I want to know how to load/import a CSV file in to mongodb using pyspark. I have a csv file named cal.csv placed in the desktop. Can somebody share the code snippet.
First read the csv as pyspark dataframe.
from pyspark import SparkConf,SparkContext
from pyspark.sql import SQLContext
sc = SparkContext(conf = conf)
sql = SQLContext(sc)
df = sql.read.csv("cal.csv", header=True, mode="DROPMALFORMED")
Then write it to mongodb,
df.write.format('com.mongodb.spark.sql.DefaultSource').mode('append')\
.option('database',NAME).option('collection',COLLECTION_MONGODB).save()
Specify the NAME and COLLECTION_MONGODB as created by you.
Also, you need to give conf and packages alongwith spark-submit according to your version,
/bin/spark-submit --conf "spark.mongodb.inuri=mongodb://127.0.0.1/DATABASE.COLLECTION_NAME?readPreference=primaryPreferred"
--conf "spark.mongodb.output.uri=mongodb://127.0.0.1/DATABASE.COLLECTION_NAME"
--packages org.mongodb.spark:mongo-spark-connector_2.11:2.2.0
tester.py
Specify COLLECTION_NAME and DATABASE above. tester.py assuming name of the code file. For more information, refer this.
This worked for me. database:people Collection:con
pyspark --conf "spark.mongodb.input.uri=mongodb://127.0.0.1/people.con?readPreference=primaryPreferred" \
--conf "spark.mongodb.output.uri=mongodb://127.0.0.1/people.con" \
--packages org.mongodb.spark:mongo-spark-connector_2.11:2.3.0
from pyspark.sql import SparkSession
my_spark = SparkSession \
.builder \
.appName("myApp") \
.config("spark.mongodb.input.uri", "mongodb://127.0.0.1/people.con") \
.config("spark.mongodb.output.uri", "mongodb://127.0.0.1/people.con") \
.getOrCreate()
df = spark.read.csv(path = "file:///home/user/Desktop/people.csv", header=True, inferSchema=True)
df.printSchema()
df.write.format("com.mongodb.spark.sql.DefaultSource").mode("append").option("database","people").option("collection", "con").save()
Next go to mongo and check if collection is wrtten by following below steps
mongo
show dbs
use people
show collections
db.con.find().pretty()
Related
I am encountering problem with printing the data to console from kafka topic.
The error message I get is shown in below image.
As you can see in the above image that after batch 0 , it doesn't process further.
All this are snapshots of the error messages. I don't understand the root cause of the errors occurring. Please help me.
Following are kafka and spark version:
spark version: spark-3.1.1-bin-hadoop2.7
kafka version: kafka_2.13-2.7.0
I am using the following jars:
kafka-clients-2.7.0.jar
spark-sql-kafka-0-10_2.12-3.1.1.jar
spark-token-provider-kafka-0-10_2.12-3.1.1.jar
Here is my code:
spark = SparkSession \
.builder \
.appName("Pyspark structured streaming with kafka and cassandra") \
.master("local[*]") \
.config("spark.jars","file:///C://Users//shivani//Desktop//Spark//kafka-clients-2.7.0.jar,file:///C://Users//shivani//Desktop//Spark//spark-sql-kafka-0-10_2.12-3.1.1.jar,file:///C://Users//shivani//Desktop//Spark//spark-cassandra-connector-2.4.0-s_2.11.jar,file:///D://mysql-connector-java-5.1.46//mysql-connector-java-5.1.46.jar,file:///C://Users//shivani//Desktop//Spark//spark-token-provider-kafka-0-10_2.12-3.1.1.jar")\
.config("spark.executor.extraClassPath","file:///C://Users//shivani//Desktop//Spark//kafka-clients-2.7.0.jar,file:///C://Users//shivani//Desktop//Spark//spark-sql-kafka-0-10_2.12-3.1.1.jar,file:///C://Users//shivani//Desktop//Spark//spark-cassandra-connector-2.4.0-s_2.11.jar,file:///D://mysql-connector-java-5.1.46//mysql-connector-java-5.1.46.jar,file:///C://Users//shivani//Desktop//Spark//spark-token-provider-kafka-0-10_2.12-3.1.1.jar")\
