Scala dot product is very slow compared to Java

Question

I am very new to Scala and I would like to translate my Java code with the same level of performance.

Given n float vectors and an extra vector, I have to calculate all n-point products and get the maximum.

Using Java is pretty simple for me

public static void main(String[] args) {

    int N = 5000000;
    int R = 200;
    float[][] t = new float[N][R];
    float[] u = new float[R];

    Random r = new Random();

    for (int i = 0;i<N;i++) {
        for (int j = 0;j<R;j++) {
            if (i == 0) {
                u[j] = r.nextFloat();
            }
            t[i][j] = r.nextFloat();
        }
    }

    long ts = System.currentTimeMillis();
    float maxScore = -1.0f;

    for (int i = 0;i < N;i++) {
        float score = 0.0f;
        for (int j = 0; i < R;i++) {
            score += u[j] * t[i][j];
        }
        if (score > maxScore) {
            maxScore = score;
        }

    }

    System.out.println(System.currentTimeMillis() - ts);
    System.out.println(maxScore);

}

The calculation time is 6 ms on my machine.

Now I have to do it with Scala

val t = Array.ofDim[Float](N,R)
val u = Array.ofDim[Float](R)

// Filling with random floats like in Java

val ts = System.currentTimeMillis()
var maxScore: Float = -1.0f

for ( i <- 0 until N) {
  var score = 0.0f
  for (j <- 0 until R) {
    score += u(j) * t(i)(j)
  }
  if (score > maxScore) {
    maxScore = score
  }

}

println(System.currentTimeMillis() - ts)
println(maxScore);

The above code takes more than a second on my machine. My thought is that Scala does not have a primitive array structure, such as float [] in Java, and is replaced with a collection. Index access seems to be slower than the one that has a primitive array in Java.

The following code is even slower:

val maxScore = t.map( r => r zip u map Function.tupled(_*_) reduceLeft (_+_)).max

which takes 26s

How can I efficiently iterate over my 2 arrays to calculate this?

Thank you so much

Answer 1

, , , Java, , Scala - 6 10 (!) , - - Java, :

for (int j = 0; j < R;j++), for (int j = 0; i < R;i++) - 200 10 ...

- Scala Java .

, BTW, Scala - for (j <- 0 until R) :)

Answer 2

( ), , :

var i = 0
while (i < N) {
  var j = 0
  var score = 0.0f
  val t1: Array[Float] = t(i)
  while (j < R) {
    score += u(j) * t1(j)
    j += 1
  }
  if (score > maxScore) {
    maxScore = score
  }

  i += 1
}

10-20% , .

! "par", :

val maxScore = t.par.map({
  arr =>
    var score = 0.0f
    var j = 0
    while (j < R) {
      score += u(j) * arr(j)
      j += 1
    }
    score
}).max

2-3 , java! !:) .