since updating to Spark 2.3.0, tests which are run in my CI (Semaphore) fail due to a allegedly invalid spark url when creating the (local) spark context:
18/03/07 03:07:11 ERROR SparkContext: Error initializing SparkContext.
org.apache.spark.SparkException: Invalid Spark URL: spark://HeartbeatReceiver#LXC_trusty_1802-d57a40eb:44610
at org.apache.spark.rpc.RpcEndpointAddress$.apply(RpcEndpointAddress.scala:66)
at org.apache.spark.rpc.netty.NettyRpcEnv.asyncSetupEndpointRefByURI(NettyRpcEnv.scala:134)
at org.apache.spark.rpc.RpcEnv.setupEndpointRefByURI(RpcEnv.scala:101)
at org.apache.spark.rpc.RpcEnv.setupEndpointRef(RpcEnv.scala:109)
at org.apache.spark.util.RpcUtils$.makeDriverRef(RpcUtils.scala:32)
at org.apache.spark.executor.Executor.<init>(Executor.scala:155)
at org.apache.spark.scheduler.local.LocalEndpoint.<init>(LocalSchedulerBackend.scala:59)
at org.apache.spark.scheduler.local.LocalSchedulerBackend.start(LocalSchedulerBackend.scala:126)
at org.apache.spark.scheduler.TaskSchedulerImpl.start(TaskSchedulerImpl.scala:164)
at org.apache.spark.SparkContext.<init>(SparkContext.scala:500)
at org.apache.spark.SparkContext$.getOrCreate(SparkContext.scala:2486)
at org.apache.spark.sql.SparkSession$Builder$$anonfun$7.apply(SparkSession.scala:930)
at org.apache.spark.sql.SparkSession$Builder$$anonfun$7.apply(SparkSession.scala:921)
at scala.Option.getOrElse(Option.scala:121)
at org.apache.spark.sql.SparkSession$Builder.getOrCreate(SparkSession.scala:921)
The spark session is created as following:
val sparkSession: SparkSession = SparkSession
.builder
.appName(s"LocalTestSparkSession")
.config("spark.broadcast.compress", "false")
.config("spark.shuffle.compress", "false")
.config("spark.shuffle.spill.compress", "false")
.master("local[3]")
.getOrCreate
Before updating to Spark 2.3.0, no problems were encountered in version 2.2.1 and 2.1.0. Also, running the tests locally works fine.
Change the SPARK_LOCAL_HOSTNAME to localhost and try.
export SPARK_LOCAL_HOSTNAME=localhost
This has been resolved by setting sparkSession config "spark.driver.host" to the IP address.
It seems that this change is required from 2.3 onwards.
If you don't want to change the environment variable, you can change the code to add the config in the SparkSession builder (like Hanisha said above).
In PySpark:
spark = SparkSession.builder.config("spark.driver.host", "localhost").getOrCreate()
As mentioned in above answers, You need to change SPARK_LOCAL_HOSTNAME to localhost. In windows, you have to use SET command, SET SPARK_LOCAL_HOSTNAME=localhost
but this SET command is temporary. you may have to run it again and again in every new terminal. but instead, you can use SETX command, which is permanent.
SETX SPARK_LOCAL_HOSTNAME localhost
You can type above command in any place. just open a command prompt and run above command. Notice that unlike SET command, SETX command do not allow equation mark. you need to separate environment variable and the value by a Space.
if Success, you will see a message like "SUCCESS: Specified value was saved"
you can also verify that your variable is successfully added by just typing SET in a different command prompt. (or type SET s , which gives variables, starting with the letter 'S'). you can see that SPARK_LOCAL_HOSTNAME=localhost in results, which will not happen if you use SET command instead of SETX
Change your hostname to have NO underscore.