.config("spark.executor.extraLibrary","file:///C://Users//shivani//Desktop//Spark//kafka-clients-2.7.0.jar,file:///C://Users//shivani//Desktop//Spark//spark-sql-kafka-0-10_2.12-3.1.1.jar,file:///C://Users//shivani//Desktop//Spark//spark-cassandra-connector-2.4.0-s_2.11.jar,file:///D://mysql-connector-java-5.1.46//mysql-connector-java-5.1.46.jar,file:///C://Users//shivani//Desktop//Spark//spark-token-provider-kafka-0-10_2.12-3.1.1.jar")\
.config("spark.driver.extraClassPath","file:///C://Users//shivani//Desktop//Spark//kafka-clients-2.7.0.jar,file:///C://Users//shivani//Desktop//Spark//spark-sql-kafka-0-10_2.12-3.1.1.jar,file:///C://Users//shivani//Desktop//Spark//spark-cassandra-connector-2.4.0-s_2.11.jar,file:///D://mysql-connector-java-5.1.46//mysql-connector-java-5.1.46.jar,file:///C://Users//shivani//Desktop//Spark//spark-token-provider-kafka-0-10_2.12-3.1.1.jar")\
.getOrCreate()
spark.sparkContext.setLogLevel("ERROR")
#streaming dataframe that reads from kafka topic
df_kafka=spark.readStream\
.format("kafka")\
.option("kafka.bootstrap.servers",kafka_bootstrap_servers)\
.option("subscribe",kafka_topic_name)\
.option("startingOffsets", "latest") \
.load()
print("Printing schema of df_kafka:")
df_kafka.printSchema()
#converting data from kafka broker to string type
df_kafka_string=df_kafka.selectExpr("CAST(value AS STRING) as value")
# schema to read json format data
ts_schema = StructType() \
.add("id_str", StringType()) \
.add("created_at", StringType()) \
.add("text", StringType())
#parse json data
df_kafka_string_parsed=df_kafka_string.select(from_json(col("value"),ts_schema).alias("twts"))
df_kafka_string_parsed_format=df_kafka_string_parsed.select("twts.*")
df_kafka_string_parsed_format.printSchema()
df=df_kafka_string_parsed_format.writeStream \
.trigger(processingTime="1 seconds") \
.outputMode("update")\
.option("truncate","false")\
.format("console")\
.start()
df.awaitTermination()
The error (NoClassDefFound, followed by the kafka010 package) is saying that spark-sql-kafka-0-10 is missing its transitive dependency on org.apache.commons:commons-pool2:2.6.2, as you can see here
You can either download that JAR as well, or you can change your code to use --packages instead of spark.jars option, and let Ivy handle downloading transitive dependencies
import os
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages org.apache...'
spark = SparkSession.bulider...
How do I use all the workers in the cluster when I run PySpark in a notebook?
I'm running on Google Dataproc with YARN.
I use this configuration:
import pyspark
from pyspark.sql import SparkSession
conf = pyspark.SparkConf().setAll([
('spark.jars', 'gs://spark-lib/bigquery/spark-bigquery-latest.jar'),
('spark.jars.packages', 'graphframes:graphframes:0.7.0-spark2.3-s_2.11'),
('spark.executor.heartbeatInterval', "1000s"),
("spark.network.timeoutInterval", "1000s"),
("spark.network.timeout", "10000s"),
("spark.network.timeout", "1001s")
])
spark = SparkSession.builder \
.appName('testing bq v04') \
.config(conf=conf) \
.getOrCreate()
But it doesn't look like it is using all the available resources:
Here I provide some more context. The problem arises when I run label propagation algorithm with GraphFrame:
g_df = GraphFrame(vertices_df, edges_df)
result_iteration_2 = g_df.labelPropagation(maxIter=5)
Can anyone let me know without converting xlsx or xls files how can we read them as a spark dataframe
I have already tried to read with pandas and then tried to convert to spark dataframe but got the error and the error is
Error:
Cannot merge type <class 'pyspark.sql.types.DoubleType'> and <class 'pyspark.sql.types.StringType'>
Code:
import pandas
import os
df = pandas.read_excel('/dbfs/FileStore/tables/BSE.xlsx', sheet_name='Sheet1',inferSchema='')
sdf = spark.createDataFrame(df)
I try to give a general updated version at April 2021 based on the answers of #matkurek and #Peter Pan.
