Related
I understand how to use if in map. For example, val result = list.map(x => if (x % 2 == 0) x * 2 else x / 2).
However, I want to use if for only part of the arguments.
val inputColumns = List(
List(1, 2, 3, 4, 5, 6), // first "column"
List(4, 6, 5, 7, 12, 15) // second "column"
)
inputColumns.zipWithIndex.map{ case (col, idx) => if (idx == 0) col * 2 else col / 10}
<console>:1: error: ';' expected but integer literal found.
inputColumns.zipWithIndex
res4: List[(List[Int], Int)] = List((List(1, 2, 3, 4, 5, 6),0), (List(4, 6, 5, 7, 12, 15),1))
I have searched the error info but have not found a solution.
Why my code is not 'legal' in Scala? Is there a better way to write it? Basically, I want to do a pattern matching and then do something on other arguments.
To explain your problem another way, inputColumns has type List[List[Int]]. You can verify this in the Scala REPL:
$ scala
Welcome to Scala 2.12.4 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_161).
Type in expressions for evaluation. Or try :help.
scala> val inputColumns = List(
| List(1, 2, 3, 4, 5, 6), // first "column"
| List(4, 6, 5, 7, 12, 15) // second "column"
| )
inputColumns: List[List[Int]] = List(List(1, 2, 3, 4, 5, 6), List(4, 6, 5, 7, 12, 15))
Now, when you call .zipWithIndex on that list, you end up with a List[(List[Int], Int)] - that is, a list of a tuple, in which the first tuple type is a List[Int] (the column) and the second is an Int (the index):
scala> inputColumns.zipWithIndex
res0: List[(List[Int], Int)] = List((List(1, 2, 3, 4, 5, 6),0), (List(4, 6, 5, 7, 12, 15),1))
Consequently, when you try to apply a map function to this list, col is a List[Int] and not an Int, and so col * 2 makes no sense - you're multiplying a List[Int] by 2. You then also try to divide the list by 10, obviously.
scala> inputColumns.zipWithIndex.map{ case(col, idx) => if(idx == 0) col * 2 else col / 10 }
<console>:13: error: value * is not a member of List[Int]
inputColumns.zipWithIndex.map{ case(col, idx) => if(idx == 0) col * 2 else col / 10 }
^
<console>:13: error: value / is not a member of List[Int]
inputColumns.zipWithIndex.map{ case(col, idx) => if(idx == 0) col * 2 else col / 10 }
^
In order to resolve this, it depends what you're trying to achieve. If you want a single list of integers, and then zip those so that each value has an associated index, you should call flatten on inputColumns before calling zipWithIndex. This will result in List[(Int, Int)], where the first value in the tuple is the column value, and the second is the index. Your map function will then work correctly without modification:
scala> inputColumns.flatten.zipWithIndex.map{ case(col, idx) => if(idx == 0) col * 2 else col / 10 }
res3: List[Int] = List(2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1)
Of course, you no longer have separate columns.
If you wish each value in each list to have an associated index, you need to firstly map inputColumns into two zipped lists, using inputColumns.map(_.zipWithIndex) to create a List[List[(Int, Int)]] - a list of a list of (Int, Int) tuples:
scala> inputColumns.map(_.zipWithIndex)
res4: List[List[(Int, Int)]] = List(List((1,0), (2,1), (3,2), (4,3), (5,4), (6,5)), List((4,0), (6,1), (5,2), (7,3), (12,4), (15,5)))
We can now apply your original map function to the result of the zipWithIndex operation:
scala> inputColumns.map(_.zipWithIndex.map { case (col, idx) => if(idx == 0) col * 2 else col / 10 })
res5: List[List[Int]] = List(List(2, 0, 0, 0, 0, 0), List(8, 0, 0, 0, 1, 1))
The result is another List[List[Int]] with each internal list being the results of your map operation on the original two input columns.
On the other hand, if idx is meant to be the index of the column, and not of each value, and you want to multiply all of the values in the first column by 2 and divide all of the values in the other columns by 10, then you need to change your original map function to map across each column, as follows:
scala> inputColumns.zipWithIndex.map {
| case (col, idx) => {
| if(idx == 0) col.map(_ * 2) // Multiply values in first column by 1
| else col.map(_ / 10) // Divide values in all other columns by 10
| }
| }
res5: List[List[Int]] = List(List(2, 4, 6, 8, 10, 12), List(0, 0, 0, 0, 1, 1))
Let me know if you require any further clarification...
