Refactoring a small Scala function - scala

Question

I have this function to compute the distance between two n-dimensional points using Pythagoras' theorem.
def computeDistance(neighbour: Point) = math.sqrt(coordinates.zip(neighbour.coordinates).map {
case (c1: Int, c2: Int) => math.pow(c1 - c2, 2)
}.sum)
The Point class (simplified) looks like:
class Point(val coordinates: List[Int])
I'm struggling to refactor the method so it's a little easier to read, can anybody help please?

Answer 1

Here's another way that makes the following three assumptions:
The length of the list is the number of dimensions for the point
Each List is correctly ordered, i.e. List(x, y) or List(x, y, z). We do not know how to handle List(x, z, y)
All lists are of equal length
def computeDistance(other: Point): Double = sqrt(
coordinates.zip(other.coordinates)
.flatMap(i => List(pow(i._2 - i._1, 2)))
.fold(0.0)(_ + _)
)
The obvious disadvantage here is that we don't have any safety around list length. The quick fix for this is to simply have the function return an Option[Double] like so:
def computeDistance(other: Point): Option[Double] = {
if(other.coordinates.length != coordinates.length) {
return None
}
return Some(sqrt(coordinates.zip(other.coordinates)
.flatMap(i => List(pow(i._2 - i._1, 2)))
.fold(0.0)(_ + _)
))
I'd be curious if there is a type safe way to ensure equal list length.
EDIT
It was politely pointed out to me that flatMap(x => List(foo(x))) is equivalent to map(foo) , which I forgot to refactor when I was originally playing w/ this. Slightly cleaner version w/ Map instead of flatMap :
def computeDistance(other: Point): Double = sqrt(
coordinates.zip(other.coordinates)
.map(i => pow(i._2 - i._1, 2))
.fold(0.0)(_ + _)
)

Answer 2

Most of your problem is that you're trying to do math with really long variable names. It's almost always painful. There's a reason why mathematicians use single letters. And assign temporary variables.
Try this:
class Point(val coordinates: List[Int]) { def c = coordinates }
import math._
def d(p: Point) = {
val delta = for ((a,b) <- (c zip p.c)) yield pow(a-b, dims)
sqrt(delta.sum)
}

Answer 3

Consider type aliases and case classes, like this,
type Coord = List[Int]
case class Point(val c: Coord) {
def distTo(p: Point) = {
val z = (c zip p.c).par
val pw = z.aggregate(0.0) ( (a,v) => a + math.pow( v._1-v._2, 2 ), _ + _ )
math.sqrt(pw)
}
}
so that for any two points, for instance,
val p = Point( (1 to 5).toList )
val q = Point( (2 to 6).toList )
we have that
p distTo q
res: Double = 2.23606797749979
Note method distTo uses aggregate on a parallelised collection of tuples, and combines the partial results by the last argument (summation). For high dimensional points this may prove more efficient than the sequential counterpart.
For simplicity of use, consider also implicit classes, as suggested in a comment above,
implicit class RichPoint(val c: Coord) extends AnyVal {
def distTo(d: Coord) = Point(c) distTo Point(d)
}
Hence
List(1,2,3,4,5) distTo List(2,3,4,5,6)
res: Double = 2.23606797749979