Preferred way of processing this data with parallel arrays - scala

Imagine a sequence of java.io.File objects. The sequence is not in any particular order, it gets populated after a directory traversal. The names of the files can be like this:
/some/file.bin
/some/other_file_x1.bin
/some/other_file_x2.bin
/some/other_file_x3.bin
/some/other_file_x4.bin
/some/other_file_x5.bin
...
/some/x_file_part1.bin
/some/x_file_part2.bin
/some/x_file_part3.bin
/some/x_file_part4.bin
/some/x_file_part5.bin
...
/some/x_file_part10.bin
Basically, I can have 3 types of files. First type is the simple ones, which only have a .bin extension. The second type of file is the one formed from _x1.bin till _x5.bin. And the third type of file can be formed of 10 smaller parts, from _part1 till _part10.
I know the naming may be strange, but this is what I have to work with :)
I want to group the files together ( all the pieces of a file should be processed together ), and I was thinking of using parallel arrays to do this. The thing I'm not sure about is how can I perform the reduce/acumulation part, since all the threads will be working on the same array.
val allBinFiles = allBins.toArray // array of java.io.File
I was thinking of handling something like that:
val mapAcumulator = java.util.Collections.synchronizedMap[String,ListBuffer[File]](new java.util.HashMap[String,ListBuffer[File]]())
allBinFiles.par.foreach { file =>
file match {
// for something like /some/x_file_x4.bin nameTillPart will be /some/x_file
case ComposedOf5Name(nameTillPart) => {
mapAcumulator.getOrElseUpdate(nameTillPart,new ListBuffer[File]()) += file
}
case ComposedOf10Name(nameTillPart) => {
mapAcumulator.getOrElseUpdate(nameTillPart,new ListBuffer[File]()) += file
}
// simple file, without any pieces
case _ => {
mapAcumulator.getOrElseUpdate(file.toString,new ListBuffer[File]()) += file
}
}
}
I was thinking of doing it like I've shown in the above code. Having extractors for the files, and using part of the path as key in the map. Like for example, /some/x_file can hold as values /some/x_file_x1.bin to /some/x_file_x5.bin. I also think there could be a better way of handling this. I would be interested in your opinions.

The alternative is to use groupBy:
val mp = allBinFiles.par.groupBy {
case ComposedOf5Name(x) => x
case ComposedOf10Name(x) => x
case f => f.toString
}
This will return a parallel map of parallel arrays of files (ParMap[String, ParArray[File]]). If you want a sequential map of sequential sequences of files from this point:
val sqmp = mp.map(_.seq).seq
To ensure that the parallelism kicks in, make sure you have enough elements in you parallel array (10k+).

Related

Creating Seq after waiting for all results from map/foreach in Scala

I am trying to loop over inputs and process them to produce scores.
Just for the first input, I want to do some processing that takes a while.
The function ends up returning just the values from the 'else' part. The 'if' part is done executing after the function returns the value.
I am new to Scala and understand the behavior but not sure how to fix it.
I've tried inputs.zipWithIndex.map instead of foreach but the result is the same.
def getscores(
inputs: inputs
): Future[Seq[scoreInfo]] = {
var scores: Seq[scoreInfo] = Seq()
inputs.zipWithIndex.foreach {
case (f, i) => {
if (i == 0) {
// long operation that returns Future[Option[scoreInfo]]
getgeoscore(f).foreach(gso => {
gso.foreach(score => {
scores = scores.:+(score)
})
})
} else {
scores = scores.:+(
scoreInfo(
id = "",
score = 5
)
)
}
}
}
Future {
scores
}
}
For what you need, I would drop the mutable variable and replace foreach with map to obtain an immutable list of Futures and recover to handle exceptions, followed by a sequence like below:
def getScores(inputs: Inputs): Future[List[ScoreInfo]] = Future.sequence(
inputs.zipWithIndex.map{ case (input, idx) =>
if (idx == 0)
getGeoScore(input).map(_.getOrElse(defaultScore)).recover{ case e => errorHandling(e) }
else
Future.successful(ScoreInfo("", 5))
})
To capture/print the result, one way is to use onComplete:
getScores(inputs).onComplete(println)
The part your missing is understanding a tricky element of concurrency, and that is that the order of execution when using multiple futures is not guaranteed.
If your block here is long running, it will take a while before appending the score to scores
// long operation that returns Future[Option[scoreInfo]]
getgeoscore(f).foreach(gso => {
gso.foreach(score => {
// stick a println("here") in here to see what happens, for demonstration purposes only
scores = scores.:+(score)
})
})
Since that executes concurrently, your getscores function will also simultaneously continue its work iterating over the rest of inputs in your zipWithindex. This iteration, especially since it's trivial work, likely finishes well before the long-running getgeoscore(f) completes the execution of the Future it scheduled, and the code will exit the function, moving on to whatever code is next after you called getscores
val futureScores: Future[Seq[scoreInfo]] = getScores(inputs)
futureScores.onComplete{
case Success(scoreInfoSeq) => println(s"Here's the scores: ${scoreInfoSeq.mkString(",")}"
}
//a this point the call to getgeoscore(f) could still be running and finish later, but you will never know
doSomeOtherWork()
Now to clean this up, since you can run a zipWithIndex on your inputs parameter, I assume you mean it's something like a inputs:Seq[Input]. If all you want to do is operate on the first input, then use the head function to only retrieve the first option, so getgeoscores(inputs.head) , you don't need the rest of the code you have there.
Also, as a note, if using Scala, get out of the habit of using mutable vars, especially if you're working with concurrency. Scala is built around supporting immutability, so if you find yourself wanting to use a var , try using a val and look up how to work with the Scala's collection library to make it work.
In general, that is when you have several concurrent futures, I would say Leo's answer describes the right way to do it. However, you want only the first element transformed by a long running operation. So you can use the future return by the respective function and append the other elements when the long running call returns by mapping the future result:
def getscores(inputs: Inputs): Future[Seq[ScoreInfo]] =
getgeoscore(inputs.head)
.map { optInfo =>
optInfo ++ inputs.tail.map(_ => scoreInfo(id = "", score = 5))
}
So you neither need zipWithIndex nor do you need an additional future or join the results of several futures with sequence. Mapping the future just gives you a new future with the result transformed by the function passed to .map().

