CODEWITHSUNDEEP
python-3.xcsvapache-sparkhivepyspark
user15051990
user15051990

Reputation: 1895

Read csv using pyspark

I am new to spark. And I am trying to read csv file using pyspark. And I referred to PySpark How to read CSV into Dataframe, and manipulate it, Get CSV to Spark dataframe and many more. I tried to read it two ways:

1 

from  pyspark.sql import SparkSession
from pyspark.sql import SQLContext
from pyspark.conf import SparkConf
sc = SparkContext.getOrCreate()
df = spark.read.csv('D:/Users/path/csv/test.csv')
df.show()

2

import pyspark
sc = pyspark.SparkContext()
sql = SQLContext(sc)

df = (sql.read
         .format("com.databricks.spark.csv")
         .option("header", "true")
         .load("D:/Users/path/csv/test.csv"))
df.show()

Neither of the codes are working. I am getting the following error:

Py4JJavaError                             Traceback (most recent call last)
<ipython-input-28-c6263cc7dab9> in <module>()
      4 
      5 sc = SparkContext.getOrCreate()
----> 6 df = spark.read.csv('D:/Users/path/csv/test.csv')
      7 df.show()
      8 

~\opt\spark\spark-2.1.0-bin-hadoop2.7\python\pyspark\sql\readwriter.py in csv(self, path, schema, sep, encoding, quote, escape, comment, header, inferSchema, ignoreLeadingWhiteSpace, ignoreTrailingWhiteSpace, nullValue, nanValue, positiveInf, negativeInf, dateFormat, timestampFormat, maxColumns, maxCharsPerColumn, maxMalformedLogPerPartition, mode)
    378         if isinstance(path, basestring):
    379             path = [path]
--> 380         return self._df(self._jreader.csv(self._spark._sc._jvm.PythonUtils.toSeq(path)))
    381 
    382     @since(1.5)

~\opt\spark\spark-2.1.0-bin-hadoop2.7\python\lib\py4j-0.10.4-src.zip\py4j\java_gateway.py in __call__(self, *args)
   1131         answer = self.gateway_client.send_command(command)
   1132         return_value = get_return_value(
-> 1133             answer, self.gateway_client, self.target_id, self.name)
   1134 
   1135         for temp_arg in temp_args:

~\opt\spark\spark-2.1.0-bin-hadoop2.7\python\pyspark\sql\utils.py in deco(*a, **kw)
     61     def deco(*a, **kw):
     62         try:
---> 63             return f(*a, **kw)
     64         except py4j.protocol.Py4JJavaError as e:
     65             s = e.java_exception.toString()

~\opt\spark\spark-2.1.0-bin-hadoop2.7\python\lib\py4j-0.10.4-src.zip\py4j\protocol.py in get_return_value(answer, gateway_client, target_id, name)
    317                 raise Py4JJavaError(
    318                     "An error occurred while calling {0}{1}{2}.\n".
--> 319                     format(target_id, ".", name), value)
    320             else:
    321                 raise Py4JError(

Py4JJavaError: An error occurred while calling o663.csv.
: java.util.ServiceConfigurationError: org.apache.spark.sql.sources.DataSourceRegister: Provider org.apache.spark.sql.hive.execution.HiveFileFormat not found
    at java.util.ServiceLoader.fail(ServiceLoader.java:239)
    at java.util.ServiceLoader.access$300(ServiceLoader.java:185)
    at java.util.ServiceLoader$LazyIterator.nextService(ServiceLoader.java:372)
    at java.util.ServiceLoader$LazyIterator.next(ServiceLoader.java:404)
    at java.util.ServiceLoader$1.next(ServiceLoader.java:480)
    at scala.collection.convert.Wrappers$JIteratorWrapper.next(Wrappers.scala:43)
    at scala.collection.Iterator$class.foreach(Iterator.scala:893)

I don't why it is throwing some hive exception Py4JJavaError: An error occurred while calling o663.csv. : java.util.ServiceConfigurationError: org.apache.spark.sql.sources.DataSourceRegister: Provider org.apache.spark.sql.hive.execution.HiveFileFormat not found. How to resolve this error HiveFileFormat not found.

Can anyone guide me to resolve this error?

Upvotes: 1

Views: 15868

Answers (5)

MMSA
MMSA

Reputation: 948

Since in PySpark 3.0.1 SQLContext is Deprectaed - to import a CSV file into PySpark.

from pyspark.sql import SparkSession
spark = SparkSession \
    .builder \
    .appName("Python") \
    .getOrCreate()

df = spark.read.csv("/path/to/file/csv")
df.show()

Upvotes: 1

Kudakwashe Nyatsanza
Kudakwashe Nyatsanza

Reputation: 41

Error most likely occurs because you are trying to access a local file.
See below how you should access it:

#Local File
spark.read.option("header","true").option("inferSchema","true").csv("file:///path")

#HDFS file
spark.read.option("header","true").option("inferSchema","true").csv("/path")

.csv(<path>) comes last.

Upvotes: 0

Mohammad Heydari
Mohammad Heydari

Reputation: 4290

First of all, the system needs to recognize Spark Session as the following commands:

from pyspark import SparkConf, SparkContext
sc = SparkContext()

after that, SQL library has to be introduced to the system like this:

from pyspark.sql import SQLContext
sqlContext = SQLContext(sc)

and finally you can read your CSV by the following command: 

df = sqlContext.read.format('com.databricks.spark.csv').options(header='true', inferschema='true').load('path/to/your/file.csv')

Upvotes: 2

secretive
secretive

Reputation: 2104

Try to specify to use local master by making a configuration object. This would remove some doubts of spark trying to access on hadoop or anywhere as mentioned by somebody in comment.

sc.stop()
conf = (conf.setMaster('local[*]'))
sc = SparkContext(conf = conf)
sqlContext = SQLContext(sc)

if this is not working, then dont use sqlcontext for reading the file. Try spark.read.csv("path/filename.csv") by creating sparksession.

Also, it is best if you use Spark/Hadoop with a Linux operating system as it is a lot simpler in those systems.

Upvotes: 0

mjrice04
mjrice04

Reputation: 164

Have you tried to use sqlContext.read.csv? This is how I read csvs in Spark 2.1

from pyspark import sql, SparkConf, SparkContext

conf = SparkConf().setAppName("Read_CSV")
sc = SparkContext(conf=conf)
sqlContext = sql.SQLContext(sc)

df = sqlContext.read.csv("path/to/data")
df.show()

Upvotes: 2

Related Questions