I am using python 2.7 with spark standalone cluster on client mode.
I want to use jdbc for mysql and found that i need to load it using --jars argument, I have the jdbc on my local, and manage to load it with pyspark console like here
When I write a python script inside my ide, using pyspark, I don't manage to load the additional jar mysql-connector-java-5.1.26.jar and keep get
no suitable driver
error
How can I load additional jar files when running a python script in client mode, using a standalone cluster on client mode and refering to a remote master?
edit: added some code #########################################################################
this is the basic code that i am using, i use pyspark with spark context in python e.g i do not use spark submit directly and don't understand how to use spark submit parameters in this case...
def createSparkContext(masterAdress = algoMaster):
"""
:return: return a spark context that is suitable for my configs
note the ip for the master
app name is not that important, just to show off
"""
from pyspark.mllib.util import MLUtils
from pyspark import SparkConf
from pyspark import SparkContext
import os
SUBMIT_ARGS = "--driver-class-path /var/nfs/general/mysql-connector-java-5.1.43 pyspark-shell"
#SUBMIT_ARGS = "--packages com.databricks:spark-csv_2.11:1.2.0 pyspark-shell"
os.environ["PYSPARK_SUBMIT_ARGS"] = SUBMIT_ARGS
conf = SparkConf()
#conf.set("spark.driver.extraClassPath", "var/nfs/general/mysql-connector-java-5.1.43")
conf.setMaster(masterAdress)
conf.setAppName('spark-basic')
conf.set("spark.executor.memory", "2G")
#conf.set("spark.executor.cores", "4")
conf.set("spark.driver.memory", "3G")
conf.set("spark.driver.cores", "3")
#conf.set("spark.driver.extraClassPath", "/var/nfs/general/mysql-connector-java-5.1.43")
sc = SparkContext(conf=conf)
print sc._conf.get("spark.executor.extraClassPath")
return sc
sql = SQLContext(sc)
df = sql.read.format('jdbc').options(url='jdbc:mysql://ip:port?user=user&password=pass', dbtable='(select * from tablename limit 100) as tablename').load()
print df.head()
Thanks
Your SUBMIT_ARGS is going to be passed to the spark-submit when creating a sparkContext from python. You should use --jars instead of --driver-class-path.
EDIT
Your problem is actually a lot simpler than it seems: you're missing the parameter driver in the options:
sql = SQLContext(sc)
df = sql.read.format('jdbc').options(
url='jdbc:mysql://ip:port',
user='user',
password='pass',
driver="com.mysql.jdbc.Driver",
dbtable='(select * from tablename limit 100) as tablename'
).load()
You can also put userand password in separate arguments.
Related
My pySpark version is 2.4 and python version is 2.7. I have multiple line sql file which needs to run in spark. Instead of running line by line, is it possible to keep the sql file in python (which initialize spark) and execute it using spark-submit ? I am trying to write a generic script in python so that we just need to replace sql file from hdfs folder later. Below is my code snippet.
import sys
from pyspark.sql import SparkSession
import pyspark.sql.functions as F
args = str(sys.argv[1]).split(',')
fd = args[0]
ld = args[1]
sd = args[2]
#Below line does not work
df = open("test.sql")
query = df.read().format(fd,ld,sd)
#Initiating SparkSession.
spark = SparkSession.builder.appName("PC").enableHiveSupport().getOrCreate()
#Below line works fine
df_s=spark.sql("""select * from test_tbl where batch_date='2021-08-01'""")
#Execute the sql (Does not work now)
df_s1=spark.sql(query)
spark-submit throws following error for the above code.
Exception in thread "main" org.apache.spark.SparkException:
Application application_1643050700073_7491 finished with failed status
at org.apache.spark.deploy.yarn.Client.run(Client.scala:1158)
at org.apache.spark.deploy.yarn.YarnClusterApplication.start(Client.scala:1606)
at org.apache.spark.deploy.SparkSubmit.org$apache$spark$deploy$SparkSubmit$$runMain(SparkSubmit.scala:847)
at org.apache.spark.deploy.SparkSubmit.doRunMain$1(SparkSubmit.scala:161)
at org.apache.spark.deploy.SparkSubmit.submit(SparkSubmit.scala:184)
at org.apache.spark.deploy.SparkSubmit.doSubmit(SparkSubmit.scala:86)
at org.apache.spark.deploy.SparkSubmit$$anon$2.doSubmit(SparkSubmit.scala:922)
at org.apache.spark.deploy.SparkSubmit$.main(SparkSubmit.scala:931)
at org.apache.spark.deploy.SparkSubmit.main(SparkSubmit.scala) 22/02/10 01:24:52 INFO util.ShutdownHookManager: Shutdown hook called
I am relatively new in pyspark. Can anyone please guide me what I am missing here ?
You cant run pyspark on your local directory. If you want to do an sql statement on a File in HDFS, you have to put your file from HDFS, first on your local directory.
