Sorting an array and changed a second one using the sorting indices of the first [swift]

Question

I have two arrays A and B. I was wondering what is the most efficient implementation for sorting array A and changing B according to the sorting indices of A.

I've worked out the following so far:

 var A = [5.0, 3.0, 7.0, 0.0, 2.0]
 var B = [3.0, 4.0, 5.0, 6.0, 7.0]
 var C = B
 let idxs = sorted(indices(A)) { A[0ドル] < A[1ドル] }
 println(idxs)
 for i in 0...A.count-1 {
 C[i] = B[idxs[i]]
 }
 println(C)

Answer 1

You could zip the two arrays together and sort by the first item:

var a = [5.0, 3.0, 7.0, 0.0, 2.0]
var b = [3.0, 4.0, 5.0, 6.0, 7.0]
var c = Array(zip(a, b)) // [(5.0, 3.0), (3.0, 4.0), (7.0, 5.0), (0.0, 6.0), (2.0, 7.0)]
c.sort { 0ドル.0 < 1ドル.0 } // [(0.0, 6.0), (2.0, 7.0), (3.0, 4.0), (5.0, 3.0), (7.0, 5.0)]

• Nicholas: can this work zipping more than two arrays? How can I unzip everything back to normal?
• MirekE: No, zip works with two arrays. Obviously, you could create array of tuples with more than two elements manually. You could unzip the array with map, but the idea here was to keep all arrays together, if possible.
• Nicholas: I'm working with more than two arrays. Your solution is nice but does not suit my problem. Thanks anyway.

Answer 2

Here is the solution I proposed for a similar question :

How to sort 1 array in Swift / Xcode and reorder multiple other arrays by the same keys changes

var names = [ "Paul", "John", "David" ]
var ages = [ 35, 42, 27 ]
let newOrder = names.enumerate().sort({0ドル.1<1ドル.1}).map({0ドル.0})
names = newOrder.map({names[0ドル]})
ages = newOrder.map({ages[0ドル]})