How to create a collection of unique values based on existing array values

Question

Below code prints an Array of fileNames.

  val pdfFileArray = getFiles()
  for(fileName <- pdfFileArray){
    println(fileName)
  }

I'm trying to convert this Array (pdfFileArray) into an array which contains unique file name extensions.

Is something like below the correct way of doing this in scala ?

  Set<String> fileNameSet = new HashSet<String>
  val pdfFileArray = getFiles()
  for(fileName <- pdfFileArray){
    String extension = fileName.substring(fileName.lastIndexOf('.'));
    fileNameSet.add(extension)
  }

Dan Simon · Accepted Answer · 2012-10-16 20:57:49Z

2

This will properly handle files with no extension (by ignoring them)

val extensions = getFiles().map{_.split('.').tail.lastOption}.flatten.distinct

so

Array("foo.jpg", "bar.jpg", "baz.png", "foobar")

becomes

Array("jpg", "png")

edited Oct 16, 2012 at 20:57

answered Oct 16, 2012 at 20:51

Dan Simon

13.2k3 gold badges52 silver badges55 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Ratan Sebastian Over a year ago

I think this wouldn't work. split returns an array of one element (the whole string) if it doesn't find the separator. So lastOption will always be a Some. That means you'd get Array("jpg", "png", "foobar")

Dan Simon Over a year ago

ah you're right, however .tail.lastOption does work, I'll edit my answer.

dhg Over a year ago

@Dan, better to use flatMap: files.flatMap(_.split('.').tail.lastOption).distinct

Ratan Sebastian · Accepted Answer · 2012-10-16 20:50:41Z

1

You could do this:

val fileNameSet = pdfFileArray.groupBy(_.split('.').last).keys

This assumes that all you filenames will have an extension and you only want the last extension. i.e. something.html.erb has the extension 'erb'

edited Oct 16, 2012 at 20:50

answered Oct 16, 2012 at 20:45

Ratan Sebastian

1,90214 silver badges23 bronze badges

1 Comment

blue-sky Over a year ago

thanks but I think your code needs to be amended slightly : pdfFileArray.groupBy(_.getName().split('.').last).keys

Jens Egholm · Accepted Answer · 2012-10-16 20:41:40Z

1

There's a method in scala's collection called distinct, which takes away all duplicate entries in the collection. So for instance:

scala> List(1, 2, 3, 1, 2).distinct
res3: List[Int] = List(1, 2, 3)

Is that what you're looking for?

answered Oct 16, 2012 at 20:41

Jens Egholm

2,7103 gold badges24 silver badges35 bronze badges

2 Comments

om-nom-nom Over a year ago

That will give you distinct by whole filenames, how about extensions?

Jens Egholm Over a year ago

Map the array by taking the extensions and then distinct: array.map(_.lastIndexOf('.')).distinct.

om-nom-nom · Accepted Answer · 2012-10-16 20:59:54Z

1

For a sake of completeness:

List("foo.jpg", "bar.jpg").map(_.takeRight(3)).toSet

Here I'm assuming that all extensions are 3 chars long. Conversion to Set, just like .distinct method (which uses mutable set underneath, by the way) in other answers gives you unique items.

answered Oct 16, 2012 at 20:59

om-nom-nom

63k13 gold badges186 silver badges231 bronze badges

2 Comments

blue-sky Over a year ago

why does it matter if Set "uses mutable set underneath" ?

om-nom-nom Over a year ago

@user470184, well, this part actually relates to .distinct, not Set, sorry, if I confused you

Luigi Plinge · Accepted Answer · 2012-10-17 02:17:29Z

1

You can also do it with regex, which gives a more general solution because you can redefine the expression to match anything you want:

val R = """.*\.(.+)""".r
getFiles.collect{ case R(x) => x }.distinct

answered Oct 17, 2012 at 2:17

Luigi Plinge

51.2k21 gold badges117 silver badges186 bronze badges

Collectives™ on Stack Overflow

How to create a collection of unique values based on existing array values

5 Answers 5

3 Comments

1 Comment

2 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

3 Comments

1 Comment

2 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related