is there any way to get list of filtered indexes instead of objects.
class Object
{
var name
var goal
}
var array<Object> = Array<Object>()
var filteredIndexes = array.filter{$0.name = "Sane"} // How to implement this?
3 Answers
There are several ways to achieve your goal. For instance, you can filter Array.indices instead of the array itself.
Self contained example:
struct Object {
let name:String
let goal:String
}
let objects = [Object(name: "John", goal: "a"),Object(name: "Jane", goal: "a"),Object(name: "John", goal: "c"),Object(name: "Pete", goal: "d")]
let nameToBeFound = "John"
let filteredIndices = objects.indices.filter({objects[$0].name == nameToBeFound}) //[0,2]
Comments
let animals = ["cat", "dog", "cat", "dog", "cat"]
let indexes = animals.enumerated().filter({ return $1 == "cat" }).map { return $0.offset }
print(indexes)
prints out [0, 2, 4]
Comments
As Lame said above, the solution from Return an array of index values from array of Bool where true can be adapted to this case:
let filteredIndices = objects.enumerated().flatMap { $0.element.name == "Sane" ? $0.offset : nil }
One could also define a custom extension
extension Sequence {
func indices(where predicate: (Element) -> Bool ) -> [Int] {
return enumerated().flatMap { predicate($0.element) ? $0.offset : nil }
}
}
which is then used as
let filteredIndices = objects.indices(where: { $0.name == "Sane" })
Remark: In Swift 4.1, this flatMap() method has been renamed to compactMap()
let indexes = array.enumerated().flatMap { (index, element) -> Int? in return element.name == "Sane" ? index: nil}? Adapted from stackoverflow.com/questions/38225121/…