UPDATE:
The use of case in map is a common Scala shorthand. If a higher-order function takes a single argument, something such as this:
def someHOF[A, B](x: A => B) = //...
and you call that function like this (with what Scala terms a partial function - a function consisting solely of a list of case statements):
someHOF {
case expr1 => //...
case expr2 => //...
...
}
then Scala treats it as a kind-of shorthand for:
someHOF {a =>
a match {
case expr1 => //...
case expr2 => //...
...
}
}
or, being slightly more terse,
someHOF {
_ match {
case expr1 => //...
case expr2 => //...
...
}
}
For a List, for example, you can use it with functions such as map, flatMap, filter, etc.
In the case of your map function, the sole argument is a tuple, and the sole case statement acts to break open the tuple and expose its contents. That is:
val l = List((1, 2), (3, 4), (5, 6))
l.map { case(a, b) => println(s"First is $a, second is $b") }
is equivalent to:
l.map {x =>
x match {
case (a, b) => println(s"First is $a, second is $b")
}
}
and both will output:
First is 1, second is 2
First is 3, second is 4
First is 5, second is 6
Note: This latter is a bit of a dumb example, since map is supposed to map (i.e. change) the values in the list into new values in a new list. If all you were doing was printing the values, this would be better:
val l = List((1, 2), (3, 4), (5, 6))
l.foreach { case(a, b) => println(s"First is $a, second is $b") }
You are trying to multiply a list by 2 when you do col * 2 as col is List(1, 2, 3, 4, 5, 6) when idx is 0, which is not possible and similar is the case with else part col / 10
If you are trying to multiply the elements of first list by 2 and devide the elements of rest of the list by 10 then you should be doing the following
inputColumns.zipWithIndex.map{ case (col, idx) => if (idx == 0) col.map(_*2) else col.map(_/10)}
Even better approach would be to use match case
inputColumns.zipWithIndex.map(x => x._2 match {
case 0 => x._1.map(_*2)
case _ => x._1.map(_/10)
})
Imagine that I have the following List l. Is it possible by using map to return a list similar to the result below but if let's say the number is 2 to return the result twice? e.g.
l.map( x => if (x=2) (return twice) x*2 )
so the resulted list should be
List(2, 4, 4, 6, 8, 10)
instead of the one presented below.
scala> val l = List(1,2,3,4,5)
scala> l.map( x => x*2 )
res60: List[Int] = List(2, 4, 6, 8, 10)
You are looking for .flatMap
l.flatMap {
case 2 => Seq(4,4)
case x => Seq(x*2)
}
I can do this quite easily, and cleanly, using a for loop. For instance, if I wanted to traverse a Seq from every element back to itself I would do the following:
val seq = Seq(1,2,3,4,5)
for (i <- seq.indices) {
for (j <- seq.indices) {
print(seq(i + j % seq.length))
}
}
But as I'm looking to fold over the collection, I'm wondering if there is a more idiomatic approach. A recursive approach would allow me to avoid any vars. But basically, I'm wondering if something like the following is possible:
seq.rotatedView(i)
Which would create a rotated view, like rotating bits (or circular shift).
Is it like below:
scala> def rotatedView(i:Int)=Seq(1,2,3,4,5).drop(i)++Seq(1,2,3,4,5).take(i)
rotatedView: (i: Int)Seq[Int]
scala> rotatedView(1)
res48: Seq[Int] = List(2, 3, 4, 5, 1)
scala> rotatedView(2)
res49: Seq[Int] = List(3, 4, 5, 1, 2)
This ought to do it in a fairly generic way, and allow for arbitrary rotations:
def rotateLeft[A](seq: Seq[A], i: Int): Seq[A] = {
val size = seq.size
seq.drop(i % size) ++ seq.take(i % size)
}
def rotateRight[A](seq: Seq[A], i: Int): Seq[A] = {
val size = seq.size
seq.drop(size - (i % size)) ++ seq.take(size - (i % size))
}
The idea is simple enough, to rotate left, drop the first i elements from the left, and take them again from the left to concatenate them in the opposite order. If you don't mind calculating the size of the collection, you can do your operations modulo the size, to allow i to be arbitrary.
