Filter Swift Array and Give Indices (Not Values)

Question

I need to filter an array by some condition on its elements, but I need to obtain the indices of the elements that passed the test, not the elements themselves.

For example: given an array of Bool, I want to transform that into an array of Int that contains only the indices of the elements of the orginal array that are true.

I could do this:

// INPUT array:
let flags = [true, false, true, false]

// OUTPUT array:
var trueIndices = [Int]()

for (index, value) in flags.enumerated() where value == true {
    trueIndices.append(index)
}

...but it isn't "swifty" at all.

Is there a more elegant way? Something akin to filter(), but that returns the indices instead of the elements.

Thank you very much everyone. The "swiftiness" in all your answers is TOO DAMN HIGH — Nicolas Miari
– Nicolas Miari, Commented Sep 20, 2017 at 15:46

vadian · Accepted Answer · 2017-09-20 14:16:10Z

5

You can directly filter the indices

let flags = [true, false, true, false]    
let trueIndices = flags.indices.filter{ flags[$0] }

and

let falseIndices = flags.indices.filter{ !flags[$0] }

edited Sep 20, 2017 at 14:16

answered Sep 20, 2017 at 14:07

vadian

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5 Comments

Nicolas Miari Over a year ago

Oh, it makes sense now. You are filtering the array (?) of all indices with a closure that captures the outer flags array, and tests its contents at the current index ($0).

NSGangster Over a year ago

This is "swiftier"

Nicolas Miari Over a year ago

It's a long way to "think in portals"!

Martin R Over a year ago

Of course one could define a func indices(where isIncluded: (Element) -> Bool) -> [Index] extension method for collections ...

Alexander Over a year ago

Just a reminder: this only works well when subscripting is possible, and possible in O(1). Won't work for sequences

David Pasztor · Accepted Answer · 2017-09-20 14:12:39Z

4

Not sure how much "swiftier" it is, but you can also use a consecutive filter and map operation or just a single flatMap instead. For sure, you can declare the array immutable and you don't have to write any explicit loops, both of which are usually considered a more functional approach than your current one.

let trueIndices = flags.enumerated().filter{$0.element}.map{$0.offset}
print(trueIndices)

Output:

[0,2]

Using a single flatMap:

let trueIndices = flags.enumerated().flatMap { (offset, flag) in flag ? offset : nil }

edited Sep 20, 2017 at 14:12

answered Sep 20, 2017 at 14:03

David Pasztor

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5 Comments

Nicolas Miari Over a year ago

Of course! enumerated() gives an array of tuples, I can filter/map that. Guess I’ve been coding for too many hours now...

Martin R Over a year ago

The flatMap version does not compile in Xcode 9. My suggestion: flags.enumerated().flatMap { (offset, flag) in flag ? offset : nil }

David Pasztor Over a year ago

@MartinR thanks for the tip, I've only tested in Xcode 8, but updated my answer, since your solution is more clear anyways

Martin R Over a year ago

You probably meant "declare the array immutable"

David Pasztor Over a year ago

Yeah, I was editing my wording and forgot to change mutable to immutable, thanks.

Alexander · Accepted Answer · 2017-09-20 15:30:01Z

1

Here's what I would do. It's generalizable to sequences, or even collections whose subscripting isn't O(1).

let indicesOfTrue = Array(flags.enumerated().lazy.filter{ $0.element }.map{ $0.offset })

answered Sep 20, 2017 at 15:30

Alexander

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