Get indexes from one array and update elements on same indexes in second array

Question

I have two arrays, upper and lower bound.

val solution : Array[Double] = Array(1.0, 33.0, 31.0, 82.0, 51.0, 25.0, 39.0, 96.0, 95.0, 49.0)
val original : Array[Double] = Array(3.0, 45.0, 18.0, 25.0, 99.0, 6.0, 36.0, 61.0, 82.0, 62.0)
val lower = 30
val upper = 90

Array of type Boolean is created using solution array.

val idx_lowest : Array[Boolean]  = solution.map ( x => x < lower )
idx_lowest: Array[Boolean] = Array(true, false, false, false, false, true, false, false, false, false)

Find indexes in idx_lowest where value is true

val trueIndexes = idx_lowest .zipWithIndex filter(x => x._1  ) map(_._2)
trueIndexes: Array[Int] = Array(0, 5)

Find values against trueIndexes in original array.

val tmp = trueIndexes map original
Array[Double] = Array(3.0, 6.0)

Perform operation on elements of tmp array.

val tmp1 = tmp  map (x => (x+lower)/2.0)
Array[Double] = Array(16.5, 18.0)

Update elements of solution array. Indexes are specified by trueIndexes. Elements those are less than lower bound are updated.

for(i <- tmp1.indices)  {
    solution(trueIndexes(i)) = tmp1(i)
  }

The updated solution array is:

Array[Double] = Array(16.5, 33.0, 31.0, 82.0, 51.0, 18.0, 39.0, 96.0, 95.0, 49.0)

Same steps have to be performed to update elements that are greater than upper bound. Here is code.

val idx_false : Array[Boolean]  = solution.map ( x => x > upper )
val falseIndexes = idx_false .zipWithIndex filter(x => x._1  ) map(_._2)
val tmp2 = falseIndexes map original 
val tmp3 = tmp2 map (x => (x+upper)/2.0)
for(i <- tmp3.indices)  {
   solution(falseIndexes(i)) = tmp3(i)
}
solution

This code doing exactly what I needed but have to perform lot of operations. In iterative algorithm I have to perform these operation in each iteration for each array. Is there more efficient,optimal and faster way to perform same operation?

Andronicus · Accepted Answer · 2020-08-07 07:52:41Z

4

You can zip them and map in a single operation:

solution.zip(original)
  .map { case(s, o) => if(s < lower) ((o + lower) / 2, o) else (s, o) }
  .map { case(s, o) => if(s > upper) (o + upper) / 2 else s }

answered Aug 7, 2020 at 7:52

Andronicus

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jwvh · Accepted Answer · 2020-08-07 07:53:04Z

3

Isn't this the same result?

val done = solution.indices.map{x =>
  if (solution(x) < lower)      (lower + original(x))/2.0
  else if (solution(x) > upper) (upper + original(x))/2.0
  else solution(x)
}.toArray

answered Aug 7, 2020 at 7:53

jwvh

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1 Comment

Andronicus Over a year ago

I might be nitpicky, but it's not equivalent. From the business perspective this might be wrong, but if you set lower > upper, the the result is different. But if it's not the case, then the solution is quite elegant

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