spark://HeartbeatReceiver#LXC_trusty_1802-d57a40eb:44610 to spark://HeartbeatReceiver#LXCtrusty1802d57a40eb:44610
Ubuntu AS root
#hostnamectl status
#hostnamectl --static set-hostname LXCtrusty1802d57a40eb
#nano /etc/hosts
127.0.0.1 LXCtrusty1802d57a40eb
#reboot
Try to run Spark locally, with as many worker threads as logical cores on your machine :
.master("local[*]")
I would like to complement #Prakash Annadurai answer by saying:
If you want to make the variable settlement last after exiting the terminal, add it to your shell profile (e.g. ~/.bash_profile) with the same command:
export SPARK_LOCAL_HOSTNAME=localhost
For anyone working in Jupyter Notebook. Adding %env SPARK_LOCAL_HOSTNAME=localhost to the very beginning of the cell solved it for me. Like so:
%env SPARK_LOCAL_HOSTNAME=localhost
import findspark
findspark.init()
from pyspark import SparkConf, SparkContext
conf = SparkConf().setMaster("local").setAppName("Test")
sc = SparkContext(conf = conf)
Setting .config("spark.driver.host", "localhost") fixed the issue for me.
SparkSession spark = SparkSession
.builder()
.config("spark.master", "local")
.config("fs.s3a.aws.credentials.provider", "org.apache.hadoop.fs.s3a.TemporaryAWSCredentialsProvider")
.config("spark.hadoop.fs.s3a.buffer.dir", "/tmp")
.config("spark.driver.memory", "2048m")
.config("spark.executor.memory", "2048m")
.config("spark.driver.bindAddress", "127.0.0.1")
.config("spark.driver.host", "localhost")
.getOrCreate();
Related
I am currently working on a Jupyter (Lab) and PySpark 2.1.1.
I want to change spark.yarn.queue and master from a notebook. Because of the kernel spark and sc are available when I open a notebook.
Following this question, I tried
spark.conf.set("spark.yarn.queue", "my_queue")
But according to spark.sparkContext.getConf() the above line has no affect.
spark.conf.setMaster("yarn-cluster")
is not working, because there is no such a method for spark.conf.
Question: How can I change the configuration (queue and master) from a Jupyter notebook?
(Or should I set any environment variables?)
You can try to initialize spark beforehand, not in the notebook. Run this on your terminal:
export PYSPARK_DRIVER_PYTHON=jupyter
export PYSPARK_DRIVER_PYTHON_OPTS='notebook'
pyspark --master <your master> --conf <your configuration> <or any other option that pyspark supports>.
My source
I am trying to pull in data from a SQL server to a Hive table using Spark in a Zeppelin notebook.
I am trying to run the following code:
%pyspark
from pyspark import SparkContext
from pyspark.sql import SparkSession
from pyspark.sql.dataframe import DataFrame
from pyspark.sql.functions import *
spark = SparkSession.builder \
.appName('sample') \
.getOrCreate()
#set url, table, etc.
df = spark.read.format('jdbc') \
.option('url', url) \
.option('driver', 'com.microsoft.sqlserver.jdbc.SQLServerDriver') \
.option('dbtable', table) \
.option('user', user) \
.option('password', password) \
.load()
However, I keep getting the exception:
...
Py4JJavaError: An error occurred while calling o81.load.
: java.lang.ClassNotFoundException: com.microsoft.sqlserver.jdbc.SQLServerDriver
at java.net.URLClassLoader.findClass(URLClassLoader.java:381)
at java.lang.ClassLoader.loadClass(ClassLoader.java:424)
...
I have been trying to figure this out all day and I believe something is wrong with how I am trying to set up the driver. I have a driver under /tmp/sqljdbc42.jar on the instance. Can you please explain how I can let Spark know where this driver is? I have tried many different ways both through the shell and through the interpreter editor.
Thanks!
EDIT
I also should note that I loaded the jar to my instance throug Zeppelin's shell (%sh) using
curl -o /tmp/sqljdbc42.jar http://central.maven.org/maven2/com/microsoft/sqlserver/mssql-jdbc/6.4.0.jre8/mssql-jdbc-6.4.0.jre8.jar
pyspark --driver-class-path /tmp/sqljdbc42.jar --jars /tmp/sqljdbc42.jar
Here is how I fixed this:
scp driver jar onto the cluster driver node
Go to Zeppelin interpreter and scroll to the Spark section then click edit.
Write the complete path to the jar under artifacts e.g. /home/Hadoop/mssql-jdbc.jar and nothing else.
Click save.
Then you should be good!
You can add it through Web UI in Interpreter settings as follow:
Click Interpreter in menu
Click 'edit' button in the Spark interpreter
Add the path for the jar in the artifact field
Then just save and restart interpreter.
