CODEWITHSUNDEEP
javaarraysscalamultidimensional-arrayapache-spark
Fredy Gomez
Fredy Gomez

Reputation: 15

Scala reading multidimensional array [Array[Array[Int]] from file

I need to replicate the result from this array definition using an external file.

scala> val data = Seq(Array(Array(1, 2), Array(3)),Array(Array(1), Array(3, 2), Array(1, 2)),Array(Array(1, 2), Array(5)),Array(Array(6)))

data: Seq[Array[Array[Int]]] = List(Array(Array(1, 2), Array(3)), Array(Array(1), Array(3, 2), Array(1, 2)), Array(Array(1, 2), Array(5)), Array(Array(6)))

I tried creating a testdataI.txt file but can't make it to work.

testdataI.txt -> 

1,2
3
1
3,2
1,2
1,2
5
6

Here the result when I do the conversion using io.Source:

import scala.io.Source

scala> val data = Seq(Source.fromFile("/tmp/testdataI.txt").getLines().map(_.split(",").map(_.trim.toInt)).toArray)

data: Seq[Array[Array[Int]]] = List(Array(Array(1, 2), Array(3), Array(1), Array(3, 2), Array(1, 2), Array(1, 2), Array(5), Array(6)))

The outcome should look like this (A series of Multidimensional Arrays)

data: Seq[Array[Array[Int]]] = List(Array(Array(1, 2), Array(3)), Array(Array(1), Array(3, 2), Array(1, 2)), Array(Array(1, 2), Array(5)), Array(Array(6)))

I found a lot of Multidimensional array information but nothing for this specific case.

Really appreciate,

Fredy A Gomez

Upvotes: 0

Views: 835

Answers (1)

mlg
mlg

Reputation: 1511

No idea why you want to structure the values like that, but here's how you can do it:

scala> import scala.io.Source
import scala.io.Source

scala> val take = List(2, 3, 2, 1)
take: List[Int] = List(2, 3, 2, 1)

scala> val data = Source.fromFile("/tmp/testdataI.txt").getLines().map(_.split(",").map(_.trim.toInt).toList).toList
data: List[List[Int]] = List(List(1, 2), List(3), List(1), List(3, 2), List(1, 2), List(1, 2), List(5), List(6))

scala> def awesomeGrouped(ungrouped: List[List[Int]], take: List[Int]): List[List[List[Int]]] = take match {
     |         case Nil => Nil
     |         case t :: ts => ungrouped.take(t) :: awesomeGrouped(ungrouped.drop(t), ts)
     |     }
awesomeGrouped: (ungrouped: List[List[Int]], take: List[Int])List[List[List[Int]]]

scala> def fixTypes(grouped: List[List[List[Int]]]) = grouped.map(_.map(_.toArray).toArray)
fixTypes: (sorted: List[List[List[Int]]])List[Array[Array[Int]]]

scala> fixTypes(awesomeGrouped(data, take))
res0: List[Array[Array[Int]]] = List(Array(Array(1, 2), Array(3)), Array(Array(1), Array(3, 2), Array(1, 2)), Array(Array(1, 2), Array(5)), Array(Array(6)))

The part that makes everyone uneasy is the take list distribution you've chosen; it seems arbitrary.

Note I added the fixTypes function specifically to return the exact return types you want. But arrays are not very idiomatic Scala; are you sure you need them? If not, just remove the fixTypes function and invocation.

Upvotes: 1

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