Need help - How to loop through a list and/or a map

Scala is pretty new for me and I have problems as soon as a leave the gatling dsl.
In my case I call an API (Mailhog) which responds with a lot of mails in json-format. I can’t grab all the values.
I need it with “jsonPath” and I need to “regex” as well.
That leads into a map and a list which I need to iterate through and save each value.
.check(jsonPath("$[*]").ofType[Map[String,Any]].findAll.saveAs("id_map"))
.check(regex("href=3D\\\\\"(.*?)\\\\\"").findAll.saveAs("url_list"))
At first I wanted to loop the “checks” but I did’nt find any to repeat them without repeating the “get”-request too. So it’s a map and a list.
1) I need every value of the map and was able to solve the problem with the following foreach loop.
.foreach("${id_map}", "idx") {
exec(session => {
val idMap = session("idx").as[Map[String,Any]]
val ID = idMap("ID")
session.set("ID", ID)
})
.exec(http("Test")
.get("/{ID}"))
})}
2) I need every 3rd value of the list and make a get-request on them. Before I can do this, I need to replace a part of the string. I tried to replace parts of the string while checking for them. But it won’t work with findAll.
.check(regex("href=3D\\\\\"(.*?)\\\\\"").findAll.transform(raw => raw.replace("""=\r\n""","")).saveAs("url"))
How can I replace a part of every string in my list?
also how can I make a get-request on every 3rd element in the list.
I can't get it to work with the same foreach structure above.
I was abole to solve the problem by myself. At first I made a little change to my check(regex ...) part.
.check(regex("href=3D\\\\\"(.*?)\\\\\"").findAll.transform(_.map(raw => raw.replace("""=\r\n""",""))).saveAs("url_list"))
Then I wanted to make a Get-Request only on every third element of my list (because the URLs I extracted appeared three times per Mail).
.exec(session => {
val url_list =
session("url_list").as[List[Any]].grouped(3).map(_.head).toList
session.set("url_list", url_list)
})
At the end I iterate through my final list with a foreach-loop.
foreach("${url_list}", "urls") {
exec(http("Activate User")
.get("${urls}")
)
}

Iterator[Something] to Iterator[Seq[Something]]

I need to process a "big" file (something that does not fit in memory).
I want to batch-process the data. Let's say for the example that I want to insert them into a database. But since it is too big to fit in memory, it is too slow too to process elements one-by-one.
So I'l like to go from an Iterator[Something] to an Iterator[Iterable[Something]] to batch elements.
Starting with this:
CSVReader.open(new File("big_file"))
.iteratorWithHeaders
.map(Something.parse)
.foreach(Jdbi.insertSomething)
I could do something dirty in the foreach statement with mutable sequences and flushes every x elements but I'm sure there is a smarter way to do this...
// Yuk... :-(
val buffer = ArrayBuffer[Something]()
CSVReader.open(new File("big_file"))
.iteratorWithHeaders
.map(Something.parse)
.foreach {
something =>
buffer.append(something)
if (buffer.size == 1000) {
Jdbi.insertSomethings(buffer.toList)
buffer.clear()
}
}
Jdbi.insertSomethings(buffer.toList)
If your batches can have a fixed size (as in your example), the grouped method on Scala's Iterator does exactly what you want:
val iterator = Iterator.continually(1)
iterator.grouped(10000).foreach(xs => println(xs.size))
This will run in a constant amount of memory (not counting whatever text in stored by your terminal in memory, of course).
I'm not sure what your iteratorWithHeaders returns, but if it's a Java iterator, you can convert it to a Scala one like this:
import scala.collection.JavaConverters.
val myScalaIterator: Iterator[Int] = myJavaIterator.asScala
This will remain appropriately lazy.
If I undestood correctly your problem, you can just use Iterator.grouped. So adapting a little bit your example:
val si: Iterator[Something] = CSVReader.open(new File("big_file"))
.iteratorWithHeaders
.map(Something.parse)
val gsi: GroupedIterator[Something] = si.grouped(1000)
gsi.foreach { slst: List[Something] =>
Jdbi.insertSomethings(slst)
}

