Sorting arrays based on number of matches - swift

Question

I am trying to find the number of array item matches between multiple test arrays and one control array. After finding the number of matches, I want to append the test arrays to another array, sorted by number of matches between the control array and test array. For example, a test array with 3 matches would be at index 0, 2 matches at index 1, and so on.
let controlArray = ["milk", "honey"]
let test1 = ["honey", "water"]
let test2 = ["milk", "honey", "eggs"]
var sortedArrayBasedOnMatches = [[String]]()
/*I want to append test1 and test2 to sortedArrayBasedOnMatches based on how many items
test1 and test2 have in common with controlArray*/
/*in my example above, I would want sortedArrayBasedOnMatches to equal
[test2, test1] since test 2 has two matches and test 1 only has one*/

Answer 1

This can be done in a very functional and Swiftish way by writing a pipeline to process the input arrays:
let sortedArrayBasedOnMatches = [test1, test2] // initial unsorted array
.map { arr in (arr, arr.filter { controlArray.contains($0) }.count) } // making pairs of (array, numberOfMatches)
.sorted { $0.1 > $1.1 } // sorting by the number of matches
.map { $0.0 } // getting rid of the match count, if not needed
Update As #Carpsen90 pointed out, Switf 5 comes with support for count(where:) which reduces the amount of code needed in the first map() call. A solution that makes use of this could be written along the lines of
// Swift 5 already has this, let's add it for current versions too
#if !swift(>=5)
extension Sequence {
// taken from the SE proposal
// https://github.com/apple/swift-evolution/blob/master/proposals/0220-count-where.md#detailed-design
func count(where predicate: (Element) throws -> Bool) rethrows -> Int {
var count = 0
for element in self {
if try predicate(element) {
count += 1
}
}
return count
}
}
#endif
let sortedArrayBasedOnMatches = [test1, test2] // initial unsorted array
.map { (arr: $0, matchCount: $0.count(where: controlArray.contains)) } // making pairs of (array, numberOfMatches)
.sorted { $0.matchCount > $1.matchCount } // sorting by the number of matches
.map { $0.arr } // getting rid of the match count, if not needed
Another change in style from the original solution is to use labels for the tuple components, this makes the code a little bit clearer, but also a little bit more verbose.

Answer 2

One option is to convert each array to a Set and find the count of elements in the intersection with controlArray.
let controlArray = ["milk", "honey"]
let test1 = ["honey", "water"]
let test2 = ["milk", "honey", "eggs"]
var sortedArrayBasedOnMatches = [ test1, test2 ].sorted { (arr1, arr2) -> Bool in
return Set(arr1).intersection(controlArray).count > Set(arr2).intersection(controlArray).count
}
print(sortedArrayBasedOnMatches)

Answer 3

This will cover the case where elements are not unique in your control array(such as milk, milk, honey...) and with any number of test arrays.
func sortedArrayBasedOnMatches(testArrays:[[String]], control: [String]) -> [[String]]{
var final = [[String]].init()
var controlDict:[String: Int] = [:]
var orderDict:[Int: [[String]]] = [:] // the value is a array of arrays because there could be arrays with the same amount of matches.
for el in control{
if controlDict[el] == nil{
controlDict[el] = 1
}
else{
controlDict[el] = controlDict[el]! + 1
}
}
for tArr in testArrays{
var totalMatches = 0
var tDict = controlDict
for el in tArr{
if tDict[el] != nil && tDict[el] != 0 {
totalMatches += 1
tDict[el] = tDict[el]! - 1
}
}
if orderDict[totalMatches] == nil{
orderDict[totalMatches] = [[String]].init()
}
orderDict[totalMatches]?.append(tArr)
}
for key in Array(orderDict.keys).sorted(by: >) {
for arr in orderDict[key]! {
final.append(arr)
}
}
return final
}