scala> rotateRight(seq, 1)
res34: Seq[Int] = List(5, 1, 2, 3, 4)
scala> rotateRight(seq, 7)
res35: Seq[Int] = List(4, 5, 1, 2, 3)
scala> rotateRight(seq, 70)
res36: Seq[Int] = List(1, 2, 3, 4, 5)
Similarly, you can use splitAt:
def rotateLeft[A](seq: Seq[A], i: Int): Seq[A] = {
val size = seq.size
val (first, last) = seq.splitAt(i % size)
last ++ first
}
def rotateRight[A](seq: Seq[A], i: Int): Seq[A] = {
val size = seq.size
val (first, last) = seq.splitAt(size - (i % size))
last ++ first
}
To make it even more generic, using the enrich my library pattern:
import scala.collection.TraversableLike
import scala.collection.generic.CanBuildFrom
implicit class TraversableExt[A, Repr <: TraversableLike[A, Repr]](xs: TraversableLike[A, Repr]) {
def rotateLeft(i: Int)(implicit cbf: CanBuildFrom[Repr, A, Repr]): Repr = {
val size = xs.size
val (first, last) = xs.splitAt(i % size)
last ++ first
}
def rotateRight(i: Int)(implicit cbf: CanBuildFrom[Repr, A, Repr]): Repr = {
val size = xs.size
val (first, last) = xs.splitAt(size - (i % size))
last ++ first
}
}
scala> Seq(1, 2, 3, 4, 5).rotateRight(2)
res0: Seq[Int] = List(4, 5, 1, 2, 3)
scala> List(1, 2, 3, 4, 5).rotateLeft(2)
res1: List[Int] = List(3, 4, 5, 1, 2)
scala> Stream(1, 2, 3, 4, 5).rotateRight(1)
res2: scala.collection.immutable.Stream[Int] = Stream(5, ?)
Keep in mind these are not all necessarily the most tuned for performance, and they also can't work with infinite collections (none can).
Following the OP's comment that they want to fold over it, here's a slightly different take on it that avoids calculating the length of the sequence first.
Define an iterator that will iterate over the rotated sequence
class RotatedIterator[A](seq: Seq[A], start: Int) extends Iterator[A] {
var (before, after) = seq.splitAt(start)
def next = after match {
case Seq() =>
val (h :: t) = before; before = t; h
case h :: t => after = t; h
}
def hasNext = after.nonEmpty || before.nonEmpty
}
And use it like this:
val seq = List(1, 2, 3, 4, 5)
val xs = new RotatedIterator(seq, 2)
println(xs.toList) //> List(3, 4, 5, 1, 2)
A simple method is to concatenate the sequence with itself and then take the slice that is required:
(seq ++ seq).slice(start, start + seq.length)
This is just a variant of the drop/take version but perhaps a little clearer.
Given:
val seq = Seq(1,2,3,4,5)
Solution:
seq.zipWithIndex.groupBy(_._2<3).values.flatMap(_.map(_._1))
or
seq.zipWithIndex.groupBy(_._2<3).values.flatten.map(_._1)
Result:
List(4, 5, 1, 2, 3)
If rotation is more than length of collection - we need to use rotation%length, if negative than formula (rotation+1)%length and take absolute value.
It's not efficient
Another tail-recursive approach. When I benchmarked it with JMH it was about 2 times faster than solution based on drop/take:
def rotate[A](list: List[A], by: Int): List[A] = {
#tailrec
def go(list: List[A], n: Int, acc: List[A]): List[A] = {
if(n > 0) {
list match {
case x :: xs => go(xs, n-1, x :: acc)
}
} else {
list ++ acc.reverse
}
}
if (by < 0) {
go(list, -by % list.length, Nil)
} else {
go(list, list.length - by % list.length, Nil)
}
}
//rotate right
rotate(List(1,2,3,4,5,6,7,8,9,10), 3) // List(8, 9, 10, 1, 2, 3, 4, 5, 6, 7)
//use negative number to rotate left
rotate(List(1,2,3,4,5,6,7,8,9,10), -3) // List(4, 5, 6, 7, 8, 9, 10, 1, 2, 3)
Here is one liner solution
def rotateRight(A: Array[Int], K: Int): Array[Int] = {
if (null == A || A.size == 0) A else (A drop A.size - (K % A.size)) ++ (A take A.size - (K % A.size))
}
rotateRight(Array(1,2,3,4,5), 3)
Here's a fairly simple and idiomatic Scala collections way to write it:
def rotateSeq[A](seq: Seq[A], isLeft: Boolean = false, count: Int = 1): Seq[A] =
if (isLeft)
seq.drop(count) ++ seq.take(count)
else
seq.takeRight(count) ++ seq.dropRight(count)
We can simply use foldLeft to reverse a list as below.
