Hi I have 90 GB data In CSV file I'm loading this data into one temp table and then from temp table to orc table using select insert command but for converting and loading data into orc format its taking 4 hrs in spark sql.Is there any kind of optimization technique which i can use to reduce this time.As of now I'm not using any kind of optimization technique I'm just using spark sql and loading data from csv file to table(textformat) and then from this temp table to orc table(using select insert)
using spark submit as:
spark-submit \
--class class-name\
--jar file
or can I add any extra Parameter in spark submit for improving the optimization.
scala code(sample):
All Imports
object sample_1 {
def main(args: Array[String]) {
//sparksession with enabled hivesuppport
var a1=sparksession.sql("load data inpath 'filepath' overwrite into table table_name")
var b1=sparksession.sql("insert into tablename (all_column) select 'ALL_COLUMNS' from source_table")
}
}
First of all, you don't need to store the data in the temp table to write into hive table later. You can straightaway read the file and write the output using the DataFrameWriter API. This will reduce one step from your code.
You can write as follows:
val spark = SparkSession.builder.enableHiveSupport().getOrCreate()
val df = spark.read.csv(filePath) //Add header or delimiter options if needed
inputDF.write.mode("append").format(outputFormat).saveAsTable(outputDB + "." + outputTableName)
Here, the outputFormat will be orc, the outputDB will be your hive database and outputTableName will be your Hive table name.
I think using the above technique, your write time will reduce significantly. Also, please mention the resources your job is using and I may be able to optimize it further.
Another optimization you can use is to partition your dataframe while writing. This will make the write operation faster. However, you need to decide the columns on which to partition carefully so that you don't end up creating a lot of partitions.
Related
In spark-shell, how do I load an existing Hive table, but only one of its partitions?
val df = spark.read.format("orc").load("mytable")
I was looking for a way so it only loads one particular partition of this table.
Thanks!
There is no direct way in spark.read.format but you can use where condition
val df = spark.read.format("orc").load("mytable").where(yourparitioncolumn)
unless until you perform an action nothing is loaded, since load (pointing to your orc file location ) is just a func in DataFrameReader like below it doesnt load until actioned.
see here DataFrameReader
def load(paths: String*): DataFrame = {
...
}
In above code i.e. spark.read.... where is just where condition when you specify this, again data wont be loaded immediately :-)
when you say df.count then your parition column will be appled on data path of orc.
There is no function available in Spark API to load only partition directory, but other way around this is partiton directory is nothing but column in where clause, here you can right simple sql query with partition column in where clause which will read data only from partition directoty. See if that will works for you.
val df = spark.sql("SELECT * FROM mytable WHERE <partition_col_name> = <expected_value>")
Hi I have 90 GB data In CSV file I'm loading this data into one temp table and then from temp table to orc table using select insert command but for converting and loading data into orc format its taking 4 hrs in spark sql.Is there any kind of optimization technique which i can use to reduce this time.As of now I'm not using any kind of optimization technique I'm just using spark sql and loading data from csv file to table(textformat) and then from this temp table to orc table(using select insert)
using spark submit as:
spark-submit \
--class class-name\
--jar file
or can I add any extra Parameter in spark submit for improving the optimization.
scala code(sample):
All Imports
object demo {
def main(args: Array[String]) {
//sparksession with enabled hivesuppport
var a1=sparksession.sql("load data inpath 'filepath' overwrite into table table_name")
var b1=sparksession.sql("insert into tablename (all_column) select 'ALL_COLUMNS' from source_table")
}
}
I'm just using spark sql and loading data from csv file to
table(textformat) and then from this temp table to orc table(using
select insert)
2 step process is not needed here..
Read the dataframe like below sample...
val DFCsv = spark.read.format("csv")
.option("sep", ",")
.option("inferSchema", "true")
.option("header", "true")
.load("yourcsv")
if needed you have to do repartition(may be this is cause of the actual 4hr delay since you have not done) since its large file and then...
dfcsv.repartition(90) means it will/may repartition the csv data in to 90 almost equal parts. where 90 is sample number. you can mention what ever you want.
DFCsv.write.format("orc")
.partitionBy('yourpartitioncolumns')
.saveAsTable('yourtable')
OR
DFCsv.write.format("orc")
.partitionBy('yourpartitioncolumns')
.insertInto('yourtable')
Note: 1) For large data you need to do repartition to uniformly distribute the data will increase the parllelism and hence
performance.
2) If you dont have patition columns and is
non-partition table then no need of partitionBy in the above
samples
I'm reading quite some data (2.3TB) into a spark dataframe.
All CSV files prepared for a prediction model.
Once loaded we use a temporary view to store it
dSales = spark.read.option("delimiter",",").option("header", "true").option("inferSchema", "true").csv("/mnt/" + sourceMountName + "/")
dSales.createOrReplaceTempView("dSales")
After that we produce several other tables with joins and write them all to the database. These tables are used in PowerBI.
My question is, how can I get that big Sales dataframe and the Tempview out of memory once everything are processed?
i am processing more than 1000000 records of json file i am reading file line by line and extract requried key values
(json are mix structure is not fix. so i am parsing and generate requried json element) and generate json string simillar to json_string variable and push to hive table data are store properly but at hadoop apps/hive/warehouse/jsondb.myjson_table folder contain small part files. every insert query the new (.1 to .20 kb)part file will be created. beacuse of that if i run simple query on hive as it will take more than 30 min. showing sample code of my logic this iterate multipal times for new records to inesrt in hive.
import org.apache.spark.sql.SparkSession
val spark = SparkSession.builder().appName("SparkSessionZipsExample").enableHiveSupport().getOrCreate()
var json_string = """{"name":"yogesh_wagh","education":"phd" }"""
val df = spark.read.json(Seq(json_string).toDS)
//df.write.format("orc").saveAsTable("bds_data1.newversion");
df.write.mode("append").format("orc").insertInto("bds_data1.newversion");
i have also try to add hive property to merge the files but it wont work,
i have also try to create table from existing table for combine small part file to one 256 mb files..
please share sample code to insert multipal records and append record in part file.
I think each of those individual inserts creating a new part file.
You could create dataset/dataframe of these json strings and then save it to hive table.
you could merge the existing small file using hive ddl ALTER TABLE table_name CONCATENATE;
I am trying to follow this example to save some data in parquet format and read it. If I use the write.parquet("filename"), then the iterating Spark job gives error that
"filename" already exists.
If I use SaveMode.Append option, then the Spark job gives the error
".spark.sql.AnalysisException: Specifying database name or other qualifiers are not allowed for temporary tables".
Please let me know the best way to ensure new data is just appended to the parquet file. Can I define primary keys on these parquet tables?
I am using Spark 1.6.2 on Hortonworks 2.5 system. Here is the code:
// Option 1: peopleDF.write.parquet("people.parquet")
//Option 2:
peopleDF.write.format("parquet").mode(SaveMode.Append).saveAsTable("people.parquet")
// Read in the parquet file created above
val parquetFile = spark.read.parquet("people.parquet")
//Parquet files can also be registered as tables and then used in SQL statements.
parquetFile.registerTempTable("parquetFile")
val teenagers = sqlContext.sql("SELECT * FROM people.parquet")
I believe if you use .parquet("...."), you should use .mode('append'),
not SaveMode.Append:
df.write.mode('append').parquet("....")