Swift 3 Array, remove more than one item at once, with .remove(at: i)

It's possible if the indexes are continuous using removeSubrange method.For example, if you would like to remove items at index 3 to 5:

myArray.removeSubrange(ClosedRange(uncheckedBounds: (lower: 3, upper: 5)))

For non-continuous indexes, I would recommend remove items with larger index to smaller one. There is no benefit I could think of of removing items "at the same time" in one-liner except the code could be shorter. You can do so with an extension method:

extension Array {  mutating func remove(at indexes: [Int]) {    for index in indexes.sorted(by: >) {      remove(at: index)    }  }}

Then:

myArray.remove(at: [3, 5, 8, 12])

UPDATE: using the solution above, you would need to ensure the indexes array does not contain duplicated indexes. Or you can avoid the duplicates as below:

extension Array {    mutating func remove(at indexes: [Int]) {        var lastIndex: Int? = nil        for index in indexes.sorted(by: >) {            guard lastIndex != index else {                continue            }            remove(at: index)            lastIndex = index        }    }}var myArray = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]myArray.remove(at: [5, 3, 5, 12]) // duplicated index 5// result: [0, 1, 2, 4, 6, 7, 8, 9, 10, 11, 13] only 3 elements are removed

Remove elements using indexes of an array elements:

1. Array of Strings and indexes

   let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]let indexAnimals = [0, 3, 4]let arrayRemainingAnimals = animals    .enumerated()    .filter { !indexAnimals.contains($0.offset) }    .map { $0.element }print(arrayRemainingAnimals)//result - ["dogs", "chimps", "cow"]

2. Array of Integers and indexes

   var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]let indexesToRemove = [3, 5, 8, 12]numbers = numbers    .enumerated()    .filter { !indexesToRemove.contains($0.offset) }    .map { $0.element }print(numbers)//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]

Remove elements using element value of another array

3. Arrays of integers

   var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]let elementsTobeRemoved = [3, 5, 8, 12]let arrayResult = numbers.filter { element in    return !elementsTobeRemoved.contains(element)}print(arrayResult)//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]

4. Arrays of strings

   let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]let arrayRemoveLetters = ["a", "e", "g", "h"]let arrayRemainingLetters = arrayLetters.filter {    !arrayRemoveLetters.contains($0)}print(arrayRemainingLetters)//result - ["b", "c", "d", "f", "i"]

Simple and clear solution, just Array extension:

extension Array {    mutating func remove(at indices: [Int]) {        Set(indices)            .sorted(by: >)            .forEach { rmIndex in                self.remove(at: rmIndex)            }    }}

• Set(indices) - ensures uniqueness
• .sorted(by: >) - function removes elements from last to first, so during removal we are sure that indexes are proper