How can I sort one array based on another array positions in swift?

Question

In Swift, say I have two arrays:

var array1: [Int] = [100, 40, 10, 50, 30, 20, 90, 70]
var array2: [Int] = [50, 20, 100, 10, 30]

I want to sort my aary1 as per array2

So final output of my array1 will be: // array1 = [50, 20, 100, 10, 30, 40, 90, 70]

• Martin R: Possible duplicate of Sort array by order of values in other array or Reorder array compared to another array in Swift

Answer 1

Even though your question is quite vaguely worded, judging from your example, what you actually want is the union of the two arrays, with the elements of the smaller array coming in first and after them the unique elements of the bigger array in an unchanged order at the end of this array.

The following code achieves the result of the example:

let combined = array2 + array1.filter{array2.index(of: 0ドル) == nil}

• David Pasztor: It does work, since it keeps the unique elements of both arrays. The question didn't specify what should the ordering be if none of the arrays is a subset of the other, so in this case they are just concatenated.
• David Pasztor: @LucaAngeletti it was a general comment toward the person, who actually did, hence I didn't tag you.

Answer 2

You have not defined how you want the elements in array1 but not in array2 to be sorted. This solution assumes you want to sort those not-found elements by their numeric value:

var array1 = [100, 40, 10, 50, 30, 20, 90, 70]
var array2 = [50, 20, 100, 10, 30]
array1.sort {
 let index0 = array2.index(of: 0ドル)
 let index1 = array2.index(of: 1ドル)
 switch (index0, index1) {
 case (nil, nil):
 return 0ドル < 1ドル
 case (nil, _):
 return false
 case (_, nil):
 return true
 default:
 return index0! < index1!
 }
}
print(array1) // [50, 20, 100, 10, 30, 40, 70, 90]
 // ^ order not defined in array2