How to convert NSArray to [Int] with Swift

Question

My question is simple,

let string = "1,2,3,4,5,6"

As I want to find the max value in the string, and I want to use the method in this way, so I need to do this:

let array = string.componentsSeparatedByString(",") as [AnyObject]
let numArray = array as! [Int]
let maxNum = numArray.reduce(Int.min, combine: {max($0,$1)})

But, there will be an error in "array as! [Int]"

fatal error: array cannot be bridged from Objective-C

And if it's

let numArray = array as! [String]

It will be ok but we cannot use the reduce method.

So, how to convert the array with string to [Int]?

Antonio · Accepted Answer · 2015-07-06 07:33:10Z

2

You can use reduce and optionals, preventing usage of forced unwrapping, which can cause the app to crash if the string doesn't actually contain numbers.

Here's the code:

let array = string.componentsSeparatedByString(",")

let initial: Int? = .None
let max = array.reduce(initial) {
    let num = $1.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet()).toInt()

    return num > $0 ? num : $0
}

The first line splits the string into an array of strings - notice that there's no need of a cast, because the method already returns an array of strings.

The reduce method is used by passing an optional integer as initial value, set to .None (which is a synonym for nil).

The closure passed to reduce converts the current value into an integer using toInt(), which returns an optional, to account for the string not convertible to a valid integer.

Next, the closure returns the converted number, if it's greater than the value calculated at the previous iteration (or than the initial value), otherwise it returns the value at the last iteration.

Note that comparing integers and nil is perfectly legit: any integer value is always greater than nil (even a negative value).

This method works with a "normal" list of numbers:

let string = "1, 2, 3, 4, 5, 6, 7" // The result is 7

but also with lists having numbers and non-numbers:

let string = "1, 2, three, 4, 5, 6, seven" // the result is 6

and with lists containing no numbers:

let string = "one, two, three, three, four, five, six, seven" // the result is nil

answered Jul 6, 2015 at 7:33

Antonio

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4 Comments

Martin R Over a year ago

Optionals are implicitly initialized to nil, = .None is not needed.

KaKa Over a year ago

I change let initial: Int? = .None to let initial: int! = .None cause I have to get the string, so return "\(max)" will have some optional string, but int! will be OK. Anyway, thanks for your answer!

Antonio Over a year ago

@KaKa: Making it an implicitly unwrapped doesn't necessarily mean you will get a non nil result - If I were in you, I'd limit usage of them to really necessary cases, favoring normal optionals in all other cases.

KaKa Over a year ago

@Antonio ,Great! I got what you mean! Looks like I haven't got the truly usage of optionals, I prefer to use ! than ?, looks like I have to learn swift deeply

vadian · Accepted Answer · 2015-07-06 07:08:30Z

2

you can map the array

let string = "1,2,3,4,5,6"
let numArray = string.componentsSeparatedByString(",").map {$0.toInt()!}

answered Jul 6, 2015 at 7:08

vadian

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2 Comments

KaKa Over a year ago

Got an error "AnyObject doesn't have a member named toInt()"

vadian Over a year ago

If you know that this AnyObject is a string, cast it to String with as! String

liushuaikobe · Accepted Answer · 2015-07-06 07:16:02Z

1

let string = "1,2,3,4,5,6"

func toInt(str: String) -> Int {
    return str.toInt()!
}

let numArray = map(string.componentsSeparatedByString(",") as [String], toInt)

Then, numArray is [1, 2, 3, 4, 5, 6].

edited Jul 6, 2015 at 7:16

answered Jul 6, 2015 at 7:10

liushuaikobe

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Comments

Jeffrey Neo · Accepted Answer · 2015-11-19 09:29:30Z

0

In Swift 2.0,

let string = "1,2,3,4,5,6"
let numArray = string.componentsSeparatedByString(",").map{ Int($0)! }

answered Nov 19, 2015 at 9:29

Jeffrey Neo

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Comments

Jake Bromberg · Accepted Answer · 2018-01-20 05:44:52Z

0

In Swift 4.0:

let string = "1,2,3,4,5,6"
let stringArray = string.split(separator: ",").map(String.init)
let intArray = stringArray.flatMap(Int.init)
intArray.max()

answered Jan 20, 2018 at 5:44

Jake Bromberg

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