First, apologies for the title, I wasn't sure how to eloquently describe this succinctly.
I have a spark job that parses logs into JSON, and then using spark-sql converts specific columns into ORC and writes to various paths. For example:
val logs = sc.textFile("s3://raw/logs")
val jsonRows = logs.mapPartitions(partition => {
partition.map(log => {
logToJson.parse(log)
}
}
jsonRows.foreach(r => {
val contentPath = "s3://content/events/"
val userPath = "s3://users/events/"
val contentDf = sqlSession.read.schema(contentSchema).json(r)
val userDf = sqlSession.read.schema(userSchema).json(r)
val userDfFiltered = userDf.select("*").where(userDf("type").isin("users")
// Save Data
val contentWriter = contentDf.write.mode("append").format("orc")
eventWriter.save(contentPath)
val userWriter = userDf.write.mode("append").format("orc")
userWriter.save(userPath)
When I wrote this I expected that the parsing would occur one time, and then it would write to the respective locations afterward. However, it seems that it is executing all of the code in the file twice - once for content and once for users. Is this expected? I would prefer that I don't end up transferring the data from S3 and parsing twice, as that is the largest bottleneck. I am attaching an image from the Spark UI to show the duplication of tasks for a single Streaming Window. Thanks for any help you can provide!
Okay, this kind of nested DFs is a no go. DataFrames are meant to be a data structure for big datasets that won't fit into normal data structures (like Seq or List) and that needs to be processed in a distributed way. It is not just another kind of array. What you are attempting to do here is to create a DataFrame per log line, which makes little sense.
As far as I can tell from the (incomplete) code you have posted here, you want to create two new DataFrames from your original input (the logs) which you then want to store in two different locations. Something like this:
val logs = sc.textFile("s3://raw/logs")
val contentPath = "s3://content/events/"
val userPath = "s3://users/events/"
val jsonRows = logs
.mapPartitions(partition => {
partition.map(log => logToJson.parse(log))
}
.toDF()
.cache() // Or use persist() if dataset is larger than will fit in memory
jsonRows
.write
.format("orc")
.save(contentPath)
jsonRows
.filter(col("type").isin("users"))
.write
.format("orc")
.save(userPath)
Hope this helps.
I am beginner in Scala, I have following question:
How to read more that one csv file and and assign each file to the next variable?
I know how the read one file:
val file_1.sc.textFile("/Users/data/urls_20170225")
I also know how to read many files:
val file_2.sc.textFile("/Users/data/urls_*")
But second way assign all data to one variables file_2, is something that I don't want to! I am looking for elegant way to do this in Spark Scala.
spark has no API to load multiple files into multiple RDD. What you can do is load them one by one into one List of RDD. Below is a sample code.
def main(arg: Array[String]): Unit = {
val dir = """F:\Works\SO\Scala\src\main\resource"""
val startsWith = """urls_""" // we will use this as the wildcard
val fileList:List[File] = getListOfFiles(new File(dir))
val filesRDD: List[RDD[String]] = fileList.collect({
case file: File if file.getName.startsWith(startsWith)=> spark.sparkContext.textFile(file.getPath)
})
}
//Get all the individual file paths
def getListOfFiles(dir: File):List[File] = dir.listFiles.filter(_.isFile).toList
I have a requirement where I want to write each individual records in an RDD to an individual file in HDFS.
I did it for the normal filesystem but obviously, it doesn't work for HDFS.
stream.foreachRDD{ rdd =>
if(!rdd.isEmpty()) {
rdd.foreach{
msg =>
val value = msg._2
println(value)
val fname = java.util.UUID.randomUUID.toString
val path = dir + fname
write(path, value)
}
}
}
where write is a function which writes to the filesystem.
Is there a way to do it within spark so that for each record I can natively write to the HDFS, without using any other tool like Kafka Connect or Flume??
EDIT: More Explanation
For eg:
If my DstreamRDD has the following records,
abcd
efgh
ijkl
mnop
I need different files for each record, so different file for "abcd", different for "efgh" and so on.