val input = List(1,2,3,4,5)
val res = input.foldLeft(List[Int]())((s, a) => { List(a) ++: s})
println(res) // List(5, 4, 3, 2, 1)
Another one line solution if you don't need to validate the "offset":
def rotate[T](seq: Seq[T], offset: Int): Seq[T] = Seq(seq, seq).flatten.slice(offset, offset + seq.size)
This is a simple piece of code
object tesing_it extends App
{
val one = ArrayBuffer(1,2,3,4,5,6)
val i = 2 //the number of index you want to move
for(z<-0 to i){
val y = 0
var x = one += one(y)
x = x -= x(y)
println("for seq after process " +z +" " + x)
}
println(one)
}
Result:
for seq after process 0 ArrayBuffer(2, 3, 4, 5, 6, 1)
for seq after process 1 ArrayBuffer(3, 4, 5, 6, 1, 2)
for seq after process 2 ArrayBuffer(4, 5, 6, 1, 2, 3)
ArrayBuffer(4, 5, 6, 1, 2, 3)
Is there a "for" syntax for
c flatmap ( x => d flatmap (y => f(x,y) ) )
?
Because I've used Haskell in the past, I keep expecting the "for" syntax in Scala to mimic the "do" syntax of Haskell. This is probably an unrealistic expectation. In Haskell which I could write
do x <- c
y <- d
f(x, y)
You could also map the last result on itself.
Using the same example as dhg:
val c = 1 to 3
val d = 4 to 6
def f(x: Int, y: Int) = Vector(x,y)
for {
x <- c
y <- d
z <- f(x, y)
} yield z
// Vector(1, 4, 1, 5, 1, 6, 2, 4, 2, 5, 2, 6, 3, 4, 3, 5, 3, 6)
Which corresponds to:
c flatMap ( x => d flatMap (y => f(x,y) map (identity) ) )
You can just flatten the result:
val c = 1 to 3
val d = 4 to 6
def f(x: Int, y: Int) = Vector(x,y)
c flatMap ( x => d flatMap (y => f(x,y) ) )
// Vector(1, 4, 1, 5, 1, 6, 2, 4, 2, 5, 2, 6, 3, 4, 3, 5, 3, 6)
(for { x <- c; y <- d } yield f(x,y)).flatten
// Vector(1, 4, 1, 5, 1, 6, 2, 4, 2, 5, 2, 6, 3, 4, 3, 5, 3, 6)
Presumably this is a much less frequently used case since it is necessarily less common that the output of the for is flattenable. And sticking .flatten on the end is pretty easy, so having a special syntax for it seems unnecessarily complicated.
Flattening may impact performance but i think scalac is clever enough to encode
for {
x <- c
y <- d
z <- f(x,y)
} yield z
into
c flatMap { x => d flatMap { y => f(x,y) } }
This is annoying that the 'for' syntax is not as convenient as the 'do'-notation (writing _ <- someExpression instead of just someExpression in a for feels my heart with sadness).
It's easy to initialize a 2D array (or, in fact, any multidimensional array) in Java by putting something like that:
int[][] x = new int[][] {
{ 3, 5, 7, },
{ 0, 4, 9, },
{ 1, 8, 6, },
};
It's easy to read, it resembles a 2D matrix, etc, etc.
But how do I do that in Scala?
The best I could come up with looks, well, much less concise:
val x = Array(
Array(3, 5, 7),
Array(0, 4, 9),
Array(1, 8, 6)
)
The problems I see here:
It repeats "Array" over and over again (like there could be anything else besides Array)
It requires to omit trailing , in every Array invocation
If I screw up and insert something besides Array() in the middle of array, it will go okay with compiler, but type of x would silently become Array[Any] instead of Array[Array[Int]]:
val x = Array(
Array(3, 5, 7),
Array(0, 4), 9, // <= OK with compiler, silently ruins x
Array(1, 8, 6)
)
There is a guard against it, to specify the type directly, but it looks even more overkill than in Java:
val x: Array[Array[Int]] = Array(
Array(3, 5, 7),
Array(0, 4), 9, // <= this one would trigger a compiler error
Array(1, 8, 6)
)
This last example needs Array even 3 times more than I have to say int[][] in Java.
Is there any clear way around this?