SPARK
You should install on your databricks cluster the following 2 libraries:
Clusters -> select your cluster -> Libraries -> Install New -> Maven -> in Coordinates: com.crealytics:spark-excel_2.12:0.13.5
Clusters -> select your cluster -> Libraries -> Install New -> PyPI-> in Package: xlrd
Then, you will be able to read your excel as follows:
sparkDF = spark.read.format("com.crealytics.spark.excel") \
.option("header", "true") \
.option("inferSchema", "true") \
.option("dataAddress", "'NameOfYourExcelSheet'!A1") \
.load(filePath)
PANDAS
You should install on your databricks cluster the following 2 libraries:
Clusters -> select your cluster -> Libraries -> Install New -> PyPI-> in Package: xlrd
Clusters -> select your cluster -> Libraries -> Install New -> PyPI-> in Package: openpyxl
Then, you will be able to read your excel as follows:
import pandas
pandasDF = pd.read_excel(io = filePath, engine='openpyxl', sheet_name = 'NameOfYourExcelSheet')
Note that you will have two different objects, in the first scenario a Spark Dataframe, in the second a Pandas Dataframe.
As mentioned by #matkurek you can read it from excel directly. Indeed, this should be a better practice than involving pandas since then the benefit of Spark would not exist anymore.
You can run the same code sample as defined qbove, but just adding the class needed to the configuration of your SparkSession.
spark = SparkSession.builder \
.master("local") \
.appName("Word Count") \
.config("spark.jars.packages", "com.crealytics:spark-excel_2.11:0.12.2") \
.getOrCreate()
Then, you can read your excel file.
df = spark.read.format("com.crealytics.spark.excel") \
.option("useHeader", "true") \
.option("inferSchema", "true") \
.option("dataAddress", "'NameOfYourExcelSheet'!A1") \
.load("your_file"))
There is no data of your excel shown in your post, but I had reproduced the same issue as yours.
Here is the data of my sample excel test.xlsx, as below.
You can see there are different data types in my column B: a double value 2.2 and a string value C.
So if I run the code below,
import pandas
df = pandas.read_excel('test.xlsx', sheet_name='Sheet1',inferSchema='')
sdf = spark.createDataFrame(df)
it will return a same error as yours.
TypeError: field B: Can not merge type <class 'pyspark.sql.types.DoubleType'> and class 'pyspark.sql.types.StringType'>
If we tried to inspect the dtypes of df columns via df.dtypes, we will see.
The dtype of Column B is object, the spark.createDateFrame function can not inference the real data type for column B from the real data. So to fix it, the solution is to pass a schema to help data type inference for column B, as the code below.
from pyspark.sql.types import StructType, StructField, DoubleType, StringType
schema = StructType([StructField("A", DoubleType(), True), StructField("B", StringType(), True)])
sdf = spark.createDataFrame(df, schema=schema)
To force make column B as StringType to solve the data type conflict.
