Converting binary files

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Converting binary files

Christopher Piggott
I have been searching for examples, but not finding exactly what I need.

I am looking for the paradigm for using spark 2.2 to convert a bunch of binary files into a bunch of different binary files.  I'm starting with:

   val files = spark.sparkContext.binaryFiles("hdfs://1.2.3.4/input")

then convert them:

   val converted = files.map {   case (filename, content) =>   ( filename -> convert(content) }

but I don't really want to save by 'partition', I want to save the file using the original name but in a different directory.e.g. "converted/*" 

I'm not quite sure how I'm supposed to do this within the framework of what's available to me in SparkContext.  Do I need to do it myself using the HDFS api?

It would seem like this would be a pretty normal thing to do.  Imagine for instance I were saying take a bunch of binary files and compress them, and save the compressed output to a different directory.  I feel like I'm missing something fundamental here.

--C



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Re: Converting binary files

JayeshLalwani

You can repartition your dataframe into 1 partition and all the data will land into one partition. However, doing this is perilious because you will end up with all your data on one node, and if you have too much data you will run out of memory. In fact, anytime you are thinking about putting data in a single file, you should ask yourself “Does this data fit into memory?”

 

The reason why Spark is geared towards reading and writing data in a partitioned manner is because fundamentally, partitioning data is how you scale your applications. Partitioned data allows Spark (or really any application that is designed to scale on a cluster) to read data in parallel, process it and spit out, without any bottlenecking. Humans prefer all their data in a single file/table, because humans have a limited ability of keeping track of multitude of files. Grid enabled software hate single files, simply because there is no good way for 2 nodes to read a large file without having some sort of bottlenecking

 

Imagine a data processing pipeline that starts with some sort of ingestion and transformation at one end, which feeds into several analytical processes. Usually there are humans at the end who are looking at the results of the analytics.  These humans love to get their analytics in a dashboard that gives them a high-level view of the data. However, all the data processing systems that go from input to analytics, prefer their data to be cut up into bite sized chunks

 

From: Christopher Piggott <[hidden email]>
Date: Saturday, December 30, 2017 at 3:45 PM
To: "[hidden email]" <[hidden email]>
Subject: Converting binary files

 

I have been searching for examples, but not finding exactly what I need.

 

I am looking for the paradigm for using spark 2.2 to convert a bunch of binary files into a bunch of different binary files.  I'm starting with:

 

   val files = spark.sparkContext.binaryFiles("hdfs://1.2.3.4/input")

 

then convert them:

 

   val converted = files.map {   case (filename, content) =>   ( filename -> convert(content) }

 

but I don't really want to save by 'partition', I want to save the file using the original name but in a different directory.e.g. "converted/*" 

 

I'm not quite sure how I'm supposed to do this within the framework of what's available to me in SparkContext.  Do I need to do it myself using the HDFS api?

 

It would seem like this would be a pretty normal thing to do.  Imagine for instance I were saying take a bunch of binary files and compress them, and save the compressed output to a different directory.  I feel like I'm missing something fundamental here.

 

--C

 

 

 



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