Iterate over scala list

Question

I have a list of strings in scala template of play framework.
I want to iterate over half of list at one time and then the other half of list on the second time.
I am not sure how to write efficient iterator for this.

I have tried
@for(i <- 0 until list.length/2 ) {list(i) }
and then for second loop

@for(i <- list.length/2+1 until list.length ) 
 { list(i) } 

This works but complexity becomes high.

Then later I did

  @defining(list.size) { size => 
    @for(i <- 0 until size/2) 
    {list(i) }
  }

Now it seems to work fine.

Solution 1

Here's one way.

scala> List("a","b","c","d","e")
res0: List[String] = List(a, b, c, d, e)

scala> res0.splitAt(res0.size/2)
res1: (List[String], List[String]) = (List(a, b),List(c, d, e))

scala> res1._1.foreach(println(_))
a
b

scala> res1._2.foreach(println(_))
c
d
e

Solution 2

Use sliding to create iterators,

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

scala> val step = (input.length + 1 ) / 2
step: Int = 2

scala> val sliding = input.sliding(step, step)
sliding: Iterator[List[Int]] = non-empty iterator

scala> val left = sliding.next()
left: List[Int] = List(1, 2)

scala> val right = sliding.next()
right: List[Int] = List(3)

scala> left.foreach(println)
1
2

scala> right.foreach(println)
3

or use take & drop,

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

scala> val step = (input.length + 1 ) / 2
step: Int = 2

scala> input.take(step).foreach(println)
1
2

scala> input.drop(step).foreach(println)
3

Comments

[Jeremy D]

Please, edit your question above, even if you had what seems to be a good answer.

[user2694377]

I just tried @for(i <- 0 until list.length/2 ) {list(i) } and then for second loop @for(i <- list.length/2+1 until list.length ) { list(i) } . This seems to be working fine for now.

[user2694377]

Yeah. This would work but I dont want the overhead of creating another list.I just found the syntax in for loop to iterate over only the half of list . @for(i <- 0 until list.length/2 ) {} and then from list.length/2+1 until list.length.

[senia]

@user2694377: the cost of creating 2 new collections is O(N), iterating through List using index is O(N^2). It's really not a good idea. You could use an Iterator with drop and take if you don't want to copy collection.

[user2694377]

Okay. Thanks. Sorry I dint notice that it was O(n^2). Thanks for pointing it out.