Below is my main code which I want to UnitTest
get_data.py
from pyspark.sql import SparkSession
from pyspark_llap.sql.session import HiveWarehouseSession
def get_hive_data(query):
hive_data = hive.executeQuery(query)
return hive_data
if __name__ == "__main__":
spark = SparkSession\
.builder\
.appName("HiveApp")\
.getOrCreate()
hive=HiveWarehouseSession.session(spark).build()
data = get_hive_data()
Below is my unittest code, I have written only the imports here, since i get error when i do from get_data import /*/
test.py
import unittest
import pyspark
import pyspark.sql.functions as f
from pyspark.sql import functions as F
from pyspark.sql import SparkSession
from get_data import *
ERROR
ImportError: No module named pyspark_llap
But if i run just get_data.py, it runs successfully
I am running it on edge node!
Related
I am trying to use pg8000 in my Glue Script, following are params in Glue Job
--extra-py-files s3://mybucket/pg8000libs.zip //NOTE: my zip contains __init__.py
Some Insights towards code
import sys
import os
from awsglue.transforms import *
from awsglue.utils import getResolvedOptions
from pyspark.context import SparkContext
from awsglue.context import GlueContext
from awsglue.job import Job
import boto3
from pyspark.sql import Row
from datetime import datetime, date
zip_path = os.path.join('/tmp', 'pg8000libs.zip')
sys.path.insert(0, zip_path)
def dump_python_path():
print("python path:", sys.path)
for path in sys.path:
if os.path.isdir(path):
print(f"dir: {path}")
print("\t" + str(os.listdir(path)))
print(path)
print(os.listdir('/tmp'))
dump_python_path()
# Import the library
import pg8000
Dump in cloudwatch
python path: ['/tmp/pg8000libs.zip', '/opt/amazon/bin', '/tmp/pg8000libs.zip', '/opt/amazon/spark/jars/spark-core_2.12-3.1.1-amzn-0.jar', '/opt/amazon/spark/python/lib/pyspark.zip', '/opt/amazon/spark/python/lib/py4j-0.10.9-src.zip', '/opt/amazon/lib/python3.6/site-packages', '/usr/lib64/python37.zip', '/usr/lib64/python3.7', '/usr/lib64/python3.7/lib-dynload', '/home/spark/.local/lib/python3.7/site-packages', '/usr/lib64/python3.7/site-packages', '/usr/lib/python3.7/site-packages']
I want to save pyspark dataframe directly into s3 bucket. I tried some options but getting error. Can someone help me to solve my problem?
I created one sample pyspark dataframe and tried to save in S3 bucket directly.
I tried below code-
from pyspark.context import SparkContext
from pyspark.sql import HiveContext
from pyspark.sql.functions import *
from pyspark.sql import SQLContext
from pyspark.sql.window import Window
import pyspark.sql.functions as func
from pyspark.sql.functions import last
from pyspark.sql import functions as F
from pyspark.sql.functions import lit
from pyspark.sql.functions import col
from pyspark.sql.functions import unix_timestamp
from functools import reduce
from pyspark.sql.session import SparkSession
from pyspark.sql import Row
from pyspark.sql.functions import max
from pyspark.sql.types import *
from pyspark.sql import DataFrame
from pyspark.sql.functions import broadcast
from pyspark.sql.functions import dense_rank
from pyspark.sql.window import Window
from pyspark.sql.functions import abs, lit
#from __future__ import division
import sys
import mysql.connector
import traceback
import json
#from sqlalchemy import create_engine
import os
import math
import os.path
import datetime
from os import getpid
import pymysql.cursors
import time
import signal
from bs4 import BeautifulSoup
import pandas as pd
from pyspark.context import SparkConf
from collections import OrderedDict
import multiprocessing
import multiprocessing as mp
from multiprocessing import Pool
from multiprocessing.pool import ThreadPool
from threading import Thread
from functools import partial
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
from email.mime.application import MIMEApplication
from email import encoders
import smtplib
import shutil
import glob
from datetime import datetime, date
from pyspark.sql import Row
spark = SparkSession.builder.appName("app_name").getOrCreate()
print(spark.sparkContext._gateway.jvm.org.apache.hadoop.util.VersionInfo.getVersion())