Personally I'd suck it up and type out (or cut and paste) "Array" a few times for clarity's sake. Include the type annotation for safety, of course. But if you're really running out of e-ink, a quick easy hack would be simply to provide an alias for Array, for example:
val > = Array
val x: Array[Array[Int]] = >(
>(3, 5, 7),
>(0, 4, 9),
>(1, 8, 6)
)
You could also provide a type alias for Array if you want to shorten the annotation:
type >[T] = Array[T]
val x: >[>[Int]] = ...
I suggest to use Scala 2.10 and macros:
object MatrixMacro {
import language.experimental.macros
import scala.reflect.macros.Context
import scala.util.Try
implicit class MatrixContext(sc: StringContext) {
def matrix(): Array[Array[Int]] = macro matrixImpl
}
def matrixImpl(c: Context)(): c.Expr[Array[Array[Int]]] = {
import c.universe.{ Try => _, _ }
val matrix = Try {
c.prefix.tree match {
case Apply(_, List(Apply(_, List(Literal(Constant(raw: String)))))) =>
def toArrayAST(c: List[TermTree]) =
Apply(Select(Select(Ident("scala"), newTermName("Array")), newTermName("apply")), c)
val matrix = raw split "\n" map (_.trim) filter (_.nonEmpty) map {
_ split "," map (_.trim.toInt)
}
if (matrix.map(_.length).distinct.size != 1)
c.abort(c.enclosingPosition, "rows of matrix do not have the same length")
val matrixAST = matrix map (_ map (i => Literal(Constant(i)))) map (i => toArrayAST(i.toList))
toArrayAST(matrixAST.toList)
}
}
c.Expr(matrix getOrElse c.abort(c.enclosingPosition, "not a matrix of Int"))
}
}
Usage with:
scala> import MatrixMacro._
import MatrixMacro._
scala> matrix"1"
res86: Array[Array[Int]] = Array(Array(1))
scala> matrix"1,2,3"
res87: Array[Array[Int]] = Array(Array(1, 2, 3))
scala> matrix"""
| 1, 2, 3
| 4, 5, 6
| 7, 8, 9
| """
res88: Array[Array[Int]] = Array(Array(1, 2, 3), Array(4, 5, 6), Array(7, 8, 9))
scala> matrix"""
| 1, 2
| 1
| """
<console>:57: error: rows of matrix do not have the same length
matrix"""
^
scala> matrix"a"
<console>:57: error: not a matrix of Int
matrix"a"
^
I don't think you will get it shorter. ;)
If using a mere List of List (which in itself cannot guarantee that every sub list is of the same size) is not a problem for you, and you are only concerned with easy syntax and avoiding errors at creation-time, scala has many ways to create nice syntax constructs.
One such possibility would be a simple helper:
object Matrix {
def apply[X]( elements: Tuple3[X, X, X]* ): List[List[X]] = {
elements.toList.map(_.productIterator.toList.asInstanceOf[List[X]] )
}
// Here you might add other overloads for Tuple4, Tuple5 etc if you need "matrixes" of those sizes
}
val x = Matrix(
(3, 5, 7),
(0, 4, 9),
(1, 8, 6)
)
About your concerns:
It repeats "List" over and over again (like there could be anything else besides List)
Not the case here.
It requires to omit trailing , in every List invocation
Unfortunately that is still true here, not much you can do given scala's syntactic rules.
If I screw up and insert something besides List() in the middle of array, it will go okay with compiler, but type of x would silently become List[Any] instead of List[List[Int]]:
val x = List(
List(3, 5, 7),
List(0, 4), 9, // <= OK with compiler, silently ruins x
List(1, 8, 6)
)
The equivalent code now faile to compile:
scala> val x = Matrix(
| (3, 5, 7),
| (0, 4), 9,
| (1, 8, 6)
| )
<console>:10: error: type mismatch;
found : (Int, Int)
required: (?, ?, ?)
(0, 4), 9,
And finally if you want to explicitly specify the type of elements (say that you want to protect against the possibility of inadvertently mixing Ints and Doubles), you only have to specify Matrix[Int] instead of the ugly List[List[Int]]:
val x = Matrix[Int](
(3, 5, 7),
(0, 4, 9),
(1, 8, 6)
)
EDIT: I see that you replaced List with Array in your question. To use arrays all you have to use is to replace List with Array and toList with toArray in my code above.