Referred to spark 2.4.0 Spark Documentation, you can simply use the pyspark API.
from os.path import expanduser, join, abspath
from pyspark.sql import SparkSession
from pyspark.sql import Row
spark.sql("YOUR QUERY").show()
or query files directly with:
df = spark.sql("SELECT * FROM parquet.`examples/src/main/resources/users.parquet`")
I'm trying to get JSON objects from an S3 bucket using PySpark (on Windows, using wsl2 terminal).
I can do this using boto3 as an intermediate step but, when I try to use the spark.read.json method, I get an error.
Code:
import findspark
from pyspark.sql import SparkSession
from pyspark import SparkContext, SparkConf
import os
import multiprocessing
#----------------APACHE CONFIGURATIONS--------------
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages com.amazonaws:aws-java-sdk:1.7.4,org.apache.hadoop:hadoop-aws:2.7.3 pyspark-shell'
#---------------spark--------------
conf = (
SparkConf()
.set('spark.executor.extraJavaOptions','-Dcom.amazonaws.services.s3.enableV4=true')
.set('spark.driver.extraJavaOptions','-Dcom.amazonaws.services.s3.enableV4=true')
.setAppName('pyspark_aws')
.setMaster(f"local[{multiprocessing.cpu_count()}]")
.setIfMissing("spark.executor.memory", "2g")
)
sc=SparkContext(conf=conf)
sc.setSystemProperty('com.amazonaws.services.s3.enableV4', 'true')
spark=SparkSession(sc)
#--------------hadoop--------------
accessKeyId='xxxxxxxxxxxx'
secretAccessKey='xxxxxxxxx'
hadoopConf = sc._jsc.hadoopConfiguration()
hadoopConf.set('fs.s3a.access.key', accessKeyId)
hadoopConf.set('fs.s3a.secret.key', secretAccessKey)
hadoopConf.set('fs.s3a.endpoint', 's3-eu-west-1.amazonaws.com')
hadoopConf.set('fs.s3a.impl', 'org.apache.hadoop.fs.s3a.S3AFileSystem')
hadoopConf.set('fs.s3a.multipart.size', '419430400')
hadoopConf.set('fs.s3a.multipart.threshold', '2097152000')
hadoopConf.set('fs.s3a.connection.maximum', '500')
hadoopConf.set('s3a.connection.timeout', '600000')
s3_df = spark.read.json('s3a://{bucket}/{directory}/{object}.json')
Error:
py4j.protocol.Py4JJavaError: An error occurred while call
: java.lang.NumberFormatException: For input string: "32M
at java.base/java.lang.NumberFormatException.forI
at java.base/java.lang.Long.parseLong(Long.java:6
at java.base/java.lang.Long.parseLong(Long.java:8
at org.apache.hadoop.conf.Configuration.getLong(C
at org.apache.hadoop.fs.s3a.S3AFileSystem.getDefa
at org.apache.hadoop.fs.FileSystem.getDefaultBloc
at org.apache.hadoop.fs.s3a.S3AFileSystem.getFile
at org.apache.hadoop.fs.s3a.S3AFileSystem.getFile
at org.apache.hadoop.fs.FileSystem.exists(FileSys
at org.apache.spark.sql.execution.datasources.Dat
at org.apache.spark.sql.execution.datasources.Dat
at org.apache.spark.util.ThreadUtils$.$anonfun$pa
at java.base/java.util.concurrent.ForkJoinTask$Ruava.util.coteAction.exec(ForkJoinTask.java:1426)ncurrent.Fojava.base/java.util.concurrent.ForkJoinTask.dorkJoinWorkejava.base/java.util.concurrent.ForkJoinPool$WorThread.runjava.base/java.util.concurrent.ForkJoinPool.sc(ForkJoinWojava.base/java.util.concurrent.ForkJoinPool.rurkerThread.java.base/java.util.concurrent.ForkJoinWorkerTjava:183)
I added the multipart.size, multipart.threshold, connection.maximum, connection.timeout hadoop conf settings when I was getting a similar error earlier (this earlier error had '64M' instead of '32M' and changed when I added these conf settings)
I'm new to Spark so any and all tips/pointers would be helpful!
if needed
the "32M" is the default of "fs.s3a.block.size"
try hadoopConf.set('fs.s3a.block.size', '33554432')
go to https://hadoop.apache.org/docs/current/hadoop-aws/tools/hadoop-aws/index.html
you will find the explanations of the "32M" and the "64M"
I am working with python and pyspark in a jupyter notebook. I am trying to read several parquet files from an aws s3 bucket and convert them into a single json file.
This is what I have:
from functools import reduce
from pyspark.sql import DataFrame
bucket = s3.Bucket(name='mybucket')
keys =[]
for key in bucket.objects.all():
keys.append(key.key)
print(keys[0])
from pyspark.sql import SparkSession
# initialise sparkContext
spark = SparkSession.builder \
.master('local') \
.appName('myAppName') \
.config('spark.executor.memory', '5gb') \
.config("spark.cores.max", "6") \
.getOrCreate()
sc = spark.sparkContext
But I am getting:
Exception: Java gateway process exited before sending its port number with pyspark
I am not sure how to fix this, thank you!
