Given an array of indexes I would like to get a subarray of myArray, with the items at those indexes. I'm currently iterating the indexes array to create a subarray, but I'm wondering if this can be achieved by using the .filter function.
var indexes = [3,4,9,11]
myArray.filter(...)
Assuming that
then a simple map operation would work:
let indexes = [2, 4, 7]
let myArray = ["a", "b", "c", "d", "e", "f", "g", "h"]
let filtered = indexes.map { myArray[$0] }
print(filtered) //["c", "e", "h"]
Remark: In earlier Swift releases, there was a PermutationGenerator
exactly for this purpose:
let filtered = Array(PermutationGenerator(elements: myArray, indices: indexes))
print(filtered) //["c", "e", "h"]
However, this has been deprecated in Swift 2.2 and will be removed in Swift 3. I haven't seen a Swift 3 replacement yet.
map() iterates internally as well. – Possible advantages compared to an explicit loop are: Shorter functional code, and the result is a constant.
you can try, although that's far away from efficient. But that's what you get if you insist on using filter
let myArray = ["a", "b", "c", "d", "e"]
var indexes = [1, 3]
myArray.filter { (vowel) -> Bool in
if let index = myArray.indexOf(vowel) {
return indexes.contains(index)
}
return false
}
indexOf returns the index of the first occurrence).You can also use flatMap to retrieve the filtered subArray
let indexes = [2, 4, 7]
let myArray = ["a", "b", "c", "d", "e", "f", "g", "h"]
let subArray = indexes.flatMap { (index) -> String? in
return (0 <= index && index < myArray.count) ? myArray[index] : nil
}
filterwould be inefficient because it would look at each value inmyArray. Consider what would happen ifmyArrayhad 100 thousand items. Your method of iterating the indexes array would loop 4 times, but filter would loop (internally) 100 thousand times.