Replace sequence of tuple value from another

Question

val x = seq((1, 'a'), (2, 'b'))  
val y = seq((1, 'AA'), (3, 'CC'), (4, 'DD'))

Expected result

val z = seq((1, 'AA'), (2, 'b'))    

I can do it as below

x.toMap.map(k => (k._1, y.toMap.getOrElse(k._1, k._2))

But this will not preserve the order

Is there a way to do an ordered replacement

Answer 1

First, your code would work the same without the first toMap, which is not doing anything useful and is the one that is destroying the order.
Second, you are converting y to a Map for each element of x, making this really expensive and even worse that just using find or collectFirst.
Third, we could use some pattern matching to make the code a little bit more readable.

// If you have a better name, feel free to change it.
def lookupOrPreserve[K, V](
  input: Seq[(K, V)],
  lookup: Map[K, V]
): Seq[(K, V)] =
  input.map { case (key, originalValue) =>
    key -> lookup.getOrElse(key, default = originalValue)
  }

val x = seq((1 -> 'a'), (2 -> 'b'))  
val y = seq((1 -> 'AA'), (3 -> 'CC'), (4 -> 'DD'))
val z = lookupOrPreserve(input = x, lookup = y.toMap)