You can use indexOf

times.indexOf(Some("Last month")) // returns 1

Update

You can convert the times ArrayBuffer[Option[String]] into a ArrayBuffer[String] via

val l = times.map(_.getOrElse("")) // ArrayBuffer("Last week", "Last month", ...)

And then you can search for the strings

l.indexOf("Last month")

First, use find instead of filter which will return the first one on an Option. Then use orNull to get the value or null if it is missing:

val res = selected.find(_.node == neighbour).orNull

Or you could use filter, headOption, and orNull

val res = selected.filter(_.node == neighbour).headOption.orNull

As a kind of mandatory note when dealing with null: Using Option instead of null is the scala idiomatic and recommended way.

Your ArrayBuffer is empty, so you cannot set an element at index i to a new String - index i is not a valid index if the ArrayBuffer is empty. First make sure the elements exist in the ArrayBuffer by adding them:

val strings = ArrayBufferString

strings += "abc1"
strings += "abc2"

Now the ArrayBuffer has two elements and you can modify them if you want. Note that indices start at 0, not at 1.

strings(0) = "something else"
strings(1) = "hello world"

If you want to pre-fill the ArrayBuffer you can for example use fill, as I showed you in my answer to your previous question.

// Fill with 10 empty strings (creates 10 elements in the ArrayBuffer)
val strings = ArrayBuffer.fill(10) { "" }

// Now you can set them (valid indices are 0...9)
strings(0) = "abc1"
strings(1) = "abc2"

Alternatively, use a Map instead of an ArrayBuffer, where the keys of your map are numbers and the values are strings.

import scala.collection.mutable.Map

val strings: Map[Int, String] = Map()

strings(1) = "abc1"
strings(2) = "abc2"