I am using Jupyter notebook and want to save csv file to cassandra db. There is no problem while getting data and showing it, But when I try to save this csv data to cassandra db it throws below exception.
: org.apache.spark.SparkException: Job aborted due to stage failure: Task 0 in stage 1.0 failed 1 times, most recent failure: Lost task 0.0 in stage 1.0 (TID 1, localhost, executor driver): java.lang.NoClassDefFoundError: com/twitter/jsr166e/LongAdder
I dowloaded maven package manually both 2.4.0 and 2.4.1 and none of them worked. Also stated packages at the top of code.
import sys
import uuid
import time
import os
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages com.datastax.spark:spark-cassandra-connector_2.11:2.4.0 pyspark-shell'
try:
from pyspark import SparkContext
from pyspark import SparkConf
from pyspark.sql import SparkSession
from itertools import islice
from pyspark.sql import SQLContext
from pyspark.sql.types import *
from pyspark.sql import Row
from datetime import datetime
except ImportError as e:
print("error importing spark modules", e)
sys.exit(1)
conf = SparkConf().setAppName("Stand Alone Python Script").setMaster("local[*]")\
.setAll([('spark.executor.memory', '8g'),\
('spark.executor.cores', '3'),\
('spark.cores.max', '3'),\
('spark.cassandra.connection.host', 'cassandra_ip'),\
('spark.cassandra.auth.username', 'cassandra_user_name'),\
('spark.cassandra.auth.password', 'cassandra_password'),\
('spark.driver.memory','8g')])
sc = SparkContext(conf=conf)
sql_context = SQLContext(sc)
consumer_complaints = sql_context.read.format("csv").option("header", "true").option("inferSchema", "false").load("in/Consumer_Complaints.csv")
consumer_complaints.write\
.format("org.apache.spark.sql.cassandra")\
.mode('append')\
.options(table="table_name", keyspace="space_name")\
.save()
sc.stop()
Hello I solved my problem with following steps:
downloaded twitter jsr jar and moved it to $SPARK_HOME/jars directory.
cp /home/jovyan/.m2/repository/com/twitter/jsr166e/1.1.0/jsr166e-1.1.0.jar /usr/local/spark/jars/
Also because of docker's jupyter user is jovyan not root I grant permission to this folder
I used directly below statement but you can use more restrictive way.
chmod -R 777 /usr/local/spark/jars/
Thank you
Related
I have tried the following code it gives error
import pandas as pd
from pyspark.sql import SparkSession
import os
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages com.databricks:spark-xml_2.13-0.14.0 pyspark-shell'
os.environ['PYSPARK_PYTHON'] = "--jars C:/Users/l/spark-xml_2.13-0.14.0 pyspark-shell"
import findspark
findspark.init
from pyspark.sql.functions import col, lit, sum, max
from pyspark.sql import SparkSession
from pyspark.sql.types import StringType, BooleanType, IntegerType, DateType
spark = SparkSession.builder.getOrCreate()
xmldf = spark.read.format("com.databricks.spark.xml").option("rowTag", "POSLog").load("C:/Users/l/trans.xml")
but it gives error code:
Py4JJavaError: An error occurred while calling o2978.load.
: java.lang.ClassNotFoundException:
Failed to find data source: com.databricks.spark.xml. Please find packages at
http://spark.apache.org/third-party-projects.html
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 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.