Using HDinsight to run spark and a scala script.
I'm using the example scripts provided by the Azure plugin in intellij.
It provides me with the following code:
val conf = new SparkConf().setAppName("MyApp")
val sc = new SparkContext(conf)
Fair enough. And I can do things like:
val rdd = sc.textFile("wasb:///HdiSamples/HdiSamples/SensorSampleData/hvac/HVAC.csv")
and I can save files:
rdd1.saveAsTextFile("wasb:///HVACout2")
However, I am looking to load in a parquet file. The code I have found (elsewhere) for parquet files coming in is:
val df = spark.read.parquet("resources/Parquet/MyFile.parquet/")
Line above gives an error on this in HDinsight (when I submit the jar via intellij).
Why don't you use?:
val spark = SparkSession.builder
.master("local[*]") // adjust accordingly
.config("spark.sql.warehouse.dir", "E:/Exp/") //change accordingly
.appName("MySparkSession") //change accordingly
.getOrCreate()
When I put in spark session and get rid of spark context, HD insight breaks.
What am I doing wrong?
How using HdInsight do I go about creating either a spark session or context, that allows me to read in text files, parquet and all the rest? How do I get the best of both worlds
My understanding is SparkSession, is the better and more recent way. And what we should be using. So how do I get it running in HDInsight?
Thanks in advance
Turns out if I add
val spark = SparkSession.builder().appName("Spark SQL basic").getOrCreate()
After the spark context line and before the parquet, read part, it works.
Related
I have below 2 clarifications on AWS Glue, could you please clarify. Because I need to use glue as part of my project.
I would like to load a csv/txt file into a Glue job to process it. (Like we do in Spark with dataframes). Is this possible in Glue? Or do we have to use only Crawlers to crawl the data into Glue tables and make use of them like below for further processing?
empdf = glueContext.create_dynamic_frame.from_catalog(
database="emp",
table_name="emp_json")
Below I used Spark code to load a file into Glue, but I'm getting lengthy error logs. Can we directly run Spark or PySpark code as it is without any changes in Glue?
import sys
from pyspark.context import SparkContext
from awsglue.context import GlueContext
sc = SparkContext()
glueContext = GlueContext(sc)
spark = glueContext.spark_session
job = Job(glueContext)
job.init(args['JOB_NAME'], args)
dfnew = spark.read.option("header","true").option("delimiter", ",").csv("C:\inputs\TEST.txt")
dfnew.show(2)
It's possible to load data directly from s3 using Glue:
sourceDyf = glueContext.create_dynamic_frame_from_options(
connection_type="s3",
format="csv",
connection_options={
"paths": ["s3://bucket/folder"]
},
format_options={
"withHeader": True,
"separator": ","
})
You can also do that just with spark (as you already tried):
sourceDf = spark.read
.option("header","true")
.option("delimiter", ",")
.csv("C:\inputs\TEST.txt")
However, in this case Glue doesn't guarantee that they provide appropriate Spark readers. So if your error is related to missing data source for CSV then you should add spark-csv lib to the Glue job by providing s3 path to its locations via --extra-jars parameter.
Below 2 cases i tested working fine:
To load a file from S3 into Glue.
dfnew = glueContext.create_dynamic_frame_from_options("s3", {'paths': ["s3://MyBucket/path/"] }, format="csv" )
dfnew.show(2)
To load data from Glue db and tables which are generated already through Glue Crawlers.
DynFr = glueContext.create_dynamic_frame.from_catalog(database="test_db", table_name="test_table")
DynFr is a DynamicFrame, so if we want to work with Spark code in Glue, then we need to convert it into a normal data frame like below.
df1 = DynFr.toDF()
I am using spark 2.2 version on Microsoft Windows 7. I want to load csv file in one variable to perform SQL related actions later on but unable to do so. I referred accepted answer from this link but of no use. I followed below steps for creating SparkContext object and SQLContext object:
import org.apache.spark.SparkContext
import org.apache.spark.SparkConf
val sc=SparkContext.getOrCreate() // Creating spark context object
val sqlContext = new org.apache.spark.sql.SQLContext(sc) // Creating SQL object for query related tasks
Objects are created successfully but when I execute below code it throws an error which can't be posted here.
val df = sqlContext.read.format("csv").option("header", "true").load("D://ResourceData.csv")
And when I try something like df.show(2) it says that df was not found. I tried databricks solution for loading CSV from the attached link. It downloads the packages but doesn't load csv file. So how can I rectify my problem?? Thanks in advance :)
I solved my problem for loading local file in dataframe using 1.6 version in cloudera VM with the help of below code:
1) sudo spark-shell --jars /usr/lib/spark/lib/spark-csv_2.10-1.5.0.jar,/usr/lib/spark/lib/commons-csv-1.5.jar,/usr/lib/spark/lib/univocity-parsers-1.5.1.jar
2) val df1 = sqlContext.read.format("com.databricks.spark.csv").option("header", "true").option("treatEmptyValuesAsNulls", "true" ).option("parserLib", "univocity").load("file:///home/cloudera/Desktop/ResourceData.csv")
NOTE: sc and sqlContext variables are automatically created
But there are many improvements in the latest version i.e 2.2.1 which I am unable to use because metastore_db doesn't gets created in windows 7. I ll post a new question regarding the same.