Way to Extract list of elements from Scala list

I have standard list of objects which is used for the some analysis. The analysis generates a list of Strings and i need to look through the standard list of objects and retrieve objects with same name.
case class TestObj(name:String,positions:List[Int],present:Boolean)
val stdLis:List[TestObj]
//analysis generates a list of strings
var generatedLis:List[String]
//list to save objects found in standard list
val lisBuf = new ListBuffer[TestObj]()
//my current way
generatedLis.foreach{i=>
val temp = stdLis.filter(p=>p.name.equalsIgnoreCase(i))
if(temp.size==1){
lisBuf.append(temp(0))
}
}
Is there any other way to achieve this. Like having an custom indexof method that over rides and looks for the name instead of the whole object or something. I have not tried that approach as i am not sure about it.
stdLis.filter(testObj => generatedLis.exists(_.equalsIgnoreCase(testObj.name)))
use filter to filter elements from 'stdLis' per predicate
use exists to check if 'generatedLis' has a value of ....
Don't use mutable containers to filter sequences.
Naive solution:
val lisBuf =
for {
str <- generatedLis
temp = stdLis.filter(_.name.equalsIgnoreCase(str))
if temp.size == 1
} yield temp(0)
if we discard condition temp.size == 1 (i'm not sure it is legal or not):
val lisBuf = stdLis.filter(s => generatedLis.exists(_.equalsIgnoreCase(s.name)))

How to generate a big data stream on the fly

I have to generate a big file on the fly. Reading to the database and send it to the client.
I read some documentation and i did this
val streamContent: Enumerator[Array[Byte]] = Enumerator.outputStream {
os =>
// new PrintWriter() read from database and for each record
// do some logic and write
// to outputstream
}
Ok.stream(streamContent.andThen(Enumerator.eof)).withHeaders(
CONTENT_DISPOSITION -> s"attachment; filename=someName.csv"
)
Im rather new to scala in general only a week so don't guide for my reputation.
My questions are :
1) Is this the best way? I found this if i have a big file, this will load in memory, and also don't know what is the chunk size in this case, if it will send for each write() is not to convenient.
2) I found this method Enumerator.fromStream(data : InputStream, chunkedSize : int) a little better cause it has a chunk-size, but i don't have an inputStream cause im creating the file on the fly.
There's a note in the docs for Enumerator.outputStream:
Not [sic!] that calls to write will not block, so if the iteratee that is being fed to is slow to consume the input, the OutputStream will not push back. This means it should not be used with large streams since there is a risk of running out of memory.
If this can happen depends on your situation. If you can and will generate Gigabytes in seconds, you should probably try something different. I'm not exactly sure what, but I'd start at Enumerator.generateM(). For many cases though, your method is perfectly fine. Have a look at this example by Gaëtan Renaudeau for serving a Zip file that's generated on the fly in the same way you're using it:
val enumerator = Enumerator.outputStream { os =>
val zip = new ZipOutputStream(os);
Range(0, 100).map { i =>
zip.putNextEntry(new ZipEntry("test-zip/README-"+i+".txt"))
zip.write("Here are 100000 random numbers:\n".map(_.toByte).toArray)
// Let's do 100 writes of 1'000 numbers
Range(0, 100).map { j =>
zip.write((Range(0, 1000).map(_=>r.nextLong).map(_.toString).mkString("\n")).map(_.toByte).toArray);
}
zip.closeEntry()
}
zip.close()
}
Ok.stream(enumerator >>> Enumerator.eof).withHeaders(
"Content-Type"->"application/zip",
"Content-Disposition"->"attachment; filename=test.zip"
)
Please keep in mind that Ok.stream has been replaced by Ok.chunked in newer versions of Play, in case you want to upgrade.
As for the chunk size, you can always use Enumeratee.grouped to gather a bunch of values and send them as one chunk.
val grouper = Enumeratee.grouped(
Traversable.take[Array[Double]](100) &>> Iteratee.consume()
)
Then you'd do something like
Ok.stream(enumerator &> grouper >>> Enumerator.eof)