I am trying to set the configuration of a few spark parameters inside the pyspark shell.
I tried the following
spark.conf.set("spark.executor.memory", "16g")
To check if the executor memory has been set, I did the following
spark.conf.get("spark.executor.memory")
which returned "16g".
I tried to check it through sc using
sc._conf.get("spark.executor.memory")
and that returned "4g".
Why do these two return different values and whats the correct way to set these configurations.
Also, I am fiddling with a bunch of parameters like
"spark.executor.instances"
"spark.executor.cores"
"spark.executor.memory"
"spark.executor.memoryOverhead"
"spark.driver.memory"
"spark.driver.cores"
"spark.driver.memoryOverhead"
"spark.memory.offHeap.size"
"spark.memory.fraction"
"spark.task.cpus"
"spark.memory.offHeap.enabled "
"spark.rpc.io.serverThreads"
"spark.shuffle.file.buffer"
Is there a way that will set the configurations for all the variables.
EDIT
I need to set the configuration programmatically. How do I change it after I have done spark-submit or started the pyspark shell? I am trying to reduce the runtime of my jobs for which I am going through multiple iterations changing the spark configuration and recording the runtimes.
You can set environment variables by using: (e.g. in spark-env.sh, only stand-alone)
SPARK_EXECUTOR_MEMORY=16g
You can also set the spark-defaults.conf:
spark.executor.memory=16g
But these solutions are hardcoded and pretty much static, and you want to have different parameters for different jobs, however, you might want to set up some defaults.
The best approach is to use spark-submit:
spark-submit --executor-memory 16G
The problem of defining variables programmatically is that some of them need to be defined at startup time if not precedence rules will take over and your changes after the initiation of the job will be ignored.
Edit:
The amount of memory per executor is looked up when SparkContext is created.
And
once a SparkConf object is passed to Spark, it is cloned and can no longer be modified by the user. Spark does not support modifying the configuration at runtime.
See: SparkConf Documentation
Have you tried changing the variable before the SparkContext is created, then running your iteration, stopping your SparkContext and changing your variable to iterate again?
import org.apache.spark.{SparkContext, SparkConf}
val conf = new SparkConf.set("spark.executor.memory", "16g")
val sc = new SparkContext(conf)
...
sc.stop()
val conf2 = new SparkConf().set("spark.executor.memory", "24g")
val sc2 = new SparkContext(conf2)
You can debug your configuration using: sc.getConf.toDebugString
See: Spark Configuration
Any values specified as flags or in the properties file will be passed on to the application and merged with those specified through SparkConf. Properties set directly on the SparkConf take highest precedence, then flags passed to spark-submit or spark-shell, then options in the spark-defaults.conf file.
You'll need to make sure that your variable is not defined with higher precedence.
Precedence order:
conf/spark-defaults.conf
--conf or -c - the command-line option used by spark-submit
SparkConf
I hope this helps.
In Pyspark,
Suppose I want to increase the driver memory and executor in code. I can do it as below:
conf = spark.sparkContext._conf.setAll([('spark.executor.memory', '23g'), ('spark.driver.memory','9.7g')])
To view the updated settings:
spark.sparkContext._conf.getAll()
Related
I have code to start a spark session
spark_session = SparkSession.builder.appName(app_name)
spark_session = spark_session.getOrCreate()
sc = spark_session.sparkContext
Now I want to dynamically be able to add jars and packages using PYSPARK_SUBMIT_ARGS so I added an environment variable with the following value before I hit the code
--jars /usr/share/aws/redshift/jdbc/RedshiftJDBC4.jar --packages com.databricks:spark-redshift_2.10:2.0.0,org.apache.spark:spark-avro_2.11:2.4.0,com.eclipsesource.minimal-json:minimal-json:0.9.4
But I get the following error:
Error: Missing application resource.
From looking online I know its because of the fact that I am explicitly passing jars and packages so I need to provide the path to my main jar file. But I am confused as to what that will be. I am just tring to run some code by starting a pyspark shell. I know the other way of passing these while starting the shell but my use case is such that I want it to be able to do it using the env variable and I have not been able to find the answers to this issue online
Lets say I have a python file my_python.py in which I have created a SparkSession 'spark' . I have a jar say my_jar.jar in which some spark logic is written. I am not creating SparkSession in my jar , rather I want to use the same session created in my_python.py. How to write a spark-submit command which take my python file , my jar and my sparksession 'spark' as an argument to my jar file.
Is it possible ?
If not , please share the alternative to do so.
So I feel there are two questions -
Q1. How in scala file you can reuse already created spark session?
Ans: Inside your scala code, you should use builder to get an existing session:
SparkSession.builder().getOrCreate()
Please check the Spark doc
https://spark.apache.org/docs/2.3.0/api/java/org/apache/spark/sql/SparkSession.html
Q2: How you do spark-submit with a .py file as driver and scala jar(s) as supporting jars?
And: It should be in something like this
./spark-submit --jars myjar.jar,otherjar.jar --py-files path/to/myegg.egg path/to/my_python.py arg1 arg2 arg3
So if you notice the method name, it is getOrCreate() - that means if a spark session is already created, no new session will be created rather existing session will be used.
Check this link for full implementation example:
https://www.crowdstrike.com/blog/spark-hot-potato-passing-dataframes-between-scala-spark-and-pyspark/
I want to change the default memory, executor and core settings of a spark session.
