Array1 = [1, 2, 3, 4, 5, 6]
Array2 = [1,5]
I want to get:
Array1 = [2, 3, 4, 6]
I want to do this by using Set because these arrays may get larger.
Also it is important that I maintain the order of the array.
Array1 = [1, 2, 3, 4, 5, 6]
Array2 = [1,5]
I want to get:
Array1 = [2, 3, 4, 6]
I want to do this by using Set because these arrays may get larger.
Also it is important that I maintain the order of the array.
Create a set with all elements from the second array, then filter the first array to get only the elements which are not in the set:
let array1 = [5, 4, 1, 2, 3, 4, 1, 2]
let array2 = [1, 5]
let set2 = Set(array2)
let result = array1.filter { !set2.contains($0) }
print(result) // [4, 2, 3, 4, 2]
This preserves the order (and duplicate elements) from the first array. Using a set is advantageous if the second array can be large, because the lookup is faster.
var array1 = [1, 2, 3, 4, 5, 6]
var array2 = [1,5]
var arrayResult = array1.enumerated()
.filter { !array2.contains($0.0 + 1) }
.map { $0.1 }
print(arrayResult)
[2, 3, 4, 6]
Another ways to achieve the same result:
1. User filter
let arrayResult = array1.filter { element in
return !array2.contains(element)
}
2. Use Sort
array2.sorted(by: >).forEach { if $0 < self.array1.count { self.array1.remove(at: $0) } }
Remove elements using indexes array:
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"]
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
Arrays of integers
let arrayResult = numbers.filter { element in
return !indexesToRemove.contains(element)
}
print(arrayResult)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
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"]
(Note: because you added to your question and maintaining order then my answer is not right anymore, because Set don't preserve the order. then the filter answers are a better fit)
use subtracting from a Set:
array1 = Array(Set(array1).subtracting(Set(array2)))
you can add this as an operator :
Using the Array → Set → Array method mentioned by Antonio, and with the convenience of an operator, as freytag pointed out, I've been very satisfied using this:
// Swift 3.x func - <Element: Hashable>(lhs: [Element], rhs: [Element]) -> [Element] { return Array(Set<Element>(lhs).subtracting(Set<Element>(rhs))) }
quoted from: https://stackoverflow.com/a/42679608/1930509
and maintaining order. i added a note to my answer.