In reference with your comment that you are able to access SparkSession variable, then follow below steps to process your csv file using SparkSQL.
Spark SQL is a Spark module for structured data processing.
There are mainly two abstractions - Dataset and Dataframe :
A Dataset is a distributed collection of data.
A DataFrame is a Dataset organized into named columns.
In the Scala API, DataFrame is simply a type alias of Dataset[Row].
With a SparkSession, applications can create DataFrames from an existing RDD, from a Hive table, or from Spark data sources.
You have a csv file and you can simply create a dataframe by doing one of the following:
From your spark-shell using the SparkSession variable spark:
val df = spark.read
.format("csv")
.option("header", "true")
.load("sample.csv")
After reading the file into dataframe, you can register it into a temporary view.
df.createOrReplaceTempView("foo")
SQL statements can be run by using the sql methods provided by Spark
val fooDF = spark.sql("SELECT name, age FROM foo WHERE age BETWEEN 13 AND 19")
You can also query that file directly with SQL:
val df = spark.sql("SELECT * FROM csv.'file:///path to the file/'")
Make sure that you run spark in local mode when you load data from local, or else you will get error. The error occurs when you have already set HADOOP_CONF_DIR environment variable,and which expects "hdfs://..." otherwise "file://".
Set your spark.sql.warehouse.dir (default: ${system:user.dir}/spark-warehouse).
.config("spark.sql.warehouse.dir", "file:///C:/path/to/my/")
It is the default location of Hive warehouse directory (using Derby)
with managed databases and tables. Once you set the warehouse directory, Spark will be able to locate your files, and you can load csv.
Reference : Spark SQL Programming Guide
Spark version 2.2.0 has built-in support for csv.
In your spark-shell run the following code
val df= spark.read
.option("header","true")
.csv("D:/abc.csv")
df: org.apache.spark.sql.DataFrame = [Team_Id: string, Team_Name: string ... 1 more field]
I have a spark job which runs in 5 mins on first initial runs and then takes several minutes ..more than 20-30 on subsequent runs. I'm reading a parquet file once and then creating dataframe and writing in .json format. I have not used cache(), persist() or unpersist() anywhere in the code.
This is local instance.
What could be the issue ?
configuration parameters
val spark = SparkSession
.builder()
.appName("example")
.config("spark.sql.warehouse.dir", warehouseLocation)
.config("spark.master", "local")
.config("spark.serializer",
"org.apache.spark.serializer.KryoSerializer")
.getOrCreate()
//set new runtime options
spark.conf.set("spark.sql.shuffle.partitions", 14)
spark.conf.set("spark.executor.memory", "6g")
spark.conf.set("spark.driver.host", "localhost")
spark.conf.set("spark.cores.max", "8")
spark.conf.set("spark.eventLog.enabled", true)
spark.sparkContext.setCheckpointDir("somedirectorypath")
spark.sparkContext.setLogLevel("WARN")
If you are writing parquet in "append" mode, the first write is optimised as direct write to the location. Same for "overwrite" mode. The subsequent writes use the ParquetOutputCommitter which will first write to a temporary location and then copy the files to your target directory.
Does anyone know why I can access to an existing hive table from spark-shell or zepelling notebook doing this
val df = spark.sql("select * from hive_table")
But when I submit a spark jar with a spark object created this way,
val spark = SparkSession
.builder()
.appName("Yet another spark app")
.config("spark.sql.shuffle.partitions", 18)
.config("spark.executor.memory", "2g")
.config("spark.serializer","org.apache.spark.serializer.KryoSerializer")
.getOrCreate()
I got this
Table or view not found
What I really want is to learn, understand, what the shell and the notebooks are doing for us in order to provide hive context to the SparkSession.
When working with Hive, one must instantiate SparkSession with Hive support
You need to call enableHiveSupport() on the session builder
I am beginner with Spark, Scala and Cassandra. I am working with ETL programming.
Now my project ETL POCs required Spark, Scala and Cassandra. I configured Cassandra with my ubuntu system in /usr/local/Cassandra/* and after that I installed Spark and Scala. Now I am using Scala editor to start my work, I created simply load a file in landing location, but after that I am trying to connect with cassandra in scala but I am not getting an help how we can connect and process the data in destination database?.
Any one help me Is this correct way? or some where I am wrong? please help me to how we can achieve this process with above combination.
Thanks in advance!
Add spark-cassandra-connector to your pom or sbt by reading instruction, then work this way
Import this in your file
import org.apache.spark.sql.SparkSession
import org.apache.spark.SparkConf
import org.apache.spark.sql.cassandra._
spark scala file
object SparkCassandraConnector {
def main(args: Array[String]) {
val conf = new SparkConf(true)
.setAppName("UpdateCassandra")
.setMaster("spark://spark:7077") // spark server
.set("spark.cassandra.input.split.size_in_mb","67108864")
.set("spark.cassandra.connection.host", "192.168.3.167") // cassandra host
.set("spark.cassandra.auth.username", "cassandra")
.set("spark.cassandra.auth.password", "cassandra")
// connecting with cassandra for spark and sql query
val spark = SparkSession.builder()
.config(conf)
.getOrCreate()
// Load data from node publish table
val df = spark
.read
.cassandraFormat( "table_nmae", "keyspace_name")
.load()
}
}
This will work for spark 2.2 and cassandra 2
you can perform this easly with spark-cassandra-connector