Your getting this error because your pyspark is not able to communicate with your cluster. you need to set the value of some global variable like this.
import os
import findspark
findspark.init()
os.environ['PYSPARK_SUBMIT_ARGS'] = """--name job_name --master yarn / local
--conf spark.dynamicAllocation.enabled=true
pyspark-shell"""
os.environ['PYSPARK_PYTHON'] = "python3.6" # what ever version of python your using
os.environ['python'] = "python3.6"
findspark package is optional but it's good to use in case of pyspark.
I am using python 2.7 with ubuntu and running spark via a python script using a sparkcontext
My db is a remote mysql, with a username and password.
I try to query it using this code
sc = createSparkContext()
sql = SQLContext(sc)
df = sql.read.format('jdbc').options(url='jdbc:mysql://ip:port?user=user&password=password', dbtable='(select * from tablename limit 100) as tablename').load()
print df.head()
And get this error
py4j.protocol.Py4JJavaError: An error occurred while calling o32.load.
: java.sql.SQLException: No suitable driver
I found that I need the JDBC driver for mysql.
I downloaded the platform free one from here
I tried including it using this code in starting the spark context
conf.set("spark.driver.extraClassPath", "/var/nfs/general/mysql-connector-java-5.1.43")
and tried to install it using
sudo apt-get install libmysql-java
on the master machine, on the db machine and on the machine running the python script with no luck.
edit2
#
i tried using
conf.set("spark.executor.extraClassPath", "/var/nfs/general/mysql-connector-java-5.1.43")
seems by the output of
print sc.getConf().getAll()
which is
[(u'spark.driver.memory', u'3G'), (u'spark.executor.extraClassPath',
u'file:///var/nfs/general/mysql-connector-java-5.1.43.jar'),
(u'spark.app.name', u'spark-basic'), (u'spark.app.id',
u'app-20170830'), (u'spark.rdd.compress', u'True'),
(u'spark.master', u'spark://127.0.0.1:7077'), (u'spark.driver.port',
u''), (u'spark.serializer.objectStreamReset', u'100'),
(u'spark.executor.memory', u'2G'), (u'spark.executor.id', u'driver'),
(u'spark.submit.deployMode', u'client'), (u'spark.driver.host',
u''), (u'spark.driver.cores', u'3')]
that it includes the correct path, but still i get the same "no driver" error...
What am i missing here?
Thanks
You need to set the classpath for both driver and worker nodes. Add the following to spark configuration
conf.set("spark.executor.extraClassPath", "/var/nfs/general/mysql-connector-java-5.1.43")
conf.set("spark.driver.extraClassPath", "/var/nfs/general/mysql-connector-java-5.1.43")
Or you can pass it using
import os
os.environ['SPARK_CLASSPATH'] = "/path/to/driver/mysql.jar"
For spark >=2.0.0 you can add the comma separated list of jars to spark-defaults.conf file located in spark_home/conf directory like this
spark.jars path_2_jar1,path_2_jar2
from pyspark.sql import SparkSession
spark = SparkSession\
.builder\
.appName("Word Count")\
.config("spark.driver.extraClassPath", "/home/tuhin/mysql.jar")\
.getOrCreate()
dataframe_mysql = spark.read\
.format("jdbc")\
.option("url", "jdbc:mysql://ip:port/db_name")\
.option("driver", "com.mysql.jdbc.Driver")\
.option("dbtable", "employees").option("user", "root")\
.option("password", "12345678").load()
print(dataframe_mysql.columns)
"/home/tuhin/mysql.jar" is the location of mysql jar file
I have a spark cluster I use in local mode. I want to read a csv with the databricks external library spark.csv. I start my app as follows:
import os
import sys
os.environ["SPARK_HOME"] = "/home/mebuddy/Programs/spark-1.6.0-bin-hadoop2.6"
spark_home = os.environ.get('SPARK_HOME', None)
sys.path.insert(0, spark_home + "/python")
sys.path.insert(0, os.path.join(spark_home, 'python/lib/py4j-0.8.2.1-src.zip'))
from pyspark import SparkContext, SparkConf, SQLContext
try:
sc
except NameError:
print('initializing SparkContext...')
sc=SparkContext()
sq = SQLContext(sc)
df = sq.read.format('com.databricks.spark.csv').options(header='true', inferschema='true').load("/my/path/to/my/file.csv")
When I run it, I get the following error:
java.lang.ClassNotFoundException: Failed to load class for data source: com.databricks.spark.csv.
My question: how can I load the databricks.spark.csv library INSIDE my python code. I don't want to load it from outside (using --packages) from instance.
I tried to add the following lines but it did not work:
os.environ["SPARK_CLASSPATH"] = '/home/mebuddy/Programs/spark_lib/spark-csv_2.11-1.3.0.jar'
If you create SparkContext from scratch you can for example set PYSPARK_SUBMIT_ARGS before SparkContext is intialized:
os.environ["PYSPARK_SUBMIT_ARGS"] = (
"--packages com.databricks:spark-csv_2.11:1.3.0 pyspark-shell"
)
sc = SparkContext()
If for some reason you expect that SparkContext has been already initialized, as it is suggested by your code, this won't work. In local mode you could try to use Py4J gateway and URLClassLoader but it doesn't look like a good idea and won't work in a cluster mode.