Similar to Tomas, you can add the driver (or any library) using maven in the interpreter:
Click Interpreter in menu
Click 'edit' button in the Spark interpreter
Add the path for the jar in the artifact field
Add the groupId:artifactId:version
For example, in your case, you can use com.microsoft.sqlserver:mssql-jdbc:jar:8.4.1.jre8 in artifact field.
When you restart the interpreter, it will download and add the dependency for you.
I'm using Pyspark from a Jupyter notebook and attempting to write a large parquet dataset to S3.
I get a 'no space left on device' error. I searched around and learned that it's because /tmp is filling up.
I want to now edit spark.local.dir to point to a directory that has space.
How can I set this parameter?
Most solutions I found suggested setting it when using spark-submit. However, I am not using spark-submit and just running it as a script from Jupyter.
Edit: I'm using Sparkmagic to work with an EMR backend.I think spark.local.dir needs to be set in the config JSON, but I am not sure how to specify it there.
I tried adding it in session_configs but it didn't work.
The answer depends on where your SparkContext comes from.
If you are starting Jupyter with pyspark:
PYSPARK_DRIVER_PYTHON='jupyter'\
PYSPARK_DRIVER_PYTHON_OPTS="notebook" \
PYSPARK_PYTHON="python" \
pyspark
then your SparkContext is already initialized when you receive your Python kernel in Jupyter. You should therefore pass a parameter to pyspark (at the end of the command above): --conf spark.local.dir=...
If you are constructing a SparkContext in Python
If you have code in your notebook like:
import pyspark
sc = pyspark.SparkContext()
then you can configure the Spark context before creating it:
import pyspark
conf = pyspark.SparkConf()
conf.set('spark.local.dir', '...')
sc = pyspark.SparkContext(conf=conf)
Configuring Spark from the command line:
It's also possible to configure Spark by editing a configuration file in bash. The file you want to edit is ${SPARK_HOME}/conf/spark-defaults.conf. You can append to it as follows (creating it if it doesn't exist):
echo 'spark.local.dir /foo/bar' >> ${SPARK_HOME}/conf/spark-defaults.conf
Now I have succeeded in running Pyspark in Jupyter in local mode by the second method as mentioned in this blog.
Here is the code:
import findspark
findspark.init()
from pyspark import SparkContext
sc = SparkContext("local", "First App")
I want to run it interactively in YARN-client mode,how can I do it?
Let's go futher,how to run in different modes,e.g.standalone mode and YARN-cluster mode.
Accrding to the Docs :
Master URLs accepts yarn parameter based on the HADOOP_CONF_DIR or YARN_CONF_DIR variable
So I can simply use:
sc = SparkContext("yarn-client", "First App")
I have been using spark-submit to test my codes on the multi-nodes system.
(Of course, I specified the master option as the master server address to achieve multi-nodes environment).
However, instead of using spark-submit, I would like to use spark-shell to test my codes on the cluster system. However, I don't know how to configure multi-nodes clusters settings on the spark-shell?
I think that just using spark-shell without changing any setups will results in the local mode.
I tried to search the info and followed the below commands.
scala> sc.stop()
...
scala> import org.apache.spark.{SparkContext, SparkConf}
import org.apache.spark.{SparkContext, SparkConf}
scala> val sc = new SparkContext(new SparkConf().setAppName("shell").setMaster("my server address"))
...
scala> import org.apache.spark.sql.SQLContext
import org.apache.spark.sql.SQLContext
scala> val sqlContext = new SQLContext(sc)
sqlContext: org.apache.spark.sql.SQLContext = org.apache.spark.sql.SQLContext#567a2954
However, I am quite sure that I am doing right behavior for the multi-node cluster setup using spark-shell.
Have you tried --master parameter of spark-shell? For Spark Standalone:
./spark-shell --master spark://master-ip:7077
Spark shell is just a driver, it will connect to any cluster you will write in master parameter
Edit:
For YARN use
./spark-shell --master yarn
If you used setMaster("my server address")) and "my server address" is not "local", then it won't run in local mode.
It is fine to set the master address in the code, but in production, you'd set --master parameter on the CLI to spark-shell or spark-submit
You can also write a separate .scala file, and pass that to spark-shell -i <filename>.scala