As others already said, we need to understand which are the keys.
However I am trying to provide a solution to my interpretation of your question.
struct User {
let id: String
let firstName: String
let lastName: String
}
Here I am assuming that 2 users with the same id
cannot exist
let users: [User] = ...
let dict = users.reduce([String:User]()) { (result, user) -> [String:User] in
var result = result
result[user.id] = user
return result
}
Now dict
is a dictionary where the key
is the user id
and the value
is the user value
.
To access a user via its id
you can now simply write
let user = dict["123"]
Update #1: General approach
Given an array of a given type Element
, and a closure that determine the key
of an Element
, the following generic function will generate a Dictionary
of type [Key:Element]
func createIndex<Key, Element>(elms:[Element], extractKey:(Element) -> Key) -> [Key:Element] where Key : Hashable {
return elms.reduce([Key:Element]()) { (dict, elm) -> [Key:Element] in
var dict = dict
dict[extractKey(elm)] = elm
return dict
}
}
Example
let users: [User] = [
User(id: "a0", firstName: "a1", lastName: "a2"),
User(id: "b0", firstName: "b1", lastName: "b2"),
User(id: "c0", firstName: "c1", lastName: "c2")
]
let dict = createIndex(elms: users) { $0.id }
// ["b0": {id "b0", firstName "b1", lastName "b2"}, "c0": {id "c0", firstName "c1", lastName "c2"}, "a0": {id "a0", firstName "a1", lastName "a2"}]
Update #2
As noted by Martin R the reduce will create a new dictionary for each iteration of the related closure. This could lead to huge memory consumption.
Here's another version of the createIndex
function where the space requirement is O(n) where n is the length of elms.
func createIndex<Key, Element>(elms:[Element], extractKey:(Element) -> Key) -> [Key:Element] where Key : Hashable {
var dict = [Key:Element]()
for elm in elms {
dict[extractKey(elm)] = elm
}
return dict
}