CODEWITHSUNDEEP
swiftdictionaryoption-typeunwrap
gran_profaci
gran_profaci

Reputation: 8461

Map with Optional Unwrapping in Swift

Say I had the below api : 

func paths() -> [String?] {
    return ["test", nil, "Two"]
}

And I was using this in a method where I needed [String], hence I had to unwrap it using the simple map function. I'm currently doing : 

func cleanPaths() -> [String] {
    return paths.map({$0 as! String})
}

Here, the force-cast will cause an error. So technically I need to unwrap the Strings in the paths array. I'm having some trouble doing this and seem to be getting silly errors. Can someone help me out here?

Upvotes: 18

Views: 12133

Answers (4)

Venu Gopal Tewari
Venu Gopal Tewari

Reputation: 5876

let name = obj.value
name.map { name  in print(name)}

Upvotes: -1

Brad Larson
Brad Larson

Reputation: 170317

compactMap() can do this for you in one step:

let paths:[String?] = ["test", nil, "Two"]

let nonOptionals = paths.compactMap{$0}

nonOptionals will now be a String array containing ["test", "Two"].

Previously flatMap() was the proper solution, but has been deprecated for this purpose in Swift 4.1

Upvotes: 46

Antonio
Antonio

Reputation: 72770

You should filter first, and map next:

return paths.filter { $0 != .None }.map { $0 as! String }

but using flatMap as suggested by @BradLarson is a just better

Upvotes: 5

Code Different
Code Different

Reputation: 93181

Perhaps what you want to is a filter followed by a map:

func cleanPaths() -> [String] {
    return paths()
            .filter {$0 != nil}
            .map {$0 as String!}
}

let x = cleanPaths()
println(x) // ["test", "two"]

Upvotes: 1

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