Scala: Get specific elements by a sequence of indices

Question

How can I get specific elements of an array by having the sequence of indices, I do it now by following:

val indices = Array(1,3,3,2)
val a = Array("a","b","c","d","e","f")
indices.map(a(_))

the problem is if I want to make it a safe search I have to check a.contains or modify the indices. sometimes the indices is to big and I can not modify it fast. What is the best time/space solution?

I think you can just do `indices.map(a)` – samthebest Sep 07 '14 at 03:32 — samthebest, Sep 07 '14 at 03:32

score 6 · Answer 1 · answered Sep 07 '14 at 05:18

6

Here it is:

scala> val indices = Array(1, 2, 3, 2, 100, -100, 4, 5)
indices: Array[Int] = Array(1, 2, 3, 2, 100, -100, 4, 5)

scala> val arr = Array("a", "b", "c", "d")
arr: Array[String] = Array(a, b, c, d)

scala> indices collect arr
res1: Array[String] = Array(b, c, d, c)

answered Sep 07 '14 at 05:18

Eastsun

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Do you know how is the complexity of collect? – Omid Sep 07 '14 at 17:41
@Omid `collect` traverses the collection exactly once. – Kigyo Sep 07 '14 at 23:24
By the way, what would `indices collect arr` look like desugared? – Kigyo Sep 07 '14 at 23:26

Dennis Traub · Answer 2 · 2014-09-06T22:36:14.013

0

Assuming you just want to ignore every out-of-bounds-index you could just use a filter before mapping:

indices.filter(_ < a.length).map(a(_))

edited Sep 06 '14 at 22:36

answered Sep 06 '14 at 22:31

Dennis Traub

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the first filter needs go over all data, it takes 2*size of indices time and size – Omid Sep 06 '14 at 23:04
Change `filter` to `withFilter` and you remove the `2n` iteration time at the least (not sure about the space off the top of my head though). – Sean Vieira Sep 06 '14 at 23:30
@SeanVieira What is the difference between them ? – Omid Sep 07 '14 at 01:41
See http://stackoverflow.com/a/19618241/135978 - basically, withFilter is most likely lazy and doesn't create another collection - instead it's a simplified `view`. – Sean Vieira Sep 07 '14 at 02:04

score 0 · Answer 3 · edited Nov 21 '18 at 16:01

0

Using flatMap where out of bound indexes are mapped onto None, like this,

indices.flatMap(i => if (i < a.length) Some(a(i)) else None)

Simpler forms with flatMap,

indices.flatMap(i => Some(a(i)))
res: Array(b, d, d, c)

indices.flatMap(a(_))
res: Array(b, d, d, c)

edited Nov 21 '18 at 16:01

Paul Reiners

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answered Sep 07 '14 at 05:05

elm

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Why? This is like collect only less clear. If you really don't want to use collect, then foldLeft would probably be quicker – The Archetypal Paul Sep 07 '14 at 12:38

score 0 · Answer 4 · answered Mar 03 '21 at 16:14

0

If you don't want to use collect you may try :

val indices = Array(1, 2, 3, 2)
val arr = Array("a", "b", "c", "d")
val res = for (i <- indices) yield arr(i)

answered Mar 03 '21 at 16:14

Lucas Meier

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Your solution doesn't address safety, as mentioned in the question. – jwvh Mar 03 '21 at 16:40

Scala: Get specific elements by a sequence of indices

4 Answers4