Scala: remove first column (first element in each row)

Question

Given an var x: Array[Seq[Any]], what would be the most efficient (fast, in-place if possible?) way to remove the first element from each row?

I've tried the following but it didn't work - probably because of immutability...

  for (row <- x.indices)
    x(row).drop(1)

Answer 1

First off, Arrays are mutable, and I'm guessing you're intending to change the elements of x, rather than replacing the entire array with a new array object, so it makes more sense to use val instead of var:

val x: Array[Seq[Any]]

Since you said your Seq objects are immutable, then you need to make sure you are setting the values in your array. This will work:

for (row <- x.indices)
  x(row) = x(row).drop(1)

This can be written in nicer ways. For example, you can use transform to map all the values of your array with a function:

x.transform(_.drop(1))

This updates in-place, unlike map, which will leave the old array unmodified and return a new array.

EDIT: I started to speculate on which method would be faster or more efficient, but the more I think about it, the more I realize I'm not sure. Both should have acceptable performance for most use cases.

Answer 2

this would work

scala> val x = Array(List(1,2,3),List(1,2,3))
x: Array[List[Int]] = Array(List(1, 2, 3), List(1, 2, 3))

scala> x map(_.drop(1))
res0: Array[List[Int]] = Array(List(2, 3), List(2, 3))