sc = spark.sparkContext
aws_access_key_id="*******"
aws_secret_access_key="********"
spark._jsc.hadoopConfiguration().set("fs.s3.awsAccessKeyId", aws_access_key_id)
spark._jsc.hadoopConfiguration().set("fs.s3.awsSecretAccessKey", aws_secret_access_key)
spark._jsc.hadoopConfiguration().set("fs.s3.impl", "org.apache.hadoop.fs.s3native.NativeS3FileSystem")
spark._jsc.hadoopConfiguration().set('fs.s3a.aws.credentials.provider', 'com.amazonaws.auth.DefaultAWSCredentialsProviderChain')
df = spark.createDataFrame([Row(a=1, b=4., c='GFG1', d=date(2000, 8, 1),e=datetime(2000, 8, 1, 12, 0)),
Row(a=2, b=8., c='GFG2', d=date(2000, 6, 2),e=datetime(2000, 6, 2, 12, 0)),
Row(a=4, b=5., c='GFG3', d=date(2000, 5, 3),e=datetime(2000, 5, 3, 12, 0))])
print(df.show())
print(df.printSchema())
df.write.format('csv').option('header','true').save('s3a://******/testing_s3/emp.csv',mode='overwrite')
After running this code I am getting below error-
py4j.protocol.Py4JJavaError: An error occurred while calling o48.save.
: com.amazonaws.services.s3.model.AmazonS3Exception: Status Code: 403, AWS Service: Amazon S3, AWS Request ID: RNKTVM6JMDACZ16W, AWS Error Code: null, AWS Error Message: Forbidden, S3 Extended Request ID: MS8lToBlzqSmn1YDdq6SPh7JC6aCKSROuldEz5x9LbsnQdxhKVEQriOpJz5KkCJPBnlk4KgsCkQ=
Please tell me what are the things I am missing in my script. Thanks in advance!!
after creating the spark context use these lines to set the credentials
spark.sparkContext._jsc.hadoopConfiguration().set("fs.s3a.access.key", AWS_ACCESS_KEY_ID)
spark.sparkContext._jsc.hadoopConfiguration().set("fs.s3a.secret.key", AWS_SECRET_ACCESS_KEY)
or
import pyspark
conf = (
pyspark.SparkConf()
.setAppName('app_name')
.setMaster(SPARK_MASTER)
.set('spark.hadoop.fs.s3a.access.key', AWS_ACCESS_KEY)
.set('spark.hadoop.fs.s3a.secret.key', AWS_SECRET_KEY)
)
sc = pyspark.SparkContext(conf=conf)
My PySpark code is below, and the first part, i.e, the import statements cell takes a very long time to run in Jupyter, in fact, the execution didn't happen till 5 - 6 hours, and later it shows a "Time limit exceeded error".
I have tried everything, like restarting jupyter, uninstalling anaconda, and then reinstalling, uninstalling spark and pyspark, and then re-installing both of them again. I even removed python completely and then installed it again, BUT THE PROBLEM NEVER SOLVED...!
Edit 1:- I realized that the problem is with the line spark = init_spark() This is taking a lot of time to run (in fact not running even after 4 - 5 hours)
Please help me with this...
import os
import sys
import pyspark
from pyspark.rdd import RDD
from pyspark.sql import Row
from pyspark.sql import DataFrame
from pyspark.sql import SparkSession
from pyspark.sql import SQLContext
from pyspark.sql import functions
from pyspark.sql.functions import lit, desc, col, size
import pandas as pd
from pandas import DataFrame
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.ticker as mtick
from IPython.core.interactiveshell import InteractiveShell
import matplotlib
from pylab import *
import scipy.stats as stats
# This helps auto print out the items without explixitly using 'print'
InteractiveShell.ast_node_interactivity = "all"
# Initialize a spark session.
conf = pyspark.SparkConf().setMaster("local[*]")
def init_spark():
spark = SparkSession \
.builder \
.appName("Statistical Inferences with Pyspark") \
.config(conf=conf) \
.getOrCreate()
return spark
spark = init_spark()
filename_data = 'D:\Subjects\ARTIFICIAL INTELLIGENCE\SEMESTER - 5\Big Data and DataBase Management\End Sem Project\endomondoHR_proper.json'
df = spark.read.json(filename_data, mode="DROPMALFORMED")
# Load meta data file into pyspark data frame as well
print('Data frame type: {}'.format(type(df)))
I tried converting my spark dataframes to dynamic to output as glueparquet files but I'm getting the error
'DataFrame' object has no attribute 'fromDF'"
My code uses heavily spark dataframes. Is there a way to convert from spark dataframe to dynamic frame so I can write out as glueparquet? If so could you please provide an example, and point out what I'm doing wrong below?
