Swift: shortcut unwrapping of array of optionals - swift

Question

Assume we have an array of optionals defined:
var arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
I can force unwrap it in a short way: var arrayForCrash = arrayOfOptionals.map { $0! }
But that will make app to crash, is there any other short way(without explicitly unwrapping) how I can unwrap an array of optional?

Answer 1

This solution will get you a new array with all values unwrapped and all nil's filtered away.
Swift 4.1:
let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.compactMap { $0 }
Swift 2.0:
let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.flatMap { $0 }
Swift 1.0:
let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.filter { $0 != nil }.map { $0! }

Answer 2

Since it is an array of optionals, it is possible some of the entries are nil. Instead of force unwrapping with !, use the nil coalescing operator to turns nils into empty strings.
let arrayOfOptionals: [String?] = ["This", "array", nil, "has", "some", "nils", nil]
let array:[String] = arrayOfOptionals.map{ $0 ?? "" }
// array is now ["This", "array", "", "has", "some", "nils", ""]

Answer 3

Although you can use flatMap { $0 } to remove nils, flatMap is actually a much more powerful function, and has an overloaded version which does something completely different (e.g. flatten [[Int]] to [Int]). If you're not careful, you may accidentally invoke the wrong function.
I would recommend using an extension on SequenceType to remove nils. If you use removeNils(), you'll be able to essentially do the following:
[1, nil, 2].removeNils() == [1, 2]
It works by making Optional conform to an OptionalType protocol which allows extending SequenceTypes that contain Optional values.
For more information see the original answer I posted.

Answer 4

I took #Cenny's answer and decided to make an operator out of it:
prefix operator <!> {}
prefix func <!> <T>(array: [T?]) -> [T] {
return array.filter{ $0 != nil }.map{ $0! }
}
I'm using it to parse an array of JSON objects and filter the ones that failed:
static func parse(j: JSONArray) -> [Agency]? {
return <!>j.map { self.parse($0) }
}
Update for Swift 2+:
Use flatMap operator and it'll only return non-nil objects

Answer 5

Swift 4
Easy to read and safe approach to filter nils of any sequence
protocol OptionalProtocol {
associatedtype Wrapped
var optional: Wrapped? { get }
}
extension Optional: OptionalProtocol {
var optional: Wrapped? {
return self
}
}
extension Sequence where Element: OptionalProtocol {
var removingOptionals: [Element.Wrapped] {
return self.compactMap { $0.optional }
}
}
Usage
let array: [Int?] = [1, 2, nil, 3, 4, nil]
print(array.removingOptionals) // prints [1, 2, 3, 4], has type [Int]

Answer 6

The more interesting, how to unwrap an optional array of optional values. It is important to deal when we are working with JSON, because JSON can potentially contain null value for an array of something.
Example:
{ "list": null }
// or
{ "list": [null, "hello"] }
To fill a Swift struct we may have a model:
struct MyStruct: Codable {
var list: [String?]?
}
And to avoid working with String?? as a first item we could:
var myStruct = try! JSONDecoder.init().decode(MyStruct.self, from: json.data(using: .utf8)!)
let firstItem: String? = s1.list?.compactMap { $0 }.first
With compactMap { $0 } we can avoid
let i2: String?? = s1.list?.first
compactMap { $0 } is an equivalent of filter { $0 != nil }. map { $0! }

Answer 7

How about:
import Foundation
var test: [String!] = ["this","is","a",nil,"test"]
for string in test {
if string != nil {
print(string)
}
}
Output is thisisatest.
In your case use [String!], if I understood you correctly.