Hive table or do we have to use the HiveContext in version 2 also ? Can anyone let me know what is the major difference in version with respect to HiveContext ?
You can use your SparkSession (normally called spark or ss) directly to fire a sql query (make sure hive-support is enabled when creating the spark-session):
spark.sql("insert into parttab select id, name, salary, dept from party")
But I would suggest this notation, you don't need to create a temp-table etc:
partDF
.select("id","name","salary","dept")
.write.mode("overwrite")
.insertInto("parttab")
I know how to read a csv file into spark using spark-csv (https://github.com/databricks/spark-csv), but I already have the csv file represented as a string and would like to convert this string directly to dataframe. Is this possible?
Update : Starting from Spark 2.2.x
there is finally a proper way to do it using Dataset.
import org.apache.spark.sql.{Dataset, SparkSession}
val spark = SparkSession.builder().appName("CsvExample").master("local").getOrCreate()
import spark.implicits._
val csvData: Dataset[String] = spark.sparkContext.parallelize(
"""
|id, date, timedump
|1, "2014/01/01 23:00:01",1499959917383
|2, "2014/11/31 12:40:32",1198138008843
""".stripMargin.lines.toList).toDS()
val frame = spark.read.option("header", true).option("inferSchema",true).csv(csvData)
frame.show()
frame.printSchema()
Old spark versions
Actually you can, though it's using library internals and not widely advertised. Just create and use your own CsvParser instance.
Example that works for me on spark 1.6.0 and spark-csv_2.10-1.4.0 below
import com.databricks.spark.csv.CsvParser
val csvData = """
|userid,organizationid,userfirstname,usermiddlename,userlastname,usertitle
|1,1,user1,m1,l1,mr
|2,2,user2,m2,l2,mr
|3,3,user3,m3,l3,mr
|""".stripMargin
val rdd = sc.parallelize(csvData.lines.toList)
val csvParser = new CsvParser()
.withUseHeader(true)
.withInferSchema(true)
val csvDataFrame: DataFrame = csvParser.csvRdd(sqlContext, rdd)
You can parse your string into a csv using, e.g. scala-csv:
val myCSVdata : Array[List[String]] =
myCSVString.split('\n').flatMap(CSVParser.parseLine(_))
Here you can do a bit more processing, data cleaning, verifying that every line parses well and has the same number of fields, etc ...
You can then make this an RDD of records:
val myCSVRDD : RDD[List[String]] = sparkContext.parallelize(msCSVdata)
Here you can massage your lists of Strings into a case class, to reflect the fields of your csv data better. You should get some inspiration from the creations of Persons in this example:
https://spark.apache.org/docs/latest/sql-programming-guide.html#inferring-the-schema-using-reflection
I omit this step.
You can then convert to a DataFrame:
import spark.implicits._
myCSVDataframe = myCSVRDD.toDF()
The accepted answer wasn't working for me in spark 2.2.0 but lead me to what I needed with csvData.lines.toList
val fileUrl = getClass.getResource(s"/file_in_resources.csv")
val stream = fileUrl.getContent.asInstanceOf[InputStream]
val streamString = Source.fromInputStream(stream).mkString
val csvList = streamString.lines.toList
spark.read
.option("header", "true")
.option("inferSchema", "true")
.csv(csvList.toDS())
.as[SomeCaseClass]