The first code in my pyspark notebook on HDInsight cluster in Jupyter looks like this:
from pyspark.sql import SparkSession
spark = SparkSession\
.builder\
.appName("Juanita_Smith")\
.config("spark.executor.instances", "2")\
.config("spark.executor.cores", "2")\
.config("spark.executor.memory", "2g")\
.config("spark.driver.memory", "2g")\
.getOrCreate()
On completion, I read the parameters back, which looks like the statement worked
However if I look in yarn, the setting have indeed not worked.
Which settings or commands do I need to make to let the session configuration take effect ?
Thank you for help in advance
By the time your notebook kernel has started, the SparkSession is already created with parameters defined in a kernel configuration file. To change this, you will need to update or replace the kernel configuration file, which I believe is usually somewhere like <jupyter home>/kernels/<kernel name>/kernel.json.
Update
If you have access to the machine hosting your Jupyter server, you can find the location of the current kernel configurations using jupyter kernelspec list. You can then either edit one of the pyspark kernel configurations, or copy it to a new file and edit that. For your purposes, you will need to add the following arguments to the PYSPARK_SUBMIT_ARGS:
"PYSPARK_SUBMIT_ARGS": "--conf spark.executor.instances=2 --conf spark.executor.cores=2 --conf spark.executor.memory=2g --conf spark.driver.memory=2g"
I'm running python script on Spark cluster using jupyter. I want to change driver default stack size. I found in the documentation that I can use spark.driver.extraJavaOptions to send any options to driver JVM, but there is a note in the documentation:
Note: In client mode, this config must not be set through the
SparkConf directly in your application, because the driver JVM has
already started at that point. Instead, please set this through the
--driver-java-options command line option or in your default properties file.
The question is: How to change default driver parameter when running from jupyter ?
You can customize the Java options used for the driver by passing spark.driver.extraJavaOptions as a configuration value into the SparkConf, eg:
from pyspark import SparkConf, SparkContext
conf = (SparkConf()
.setMaster("spark://spark-master:7077")
.setAppName("MyApp")
.set("spark.driver.extraJavaOptions", "-Xss4M"))
sc = SparkContext.getOrCreate(conf = conf)
Note that in http://spark.apache.org/docs/latest/configuration.html it states about spark.driver.extraJavaOptions:
Note: In client mode, this config must not be set through the SparkConf directly in your application, because the driver JVM has already started at that point. Instead, please set this through the --driver-java-options command line option or in your default properties file.
However this is talking about the JVM SparkConf class. When it’s set in the PySpark Python SparkConf, that passes it as a command-line parameter to spark-submit, which then uses it when instantiating the JVM, so that comment in the Spark docs does not apply.
How do you set a hive property like: hive.metastore.warehouse.dir at runtime? Or at least a more dynamic way of setting a property like the above, than putting it in a file like spark_home/conf/hive-site.xml
I faced the same issue and for me it worked by setting Hive properties from Spark (2.4.0). Please find below all the options through spark-shell, spark-submit and SparkConf.
Option 1 (spark-shell)
spark-shell --conf spark.hadoop.hive.metastore.warehouse.dir=some_path\metastore_db_2
Initially I tried with spark-shell with hive.metastore.warehouse.dir set to some_path\metastore_db_2. Then I get the next warning:
Warning: Ignoring non-spark config property:
hive.metastore.warehouse.dir=C:\winutils\hadoop-2.7.1\bin\metastore_db_2
Although when I create a Hive table with:
bigDf.write.mode("overwrite").saveAsTable("big_table")
The Hive metadata are stored correctly under metastore_db_2 folder.
When I use spark.hadoop.hive.metastore.warehouse.dir the warning disappears and the results are still saved in the metastore_db_2 directory.
Option 2 (spark-submit)
In order to use hive.metastore.warehouse.dir when submitting a job with spark-submit I followed the next steps.
First I wrote some code to save some random data with Hive:
import org.apache.spark.SparkConf
import org.apache.spark.sql.SparkSession
val sparkConf = new SparkConf().setAppName("metastore_test").setMaster("local")
val spark = SparkSession.builder().config(sparkConf).getOrCreate()
import spark.implicits._
var dfA = spark.createDataset(Seq(
(1, "val1", "p1"),
(2, "val1", "p2"),
(3, "val2", "p3"),
(3, "val3", "p4"))).toDF("id", "value", "p")
dfA.write.mode("overwrite").saveAsTable("metastore_test")
spark.sql("select * from metastore_test").show(false)
Next I submitted the job with:
spark-submit --class org.tests.Main \
--conf spark.hadoop.hive.metastore.warehouse.dir=C:\winutils\hadoop-2.7.1\bin\metastore_db_2
spark-scala-test_2.11-0.1.jar
The metastore_test table was properly created under the C:\winutils\hadoop-2.7.1\bin\metastore_db_2 folder.
Option 3 (SparkConf)
Via SparkSession in the Spark code.
val sparkConf = new SparkConf()
.setAppName("metastore_test")
.set("spark.hadoop.hive.metastore.warehouse.dir", "C:\\winutils\\hadoop-2.7.1\\bin\\metastore_db_2")
.setMaster("local")
This attempt was successful as well.
The question which still remains is why I have to extend the property with spark.hadoop in order to work as expected?