You can read excel file through spark's read function. That requires a spark plugin, to install it on databricks go to:
clusters > your cluster > libraries > install new > select Maven and in 'Coordinates' paste com.crealytics:spark-excel_2.12:0.13.5
After that, this is how you can read the file:
df = spark.read.format("com.crealytics.spark.excel") \
.option("useHeader", "true") \
.option("inferSchema", "true") \
.option("dataAddress", "'NameOfYourExcelSheet'!A1") \
.load(filePath)
Just open file xlsx or xlms,open file in pandas,after that in spark
import pandas as pd
df = pd.read_excel('file.xlsx', engine='openpyxl')
df = spark_session.createDataFrame(df.astype(str))
Below configuration and code works for me to read excel file into pyspark dataframe. Pre-requisites before executing python code.
Install Maven library on your databricks cluster.
Maven library name & version: com.crealytics:spark-excel_2.12:0.13.5
Databricks Runtime: 9.0 (includes Apache Spark 3.1.2, Scala 2.12)
Execute below code in your python notebook to load excel file into pyspark dataframe:
sheetAddress = "'<enter sheetname>'!A1"
filePath = "<enter excel file full path>"
df = spark.read.format("com.crealytics.spark.excel") \
.option("header", "true") \
.option("dataAddress", sheetAddress) \
.option("treatEmptyValuesAsNulls", "false") \
.option("inferSchema", "true") \
.load(filePath)
Steps to read .xls / .xlsx files from Azure Blob storage into a Spark DF
You can read the excel files located in Azure blob storage to a pyspark dataframe with the help of a library called spark-excel. (Also refered as com.crealytics.spark.excel)
Install the library either using the UI or Databricks CLI. (Cluster settings page > Libraries > Install new option. Make sure to chose maven)
Once the library is installed. You need proper credentials to access Azure blob storage. You can provide the access key in Cluster settings page > Advanced option > Spark configs
Example:
spark.hadoop.fs.azure.account.key.<storage-account>.blob.core.windows.net <access key>
Note: If you're the cluster owner you can provide it as a secret instead of giving access key as plain text as mentioned in the docs
Restart the cluster. you can use below code to read those excel files located in blob storage
filePath = "wasbs://<container-name>#<storage-account>.blob.core.windows.net/MyFile1.xls"
DF = spark.read.format("excel").option("header", "true").option("inferSchema", "true").load(filePath)
display(DF)
PS: The spark.read.format("excel") is the V2 approach. while spark.read.format("com.crealytics.spark.excel") is the V1, you can read more here
Can someone point me to a working example of saving a csv file to Hbase table using Spark 2.2
Options that I tried and failed (Note: all of them work with Spark 1.6 for me)
phoenix-spark
hbase-spark
it.nerdammer.bigdata : spark-hbase-connector_2.10
All of them finally after fixing everything give similar error to this Spark HBase
Thanks
Add below parameters to your spark job-
spark-submit \
--conf "spark.yarn.stagingDir=/somelocation" \
--conf "spark.hadoop.mapreduce.output.fileoutputformat.outputdir=/somelocation" \
--conf "spark.hadoop.mapred.output.dir=/somelocation"
Phoexin has plugin and jdbc thin client which can connect(read/write) to HBASE, example are in https://phoenix.apache.org/phoenix_spark.html
Option 1 : Connect via zookeeper url - phoenix plugin
import org.apache.spark.SparkContext
import org.apache.spark.sql.SQLContext
import org.apache.phoenix.spark._
val sc = new SparkContext("local", "phoenix-test")
val sqlContext = new SQLContext(sc)
val df = sqlContext.load(
"org.apache.phoenix.spark",
Map("table" -> "TABLE1", "zkUrl" -> "phoenix-server:2181")
)
df
.filter(df("COL1") === "test_row_1" && df("ID") === 1L)
.select(df("ID"))
.show
Option 2 : Use JDBC thin client provied by phoenix query server
more info on https://phoenix.apache.org/server.html
jdbc:phoenix:thin:url=http://localhost:8765;serialization=PROTOBUF
I am trying to follow the instructions mentioned here...
https://www.percona.com/blog/2016/08/17/apache-spark-makes-slow-mysql-queries-10x-faster/
and here...
https://www.percona.com/blog/2015/10/07/using-apache-spark-mysql-data-analysis/
I am using sparkdocker image.