Since I'm also in disgust with this trailing comma issue (i.e. I cannot simply exchange the last line with any other) I sometimes use either a fluent API or the constructor syntax trick to get the syntax I like. An example using the constructor syntax would be:
trait Matrix {
// ... and the beast
private val buffer = ArrayBuffer[Array[Int]]()
def >(vals: Int*) = buffer += vals.toArray
def build: Array[Array[Int]] = buffer.toArray
}
Which allows:
// beauty ...
val m = new Matrix {
>(1, 2, 3)
>(4, 5, 6)
>(7, 8, 9)
} build
Unfortunately, this relies on mutable data although it is only used temporarily during the construction. In cases where I want maximal beauty for the construction syntax I would prefer this solution.
In case build is too long/verbose you might want to replace it by an empty apply function.
I don't know if this is the easy way, but I've included some code below for converting nested tuples into '2D' arrays.
First, you need some boiler plate for getting the size of the tuples as well as converting the tuples into [Array[Array[Double]]. The series of steps I used were:
Figure out the number of rows and columns in the tuple
Turn the nested tuple into a one row Array
Reshape the array based on the size of the original tuple.
The code for that is:
object Matrix {
/**
* Returns the size of a series of nested tuples.
*/
def productSize(t: Product): (Int, Int) = {
val a = t.productArity
val one = t.productElement(0)
if (one.isInstanceOf[Product]) {
val b = one.asInstanceOf[Product].productArity
(a, b)
}
else {
(1, a)
}
}
/**
* Flattens out a nested tuple and returns the contents as an iterator.
*/
def flattenProduct(t: Product): Iterator[Any] = t.productIterator.flatMap {
case p: Product => flattenProduct(p)
case x => Iterator(x)
}
/**
* Convert a nested tuple to a flattened row-oriented array.
* Usage is:
* {{{
* val t = ((1, 2, 3), (4, 5, 6))
* val a = Matrix.toArray(t)
* // a: Array[Double] = Array(1, 2, 3, 4, 5, 6)
* }}}
*
* #param t The tuple to convert to an array
*/
def toArray(t: Product): Array[Double] = flattenProduct(t).map(v =>
v match {
case c: Char => c.toDouble
case b: Byte => b.toDouble
case sh: Short => sh.toDouble
case i: Int => i.toDouble
case l: Long => l.toDouble
case f: Float => f.toDouble
case d: Double => d
case s: String => s.toDouble
case _ => Double.NaN
}
).toArray[Double]
def rowArrayTo2DArray[#specialized(Int, Long, Float, Double) A: Numeric](m: Int, n: Int,
rowArray: Array[A]) = {
require(rowArray.size == m * n)
val numeric = implicitly[Numeric[A]]
val newArray = Array.ofDim[Double](m, n)
for (i <- 0 until m; j <- 0 until n) {
val idx = i * n + j
newArray(i)(j) = numeric.toDouble(rowArray(idx))
}
newArray
}
/**
* Factory method for turning tuples into 2D arrays
*/
def apply(data: Product): Array[Array[Double]] = {
def size = productSize(data)
def array = toArray(data)
rowArrayTo2DArray(size._1, size._2, array)
}
}
Now to use this, you could just do the following:
val a = Matrix((1, 2, 3))
// a: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0))
val b = Matrix(((1, 2, 3), (4, 5, 6), (7, 8, 9)))
// b: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0),
// Array(4.0, 5.0, 6.0),
// Array(7.0, 8.0, 9.0))
val c = Matrix((1L, 2F, "3")) // Correctly handles mixed types
// c: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0))
val d = Matrix((1L, 2F, new java.util.Date())) // Non-numeric types convert to NaN
// d: Array[Array[Double]] = Array(Array(1.0, 2.0, NaN))
Alternatively, if you could just call the rowArrayTo2DArray directly using the size of the array you want and a 1D array of values:
val e = Matrix.rowArrayTo2DArray(1, 3, Array(1, 2, 3))
// e: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0))
val f = Matrix.rowArrayTo2DArray(3, 1, Array(1, 2, 3))
// f: Array[Array[Double]] = Array(Array(1.0), Array(2.0), Array(3.0))
val g = Matrix.rowArrayTo2DArray(3, 3, Array(1, 2, 3, 4, 5, 6, 7, 8, 9))
// g: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0),
// Array(4.0, 5.0, 6.0),
// Array(7.0, 8.0, 9.0))
glancing through the answers, i did not find what to me seems the most obvious & simple way to do it. instead of Array, you can use a tuple.
would look something like that:
scala> val x = {(
| (3,5,7),
| (0,4,9),
| (1,8,6)
| )}
x: ((Int, Int, Int), (Int, Int, Int), (Int, Int, Int)) = ((3,5,7),(0,4,9),(1,8,6))
seems clean & elegant?
i think so :)