Remove multiple indices from array

Question

I have an array and I want to remove a bunch of indices

var arr = [0,1,2,3,4,5,6]
var rmIndices = [1,4,5]

What is the best way to remove in

Answer 1

Note that PermutationGenerator is going away in Swift 3 and also doesn't keep the ordering the same, though perhaps it did at one time. Using the accepted answer results in [2, 6, 0, 3] which may be unexpected. A couple of alternative approaches that give the expected result of [0, 2, 3, 6] are:

let flatArr = arr.enumerate().flatMap { rmIndices.contains($0.0) ? nil : $0.1 }

or

let filterArr = arr.enumerate().filter({ !rmIndices.contains($0.0) }).map { $0.1 }

Answer 2

Using lodash https://lodash.com/

var arr = [0,1,2,3,4,5,6]
var rmIndices = [1,4,5]
_.pullAt(arr, rmIndices);

Answer 3

Rather than a list of indices to remove, it may be easier to have a list of indices to keep, which you can do using the Set type:

let rmIndices = [1,4,5]
let keepIndices = Set(arr.indices).subtract([1,4,5])

Then you can use PermutationGenerator to create a fresh array of just those indices:

arr = Array(PermutationGenerator(elements: arr, indices: keepIndices))

Answer 4

For Swift 3

    var arr = [0,1,2,3,4,5,6]
    let rmIndices = [1,4,5]
    arr = arr.filter{ !rmIndices.contains($0) }
    print(arr)

if you want to produce output very fastly then you can use

    var arr = [0,1,2,3,4,5,6]
    let rmIndices = [1,4,5]
    arr = Array(Set(arr).subtracting(rmIndices))
    print(array)

But it will change order of your array

Answer 5

In Swift 4:

let newArr = arr.enumerated().compactMap {
    rmIndices.contains($0.0) ? nil : $0.1
}

• enumerated() generates (index, value) pairs
• compactMap concatenates non-nil values
• In the closure, $0.0 is the index (first element of enumerated pair) as $0.1$ is the value
• compactMap gathers values whose indices are not found in rmIndices

Answer 6

rmIndices.sort({ $1 < $0 })     

for index in rmIndices
{
    arr.removeAtIndex(index)
}

Note that I've sorted the indices in descending order. This is because everytime you remove an element E, the indices of the elements beyond E reduce by one.