docker run -it -p 8088:8088 -p 8042:8042 -p 4040:4040 -h sandbox sequenceiq/spark:1.6.0 bash
cd /usr/local/spark/
./sbin/start-master.sh
./bin/spark-shell --driver-memory 1G --executor-memory 1g --executor-cores 1 --master local
This works as expected:
scala> sc.parallelize(1 to 1000).count()
But this shows an error:
val jdbcDF = spark.read.format("jdbc").options(
Map("url" -> "jdbc:mysql://1.2.3.4:3306/test?user=dba&password=dba123",
"dbtable" -> "ontime.ontime_part",
"fetchSize" -> "10000",
"partitionColumn" -> "yeard", "lowerBound" -> "1988", "upperBound" -> "2016", "numPartitions" -> "28"
)).load()
And here is the error:
<console>:25: error: not found: value spark
val jdbcDF = spark.read.format("jdbc").options(
How do I connect to MySQL from within spark shell?
With spark 2.0.x,you can use DataFrameReader and DataFrameWriter.
Use SparkSession.read to access DataFrameReader and use Dataset.write to access DataFrameWriter.
Suppose using spark-shell.
read example
val prop=new java.util.Properties()
prop.put("user","username")
prop.put("password","yourpassword")
val url="jdbc:mysql://host:port/db_name"
val df=spark.read.jdbc(url,"table_name",prop)
df.show()
read example 2
val jdbcDF = spark.read
.format("jdbc")
.option("url", "jdbc:mysql:dbserver")
.option("dbtable", “schema.tablename")
.option("user", "username")
.option("password", "password")
.load()
from spark doc
write example
import org.apache.spark.sql.SaveMode
val prop=new java.util.Properties()
prop.put("user","username")
prop.put("password","yourpassword")
val url="jdbc:mysql://host:port/db_name"
//df is a dataframe contains the data which you want to write.
df.write.mode(SaveMode.Append).jdbc(url,"table_name",prop)
Create the spark context first
Make sure you have jdbc jar files in attached to your classpath
if you are trying to read data from jdbc. use dataframe API instead of RDD as dataframes have better performance. refer to the below performance comparsion graph.
here is the syntax for reading from jdbc
SparkConf conf = new SparkConf().setAppName("app"))
.setMaster("local[2]")
.set("spark.serializer",prop.getProperty("spark.serializer"));
JavaSparkContext sc = new JavaSparkContext(conf);
sqlCtx = new SQLContext(sc);
df = sqlCtx.read()
.format("jdbc")
.option("url", "jdbc:mysql://1.2.3.4:3306/test")
.option("driver", "com.mysql.jdbc.Driver")
.option("dbtable","dbtable")
.option("user", "dbuser")
.option("password","dbpwd"))
.load();
It looks like spark is not defined, you should use the SQLContext to connect to the driver like this:
import org.apache.spark.sql.SQLContext
val sqlcontext = new org.apache.spark.sql.SQLContext(sc)
val dataframe_mysql = sqlcontext.read.format("jdbc").option("url", "jdbc:mysql://Public_IP:3306/DB_NAME").option("driver", "com.mysql.jdbc.Driver").option("dbtable", "tblage").option("user", "sqluser").option("password", "sqluser").load()
Later you can user sqlcontext where you used spark (in spark.read etc)
This is a common problem for those migrating to Spark 2.0.0 from the earlier versions. The Spark documentation is not very good. To solve this, you have to define a SparkSession, like this:
import org.apache.spark.sql.SparkSession
val spark = SparkSession
.builder()
.appName("Spark SQL Example")
.config("spark.some.config.option", "some-value")
.getOrCreate()
This solution is hidden in the Spark SQL, Dataframes and Data Sets Guide located here. SparkSession is the new entry point to the DataFrame API and it incorporates both SQLContext and HiveContext and has some additional advantages, so there is no need to define either of those anymore. Further information about this can be found here.
Please accept this as the answer, if you find this useful.