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iOS

Posted by Leslie Chicoine
Leslie Chicoine
3,655 Points

[Hip, hip] ...(hip hip, array!) x2

I just learned this, which gives me the sum of an array:

func sumArray(array: [Int]) -> Int {
    return array.reduce(0, combine: +)
}

but next up, I have a multidimensional array that I want to sum... so I want a sum all of the values in the arrays within the parent array (ie: [[1, 2], [3, 4]] should equal 10). I'm trying:

parentArray.reduce(0, combine: .....)

but I'm getting stuck on how to unpack the child arrays. I've been researching how to work with closures, but it's super confusing given that I'm still pretty new to coding. I could also use a for loop and unpack the arrays that way, but I'm wondering if there is a better/cleaner way.

Any thoughts on how to do this?

3 Answers

David Papandrew
David Papandrew
8,386 Points

Hi Leslie,

Try performing "flatmap" on the multidimensional array first (to "flatten" them into a single collection) and then reduce.

You can try this in Playgrounds to see:

let array = [[1, 2], [3, 4]]

let sum = array.flatMap{$0}.reduce(0, +)
print(sum)

Note that the above works for Swift 3. If you are using Swift 2, use ".reduce(0, combine:+)"

Leslie Chicoine
Leslie Chicoine
3,655 Points
Leslie Chicoine
Leslie Chicoine
3,655 Points

Thanks, that's awesome!

I also got the following for loop inside a for loop working, which isn't nearly as nice as the above answer:

var parentArray = [[Int]]() 
var total = 0

func sumParentArray() -> Int {   
     for x in 0 ..< parentArray.count {
            for y in 0 ..< parentArray[x].count {
            total += parentArray[x][y]
        }
    }
    return total
}

sumParentArray()

Though I don't quite understand why I had to use a range (x in 0..< parentArray.count) instead of just having it iterate over the arrays (x in parentArray).

Leslie Chicoine
Leslie Chicoine
3,655 Points
Leslie Chicoine
Leslie Chicoine
3,655 Points

I found a thorough explanation of Map Filter Reduce: http://useyourloaf.com/blog/swift-guide-to-map-filter-reduce/ for those who are curious.