Write single CSV file using spark-csv

Question

I am using https://github.com/databricks/spark-csv , I am trying to write a single CSV, but not able to, it is making a folder.

Need a Scala function which will take

Answer 1

There is one more way to use Java

import java.io._

def printToFile(f: java.io.File)(op: java.io.PrintWriter => Unit) 
  {
     val p = new java.io.PrintWriter(f);  
     try { op(p) } 
     finally { p.close() }
  } 

printToFile(new File("C:/TEMP/df.csv")) { p => df.collect().foreach(p.println)}

Answer 2

If you are running Spark with HDFS, I've been solving the problem by writing csv files normally and leveraging HDFS to do the merging. I'm doing that in Spark (1.6) directly:

import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs._

def merge(srcPath: String, dstPath: String): Unit =  {
   val hadoopConfig = new Configuration()
   val hdfs = FileSystem.get(hadoopConfig)
   FileUtil.copyMerge(hdfs, new Path(srcPath), hdfs, new Path(dstPath), true, hadoopConfig, null) 
   // the "true" setting deletes the source files once they are merged into the new output
}


val newData = << create your dataframe >>


val outputfile = "/user/feeds/project/outputs/subject"  
var filename = "myinsights"
var outputFileName = outputfile + "/temp_" + filename 
var mergedFileName = outputfile + "/merged_" + filename
var mergeFindGlob  = outputFileName

    newData.write
        .format("com.databricks.spark.csv")
        .option("header", "false")
        .mode("overwrite")
        .save(outputFileName)
    merge(mergeFindGlob, mergedFileName )
    newData.unpersist()

Can't remember where I learned this trick, but it might work for you.

Answer 3

you can use rdd.coalesce(1, true).saveAsTextFile(path)

it will store data as singile file in path/part-00000

Answer 4

repartition/coalesce to 1 partition before you save (you'd still get a folder but it would have one part file in it)

Answer 5

I might be a little late to the game here, but using coalesce(1) or repartition(1) may work for small data-sets, but large data-sets would all be thrown into one partition on one node. This is likely to throw OOM errors, or at best, to process slowly.

I would highly suggest that you use the FileUtil.copyMerge() function from the Hadoop API. This will merge the outputs into a single file.

EDIT - This effectively brings the data to the driver rather than an executor node. Coalesce() would be fine if a single executor has more RAM for use than the driver.

EDIT 2: copyMerge() is being removed in Hadoop 3.0. See the following stack overflow article for more information on how to work with the newest version: How to do CopyMerge in Hadoop 3.0?

Answer 6

import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs._
import org.apache.spark.sql.{DataFrame,SaveMode,SparkSession}
import org.apache.spark.sql.functions._

I solved using below approach (hdfs rename file name):-

Step 1:- (Crate Data Frame and write to HDFS)

df.coalesce(1).write.format("csv").option("header", "false").mode(SaveMode.Overwrite).save("/hdfsfolder/blah/")

Step 2:- (Create Hadoop Config)

val hadoopConfig = new Configuration()
val hdfs = FileSystem.get(hadoopConfig)

Step3 :- (Get path in hdfs folder path)

val pathFiles = new Path("/hdfsfolder/blah/")

Step4:- (Get spark file names from hdfs folder)

val fileNames = hdfs.listFiles(pathFiles, false)
println(fileNames)

setp5:- (create scala mutable list to save all the file names and add it to the list)

    var fileNamesList = scala.collection.mutable.MutableList[String]()
    while (fileNames.hasNext) {
      fileNamesList += fileNames.next().getPath.getName
    }
    println(fileNamesList)

Step 6:- (filter _SUCESS file order from file names scala list)

    // get files name which are not _SUCCESS
    val partFileName = fileNamesList.filterNot(filenames => filenames == "_SUCCESS")

step 7:- (convert scala list to string and add desired file name to hdfs folder string and then apply rename)

val partFileSourcePath = new Path("/yourhdfsfolder/"+ partFileName.mkString(""))
    val desiredCsvTargetPath = new Path(/yourhdfsfolder/+ "op_"+ ".csv")
    hdfs.rename(partFileSourcePath , desiredCsvTargetPath)