[scala] Write single CSV file using spark-csv

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 parameter like path and file name and write that CSV file.

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)}

spark.sql("select * from df").coalesce(1).write.option("mode","append").option("header","true").csv("/your/hdfs/path/")

spark.sql("select * from df") --> this is dataframe

coalesce(1) or repartition(1) --> this will make your output file to 1 part file only

write --> writing data

option("mode","append") --> appending data to existing directory

option("header","true") --> enabling header

csv("") --> write as CSV file & its output location in HDFS

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)

If you are using Databricks and can fit all the data into RAM on one worker (and thus can use .coalesce(1)), you can use dbfs to find and move the resulting CSV file:

val fileprefix= "/mnt/aws/path/file-prefix"

dataset
  .coalesce(1)       
  .write             
//.mode("overwrite") // I usually don't use this, but you may want to.
  .option("header", "true")
  .option("delimiter","\t")
  .csv(fileprefix+".tmp")

val partition_path = dbutils.fs.ls(fileprefix+".tmp/")
     .filter(file=>file.name.endsWith(".csv"))(0).path

dbutils.fs.cp(partition_path,fileprefix+".tab")

dbutils.fs.rm(fileprefix+".tmp",recurse=true)

If your file does not fit into RAM on the worker, you may want to consider chaotic3quilibrium's suggestion to use FileUtils.copyMerge(). I have not done this, and don't yet know if is possible or not, e.g., on S3.

This answer is built on previous answers to this question as well as my own tests of the provided code snippet. I originally posted it to Databricks and am republishing it here.

The best documentation for dbfs's rm's recursive option I have found is on a Databricks forum.

It is creating a folder with multiple files, because each partition is saved individually. If you need a single output file (still in a folder) you can repartition (preferred if upstream data is large, but requires a shuffle):

df
   .repartition(1)
   .write.format("com.databricks.spark.csv")
   .option("header", "true")
   .save("mydata.csv")

or coalesce:

df
   .coalesce(1)
   .write.format("com.databricks.spark.csv")
   .option("header", "true")
   .save("mydata.csv")

data frame before saving:

All data will be written to mydata.csv/part-00000. Before you use this option be sure you understand what is going on and what is the cost of transferring all data to a single worker. If you use distributed file system with replication, data will be transfered multiple times - first fetched to a single worker and subsequently distributed over storage nodes.

Alternatively you can leave your code as it is and use general purpose tools like cat or HDFS getmerge to simply merge all the parts afterwards.

by using Listbuffer we can save data into single file:

import java.io.FileWriter
import org.apache.spark.sql.SparkSession
import scala.collection.mutable.ListBuffer
    val text = spark.read.textFile("filepath")
    var data = ListBuffer[String]()
    for(line:String <- text.collect()){
      data += line
    }
    val writer = new FileWriter("filepath")
    data.foreach(line => writer.write(line.toString+"\n"))
    writer.close()

spark's df.write() API will create multiple part files inside given path ... to force spark write only a single part file use df.coalesce(1).write.csv(...) instead of df.repartition(1).write.csv(...) as coalesce is a narrow transformation whereas repartition is a wide transformation see Spark - repartition() vs coalesce()

df.coalesce(1).write.csv(filepath,header=True) 

will create folder in given filepath with one part-0001-...-c000.csv file use

cat filepath/part-0001-...-c000.csv > filename_you_want.csv 

to have a user friendly filename

This answer expands on the accepted answer, gives more context, and provides code snippets you can run in the Spark Shell on your machine.

More context on accepted answer

The accepted answer might give you the impression the sample code outputs a single mydata.csv file and that's not the case. Let's demonstrate:

val df = Seq("one", "two", "three").toDF("num")
df
  .repartition(1)
  .write.csv(sys.env("HOME")+ "/Documents/tmp/mydata.csv")

Here's what's outputted:

Documents/
  tmp/
    mydata.csv/
      _SUCCESS
      part-00000-b3700504-e58b-4552-880b-e7b52c60157e-c000.csv

N.B. mydata.csv is a folder in the accepted answer - it's not a file!

How to output a single file with a specific name

We can use spark-daria to write out a single mydata.csv file.

import com.github.mrpowers.spark.daria.sql.DariaWriters
DariaWriters.writeSingleFile(
    df = df,
    format = "csv",
    sc = spark.sparkContext,
    tmpFolder = sys.env("HOME") + "/Documents/better/staging",
    filename = sys.env("HOME") + "/Documents/better/mydata.csv"
)

This'll output the file as follows:

Documents/
  better/
    mydata.csv

S3 paths

You'll need to pass s3a paths to DariaWriters.writeSingleFile to use this method in S3:

DariaWriters.writeSingleFile(
    df = df,
    format = "csv",
    sc = spark.sparkContext,
    tmpFolder = "s3a://bucket/data/src",
    filename = "s3a://bucket/data/dest/my_cool_file.csv"
)

See here for more info.

Avoiding copyMerge

copyMerge was removed from Hadoop 3. The DariaWriters.writeSingleFile implementation uses fs.rename, as described here. Spark 3 still used Hadoop 2, so copyMerge implementations will work in 2020. I'm not sure when Spark will upgrade to Hadoop 3, but better to avoid any copyMerge approach that'll cause your code to break when Spark upgrades Hadoop.

Source code

Look for the DariaWriters object in the spark-daria source code if you'd like to inspect the implementation.

PySpark implementation

It's easier to write out a single file with PySpark because you can convert the DataFrame to a Pandas DataFrame that gets written out as a single file by default.

from pathlib import Path
home = str(Path.home())
data = [
    ("jellyfish", "JALYF"),
    ("li", "L"),
    ("luisa", "LAS"),
    (None, None)
]
df = spark.createDataFrame(data, ["word", "expected"])
df.toPandas().to_csv(home + "/Documents/tmp/mydata-from-pyspark.csv", sep=',', header=True, index=False)

Limitations

The DariaWriters.writeSingleFile Scala approach and the df.toPandas() Python approach only work for small datasets. Huge datasets can not be written out as single files. Writing out data as a single file isn't optimal from a performance perspective because the data can't be written in parallel.

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?

I'm using this in Python to get a single file:

df.toPandas().to_csv("/tmp/my.csv", sep=',', header=True, index=False)

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

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

A solution that works for S3 modified from Minkymorgan.

Simply pass the temporary partitioned directory path (with different name than final path) as the srcPath and single final csv/txt as destPath Specify also deleteSource if you want to remove the original directory.

/**
* Merges multiple partitions of spark text file output into single file. 
* @param srcPath source directory of partitioned files
* @param dstPath output path of individual path
* @param deleteSource whether or not to delete source directory after merging
* @param spark sparkSession
*/
def mergeTextFiles(srcPath: String, dstPath: String, deleteSource: Boolean): Unit =  {
  import org.apache.hadoop.fs.FileUtil
  import java.net.URI
  val config = spark.sparkContext.hadoopConfiguration
  val fs: FileSystem = FileSystem.get(new URI(srcPath), config)
  FileUtil.copyMerge(
    fs, new Path(srcPath), fs, new Path(dstPath), deleteSource, config, null
  )
}

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.

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