I have a jupyter notebook running a spylon-kernel (Scala / Spark).
Currently, I try to load records from a csv into a RDD and then map each record into objects of the "Weather" class as follows:
val lines = scala.io.Source.fromFile("/path/to/nycweather.csv").mkString
println(lines)
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
//Next, you need to import a library for creating a SchemaRDD. Type this:
import sqlContext.implicits._
//Create a case class in Scala that defines the schema of the table. Type in:
case class Weather(date: String, temp: Int, precipitation: Double)
//Create the RDD of the Weather object:
val weather = sc.textFile("/path/to/nycweather.csv").map(_.split(",")). map(w => Weather(w(0), w(1).trim.toInt, w(2).trim.toDouble)).toDF()
//It all works fine until the last line above.
//But when I run this line of code:
weather.first()
It all bursts out with the following error message
the message has a couple more lines but I omitted to be more visible.
Could someone indicate why am I getting this error and suggest code changes to solve it?
You are using older RDD syntax for reading a CSV. There is an easier way to read a CSV as
val weather1 = spark.read.csv("path to nycweather.csv").toDF("date","temp","precipitation")
weather1.show()
Input file contains following data
1/1/2010,30,35.0
2/4/2015,35,27.9
Result
+--------+----+-------------+
| date|temp|precipitation|
+--------+----+-------------+
|1/1/2010| 30| 35.0|
|2/4/2015| 35| 27.9|
+--------+----+-------------+
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
When I use Spark to parse log files, I notice that if the first character of filename is _ , the result will be empty. Here is my test code:
SparkSession spark = SparkSession
.builder()
.appName("TestLog")
.master("local")
.getOrCreate();
JavaRDD<String> input = spark.read().text("D:\\_event_2.log").javaRDD();
System.out.println("size : " + input.count());
If I modify the file name to event_2.log, the code will run it correctly.
I found that the text function is defined as:
#scala.annotation.varargs
def text(paths: String*): Dataset[String] = {
format("text").load(paths : _*).as[String](sparkSession.implicits.newStringEncoder)
}
I think it could be due to _ being scala's placeholder. How can I avoid this problem?
This has nothing to do with Scala. Spark uses Hadoop Input API to read file, which ignore every file that starts with underscore(_) or dot (.)
I don't know how to disable this in Spark though.
Hi I am new to spark and scala. I am running scala code in spark scala prompt. The program is fine, it's showing "defined module MLlib" but its not printing anything on screen. What have I done wrong? Is there any other way to run this program spark in scala shell and get the output?
import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.mllib.classification.LogisticRegressionWithSGD
import org.apache.spark.mllib.feature.HashingTF
import org.apache.spark.mllib.regression.LabeledPoint
object MLlib {
def main(args: Array[String]) {
val conf = new SparkConf().setAppName(s"Book example: Scala")
val sc = new SparkContext(conf)
// Load 2 types of emails from text files: spam and ham (non-spam).
// Each line has text from one email.
val spam = sc.textFile("/home/training/Spam.txt")
val ham = sc.textFile("/home/training/Ham.txt")
// Create a HashingTF instance to map email text to vectors of 100 features.
val tf = new HashingTF(numFeatures = 100)
// Each email is split into words, and each word is mapped to one feature.
val spamFeatures = spam.map(email => tf.transform(email.split(" ")))
val hamFeatures = ham.map(email => tf.transform(email.split(" ")))
// Create LabeledPoint datasets for positive (spam) and negative (ham) examples.
val positiveExamples = spamFeatures.map(features => LabeledPoint(1, features))
val negativeExamples = hamFeatures.map(features => LabeledPoint(0, features))
val trainingData = positiveExamples ++ negativeExamples
trainingData.cache() // Cache data since Logistic Regression is an iterative algorithm.
// Create a Logistic Regression learner which uses the LBFGS optimizer.
val lrLearner = new LogisticRegressionWithSGD()
// Run the actual learning algorithm on the training data.
val model = lrLearner.run(trainingData)
// Test on a positive example (spam) and a negative one (ham).
// First apply the same HashingTF feature transformation used on the training data.
val posTestExample = tf.transform("O M G GET cheap stuff by sending money to ...".split(" "))
val negTestExample = tf.transform("Hi Dad, I started studying Spark the other ...".split(" "))
// Now use the learned model to predict spam/ham for new emails.
println(s"Prediction for positive test example: ${model.predict(posTestExample)}")
println(s"Prediction for negative test example: ${model.predict(negTestExample)}")
sc.stop()
}
}
A couple of things:
You defined your object in the the Spark shell, so the main class won't get called immediately. You'll have to call it explicitly after you define the object:
MLlib.main(Array())
In fact, if you continue to work on the shell/REPL you can do away with the object altogether; you can define the function directly. For example:
import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.mllib.classification.LogisticRegressionWithSGD
import org.apache.spark.mllib.feature.HashingTF
import org.apache.spark.mllib.regression.LabeledPoint
def MLlib {
//the rest of your code
}
However, you shouldn't initialize SparkContext it within the shell. From the documentation:
In the Spark shell, a special interpreter-aware SparkContext is
already created for you, in the variable called sc. Making your own
SparkContext will not work
So, you have to either remove that bit from your code, or compile it into a jar and run it using spark-submit
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)}