Hello Rui,

Sorry , What I meant was the resultant of the original dataframe to which a
new column was added gives a new DataFrame.

Please check this for more

https://spark.apache.org/docs/1.5.1/api/R/index.html

Check for
WithColumn


Thanks,
Vipul


On 23 November 2015 at 12:42, Sun, Rui <rui....@intel.com> wrote:

> Vipul,
>
> Not sure if I understand your question. DataFrame is immutable. You can't
> update a DataFrame.
>
> Could you paste some log info for the OOM error?
>
> -----Original Message-----
> From: vipulrai [mailto:vipulrai8...@gmail.com]
> Sent: Friday, November 20, 2015 12:11 PM
> To: user@spark.apache.org
> Subject: SparkR DataFrame , Out of memory exception for very small file.
>
> Hi Users,
>
> I have a general doubt regarding DataFrames in SparkR.
>
> I am trying to read a file from Hive and it gets created as DataFrame.
>
> sqlContext <- sparkRHive.init(sc)
>
> #DF
> sales <- read.df(sqlContext, "hdfs://sample.csv", header ='true',
>                  source = "com.databricks.spark.csv", inferSchema='true')
>
> registerTempTable(sales,"Sales")
>
> Do I need to create a new DataFrame for every update to the DataFrame like
> addition of new column or  need to update the original sales DataFrame.
>
> sales1<- SparkR::sql(sqlContext,"Select a.* , 607 as C1 from Sales as a")
>
>
> Please help me with this , as the orignal file is only 20MB but it throws
> out of memory exception on a cluster of 4GB Master and Two workers of 4GB
> each.
>
> Also, what is the logic with DataFrame do I need to register and drop
> tempTable after every update??
>
> Thanks,
> Vipul
>
>
>
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-- 
Regards,
Vipul Rai
www.vipulrai.me
+91-8892598819
<http://in.linkedin.com/in/vipulrai/>

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