Scala: Get Items from Seq by given index (all the items above given index)

StackOverflow https://stackoverflow.com/questions/18133727

  •  24-06-2022
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Question

I am new to scala, the way I understood Seq is that its an ordered list. So I was wanted to see if I can get all the items based on a given index where retrieved items index is lesser than the given items index.

Lets say I have Seq:

scala> val s = Seq(1, 2, 34 ,44 )
s: Seq[Int] = List(1, 2, 34, 44)

Given index index as 3rd item I was expecting to get all the items(values) that has a lower index position than the given index.

Keep this in mind I wrote the following and Looks like I am wrong.

scala> val x = s.map {
     | id => id < s.indexOf(3) }
x: Seq[Boolean] = List(false, false, false, false)

What what I want is Seq(1,2,34) as the output because if of those element's index is less than the index of 44.

Whats the best way to do this?

Était-ce utile?

La solution

s.take(3) will take the first three elements of the sequence, i.e. all elements whose index is smaller than 3 (index counts from zero).

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