code:
# importing libraries
import sys
from awsglue.transforms import *
from awsglue.utils import getResolvedOptions
from pyspark.context import SparkContext
from awsglue.context import GlueContext
from awsglue.job import Job
glueContext = GlueContext(SparkContext.getOrCreate())
# updated 11/19/19 for error caused in error logging function
spark = glueContext.spark_session
from pyspark.sql import Window
from pyspark.sql.functions import col
from pyspark.sql.functions import first
from pyspark.sql.functions import date_format
from pyspark.sql.functions import lit,StringType
from pyspark.sql.types import *
from pyspark.sql.functions import substring, length, min,when,format_number,dayofmonth,hour,dayofyear,month,year,weekofyear,date_format,unix_timestamp
base_pth='s3://test/'
bckt_pth1=base_pth+'test_write/glueparquet/'
test_df=glueContext.create_dynamic_frame.from_catalog(
database='test_inventory',
table_name='inventory_tz_inventory').toDF()
test_df.fromDF(test_df, glueContext, "test_nest")
glueContext.write_dynamic_frame.from_options(frame = test_nest,
connection_type = "s3",
connection_options = {"path": bckt_pth1+'inventory'},
format = "glueparquet")
error:
'DataFrame' object has no attribute 'fromDF'
Traceback (most recent call last):
File "/mnt/yarn/usercache/livy/appcache/application_1574556353910_0001/container_1574556353910_0001_01_000001/pyspark.zip/pyspark/sql/dataframe.py", line 1300, in __getattr__
"'%s' object has no attribute '%s'" % (self.__class__.__name__, name))
AttributeError: 'DataFrame' object has no attribute 'fromDF'
fromDF is a class function. Her's how you can convert Dataframe to DynamicFrame
from awsglue.dynamicframe import DynamicFrame
DynamicFrame.fromDF(test_df, glueContext, "test_nest")
Just to consolidate the answers for Scala users too, here's how to transform a Spark Dataframe to a DynamicFrame (the method fromDF doesn't exist in the scala API of the DynamicFrame) :
import com.amazonaws.services.glue.DynamicFrame
val dynamicFrame = DynamicFrame(df, glueContext)
I hope it helps !
My code in Python3, fails with the following error:
Py4JJavaError: An error occurred while calling o45.load. :
java.lang.RuntimeException: java.lang.ClassNotFoundException: Class
org.apache.hadoop.fs.s3a.S3AFileSystem not found
The code:
from pyspark.sql.functions import udf
from pyspark.sql.types import *
import os
from pandas.io.json import json_normalize
from pyspark.sql.types import *
os.environ['PYSPARK_SUBMIT_ARGS'] = '--driver-memory 10g --packages org.elasticsearch:elasticsearch-hadoop:6.5.2,org.apache.hadoop:hadoop-aws:3.1.1 pyspark-shell'
from pyspark import SparkConf
from pyspark.sql import SparkSession
spark = SparkSession.builder.master("local[*]").config(conf=SparkConf()).config("spark.local.dir","/Users/psuresh
/spark_local_dir").enableHiveSupport().getOrCreate()
spark.sparkContext.setLogLevel("ERROR")
hadoopConf = spark._jsc.hadoopConfiguration()
hadoopConf.set("fs.s3a.impl", "org.apache.hadoop.fs.s3a.S3AFileSystem")
hadoopConf.set("fs.s3a.access.key", "XXXXXX")
hadoopConf.set("fs.s3a.secret.key","XXXXXXXXXX")
from pyspark.sql.functions import input_file_name
from pyspark.sql.functions import regexp_extract, col
from pyspark.sql.functions import *
df= spark.read.format("parquet").load("s3a://a/b/c/d/dump/")