Given I have array of stucts like this:
let array = [Struct(key: "a", value: 1), Struct(key: "b", value:2)]
How can I subscript the array with the key?
array["b"] would be nice, but as expected, it doesn't work.
Edit: The reason I'm not using dictionary, is I need to preserve the order of items.
This is syntactic sugar around your solution, @AdamBardon.
You can extend Array to allow you to subscript it directly. Under the covers it is just using the same first(where:) call:
protocol HasKey {
var key: String { get }
}
struct Struct: HasKey {
var key: String
var value: Int
}
extension Array where Element: HasKey {
subscript(str: String) -> Element? {
return self.first(where: { $0.key == str })
}
}
Example:
let array = [Struct(key: "a", value: 1), Struct(key: "b", value:2)]
if let x = array["a"] {
print(x)
}
Output:
Struct(key: "a", value: 1)
Using the protocol allows you to easily extend this functionality to any class or struct that has a key: String property by having them adopt the HasKey property:
extension SomeOtherClass: HasKey { }
You can also accomplish it without the protocol by checking if Element == Struct:
extension Array where Element == Struct {
subscript(str: String) -> Element? {
return self.first(where: { $0.key == str })
}
}
You can make the dictionary of dictionaries like below code:
let myDictionaryOfDictionaries : [String : [String : String]] =
["Apples" : ["Colour" : "Red", "Type" : "Granny Smith"],
"Oranges" : ["Colour" : "Orange", "Type" : "Seville"]]
print(myDictionaryOfDictionaries["Apples"] ?? "")
Hope it will help you.
You should use a dictionary. Arrays are designed to access elements by their index and not by a property of the element. Indexes also have to be integers.
Dictionaries on the other hand are key-value pairs, so using a Dictionary seems to perfect for your use case.
let structs = ["a":Struct(key:"a",value:1),"b":Struct(key:"b",value:2)]
structs["b"] returns the struct with key "b".
This is not the exact way how I imagined the solution, but it's a good alternative which solves my problem.
I got what I needed using this:
array.first(where: { $0.key == "b" })
