Splitting string into groups
I'm trying to 'group' a string into segments, I guess this example would explain it more succintly scala> val str: String = "aaaabbcddeeeeeeffg" ... (do something) res0: List("aaaa","bb","c","dd","eeeee","ff","g") I can thnk of a few ways to do this in an imperative style (with vars and stepping through the string to find groups) but I was wondering if any better functional solution could be attai
How to force SBT to use Java 8?
How can I force SBT to compile to Java 8 class files. I added scalacOptions += "-target:jvm-1.8" but it gives the following error message: [error] 'jvm-1.8' is not a valid choice for '-target' [error] bad option: '-target:jvm-1.8' [error] (compile:compile) Compilation failed I am using SBT version 0.15.5. I know I am using Java 8 to compile as I added this to build.sbt, but I still wonder why the
How to load local file in sc.textFile, instead of HDFS
I'm following the great spark tutorial so i'm trying at 46m:00s to load the README.md but fail to what i'm doing is this: $ sudo docker run -i -t -h sandbox sequenceiq/spark:1.1.0 /etc/bootstrap.sh -bash bash-4.1# cd /usr/local/spark-1.1.0-bin-hadoop2.4 bash-4.1# ls README.md README.md bash-4.1# ./bin/spark-shell scala> val f = sc.textFile("README.md") 14/12/04 12:11:14 INFO storage.MemoryStore: e
Most elegant repeat loop in Scala
I'm looking for an equivalent of: for(_ <- 1 to n) some.code() that would be shortest and most elegant. Isn't there in Scala anything similar to this? rep(n) some.code() This is one of the most common constructs after all. PS I know it's easy to implement rep, but I'm looking for something predefined.
Why does my Scala list disappear in the following code?
val files = new File("data").list() val filtered = files.filter(name => name.contains("txn")) val masterList = new ListBuffer[String] for (file <- filtered) { val lines = Source.fromFile(new File("data\\" + file), "UTF-16").getLines val cleaned = lines.filter(!masterList.contains(_)) println("*" + cleaned.length) cleaned.foreach(println(_)) println("**" + cleaned.length) cleaned.foreach(masterList
Why does IDEA 14 report "Cannot load facet "Scala": Unknown type of facet "scala""?
I'm trying to run play framework application in Intellij IDEA 14. I installed the Scala plugin and have all jar files required. However inside the Project-Structure -> Facet, I cannot find the Scala facet in the adding list. And also it errors with: Cannot load facet "Scala": Unknown type of facet "scala" I tried to create a new project with SBT, still I cannot find the Scala facet in the adding l
Difference between method and function in Scala
Jim has got this pretty much covered in his blog post, but I'm posting a briefing here for reference. First, let's see what the Scala Specification tell us. Chapter 3 (types) tell us about Function Types (3.2.9) and Method Types (3.3.1). Chapter 4 (basic declarations) speaks of Value Declaration and Definitions (4.1), Variable Declaration and Definitions (4.2) and Functions Declarations and Defini
How to paramaterize Int as Ordered in scala
I have a class with a parameterized type that I want to do comparison operators on. I assmue I need to use the Ordered trait to achieve this but the compiler doesn't like my use of it. So say I have the following class: class Test[T <: Ordered[T]] { def someOp(t: T) if(t < 3) ... ... } However if I try to use this class as follows: val test = new Test[Int]() the compiler complains as follows: type
Why does Int not inherit/extend from Ordered[Int]
I have a question on type design. Why does Int not extend the Ordered trait. Isn't Int ordered by nature? Instead, the scala library provides implicit 'orderer' methods which convert Int to Ordered[Int]. What are the design choices being made here? Example taken from the book Programming in Scala def maxListImpParm[T <% Ordered[T]](elements:List[T]):T= ... maxListImpParm(List(1,5,10,3)) // works b
scala c integration
I would like to port some of my scala code to c and call this ported code from my current project. But I did not find any documentation about how to do that. It would be great if this would be possible just from sbt, because thats my current build system. Currently I have heared about SNA, but without documentation I am not looking for an automatic scala to c compiler or anything like that. I just
Play JSON Reads/Writes with single-parameter case classes
This creates a Writes for a case class import play.api.libs.json._ import play.api.libs.functional.syntax._ case class A(a: String, b: String, c: String) (JsPath.write[String] and JsPath.write[String] and JsPath.write[String])(unlift(A.unapply)) This can be extended to work for 2, 3, 4, 5, 6, etc. parameters...but not 1. case class B(a: String) (JsPath.write[String])(unlift(B.unapply)) Compiler er
What's the Scala way to implement a retry-able call like this one?
Still the newbie in Scala and I'm now looking for a way to implement the following code on it: @Override public void store(InputStream source, String destination, long size) { ObjectMetadata metadata = new ObjectMetadata(); metadata.setContentLength(size); final PutObjectRequest request = new PutObjectRequest( this.configuration.getBucket(), destination, source, metadata); new RetryableService(3)
Scala best way of turning a Collection into a Map-by-key?
If I have a collection c of type T and there is a property p on T (of type P, say), what is the best way to do a map-by-extracting-key? val c: Collection[T] val m: Map[P, T] One way is the following: m = new HashMap[P, T] c foreach { t => m add (t.getP, t) } But now I need a mutable map. Is there a better way of doing this so that it's in 1 line and I end up with an immutable Map? (Obviously I cou
Scala: List[Future] to Future[List] disregarding failed futures
I'm looking for a way to convert an arbitrary length list of Futures to a Future of List. I'm using Playframework, so ultimately, what I really want is a Future[Result], but to make things simpler, let's just say Future[List[Int]] The normal way to do this would be to use Future.sequence(...) but there's a twist... The list I'm given usually has around 10-20 futures in it, and it's not uncommon fo
FoldLeft using FoldRight in scala
While going through Functional Programming in Scala, I came across this question: Can you right foldLeft in terms of foldRight? How about the other way around? In solution provided by the authors they have provided an implementation as follows: def foldRightViaFoldLeft_1[A,B](l: List[A], z: B)(f: (A,B) => B): B = foldLeft(l, (b:B) => b)((g,a) => b => g(f(a,b)))(z) def foldLeftViaFoldRight[A,B](l:
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