I tried creating an RDD within the streamRDD but I learnt it's not allowed as the RDD's are not serializable.
You can forcefully repartition the rdd to no. of partitions as many no. of records and then save
val rddCount = rdd.count()
rdd.repartition(rddCount).saveAsTextFile("your/hdfs/loc")
You can do in couple of ways..
From rdd, you can get the sparkCOntext, once you got the sparkCOntext, you can use parallelize method and pass the String as List of String.
For example:
val sc = rdd.sparkContext
sc.parallelize(Seq("some string")).saveAsTextFile(path)
Also, you can use sqlContext to convert the string to DF then write in the file.
for Example:
import sqlContext.implicits._
Seq(("some string")).toDF
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.
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.
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.
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?
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.
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'm using this in Python to get a single file:
df.toPandas().to_csv("/tmp/my.csv", sep=',', header=True, index=False)
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
)
}
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)
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("<hdfs dir>") --> write as CSV file & its output location in HDFS
repartition/coalesce to 1 partition before you save (you'd still get a folder but it would have one part file in it)
you can use rdd.coalesce(1, true).saveAsTextFile(path)
it will store data as singile file in path/part-00000
Here is a helper function with which you can get a single result-file without the part-0000 and without a subdirectory on S3 and AWS EMR:
def renameSinglePartToParentFolder(directoryUrl: String): Unit = {
import sys.process._
val lsResult = s"aws s3 ls ${directoryUrl}/" !!
val partFilename = lsResult.split("\n").map(_.split(" ").last).filter(_.contains("part-0000")).last
s"aws s3 rm ${directoryUrl}/_SUCCESS" !
s"aws s3 mv ${directoryUrl}/${partFilename} ${directoryUrl}" !
}
val targetPath = "s3://my-bucket/my-folder/my-file.csv"
df.coalesce(1).write.csv(targetPath)
renameSinglePartToParentFolder(targetPath)
Write to a single part-0000... file.
Use AWS CLI to list all files and rename the single file accordingly.
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()
def export_csv(
fileName: String,
filePath: String
) = {
val filePathDestTemp = filePath + ".dir/"
val merstageout_df = spark.sql(merstageout)
merstageout_df
.coalesce(1)
.write
.option("header", "true")
.mode("overwrite")
.csv(filePathDestTemp)
val listFiles = dbutils.fs.ls(filePathDestTemp)
for(subFiles <- listFiles){
val subFiles_name: String = subFiles.name
if (subFiles_name.slice(subFiles_name.length() - 4,subFiles_name.length()) == ".csv") {
dbutils.fs.cp (filePathDestTemp + subFiles_name, filePath + fileName+ ".csv")
dbutils.fs.rm(filePathDestTemp, recurse=true)
}}}
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)}
I have scala code that takes multiple input files from HDFS using wildcards and each files goes into a function where processing takes place for each file individually.
import de.l3s.boilerpipe.extractors.KeepEverythingExtractor
val data = sc.wholeTextFiles("hdfs://localhost:port/akshat/folder/*/*")
val files = data.map { case (filename, content) => filename}
def doSomething(file: String): (String,String) = {
// logic of processing a single file comes here
val logData = sc.textFile(file);
val c = logData.toLocalIterator.mkString
val d = KeepEverythingExtractor.INSTANCE.getText(c)
val e = sc.parallelize(d.split("\n"))
val recipeName = e.take(10).last
val prepTime = e.take(18).last
(recipeName,prepTime)
}
//How transformation and action applied here?
I am stuck at how to to apply further transformations and actions so that all my input files are mapped according to function doSomething and all the output from each of the input files is stored in a single file using saveAsTextFile.
So if my understanding is correct, you have an RDD of Pairs and you wish to transform it some more, and then save the output for each key to a unique file. Transforming it some more is relatively easy, mapValue will allow you to write transformations on just the value, as well any other transformation will work on RDDs of Pairs.
Saving the output to a unique file for each key however, is a bit trickier. One option would be to try and find a hadoopoutput format which does what you want and then use saveAsHadoopFile, another option would be to use foreach and then just write the code to output each key/value pair as desired.