I would like to create an multidimensional array of arrays of String from an array of String

Question

I have an array of Strings, and I would like to group the content by three to create arrays within the array.

For instance:

var arrayOfStrings = ["a0", "a1", a2", "b0", "b1", "b2"]

And I am looking to achieve:

var multiDimensionalArray = [["a0", "a1", a2"], ["b0", "b1", "b2"]]

Therefore, I am looking for a function that could group the elements within arrayOfStrings in arrays containing 3 elements, in the right order.

Thank you,

egor.zhdan · Accepted Answer · 2015-09-08 12:33:35Z

1

Try this function:

func makeMultidimensional(a: [String]) -> [[String]] {
    var result: [[String]] = []
    for var i = 0; i < a.count; i += 3 {
        result.append([a[i], a[i + 1], a[i + 2]])
    }
    return result
}

print(makeMultidimensional(arrayOfStrings)) // prints [["a0", "a1", "a2"], ["b0", "b1", "b2"]]

It iterates over array of strings and adds new array of 3 items to the result every time.

answered Sep 8, 2015 at 12:33

egor.zhdan

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Jagveer Singh · Accepted Answer · 2015-09-08 12:40:19Z

0

very simple

NSArray *multiDimensionalArray = @[@[@"a0", @"a1", @"a2"], @[@"b0", @"b1", @"b2"]];

for swift

var multiDimensionalArray = [["a0", "a1", "a2"], ["b0", "b1", "b2"]];

runnig on my system well

edited Sep 8, 2015 at 12:40

answered Sep 8, 2015 at 12:36

Jagveer Singh

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3 Comments

egor.zhdan Over a year ago

I think OP is looking for solutions in Swift

Blind Ninja Over a year ago

Well Bhai Jagveer OP want a solution. He knows how to declare multiDimensionalArray (as you did) , look in his question.

Jagveer Singh Over a year ago

great thanks, actualy i dont understand the quetion!!

Vladimir Kofman · Accepted Answer · 2015-09-08 19:36:51Z

0

func makeMultidimensional<T>(input:[T], dimensions:Int) -> [[T]]?
{
    if dimensions <= 0 || count(input) % dimensions > 0 {
        return nil
    }

    return
        input.reduce(([[T]](), [T]())) {
            if 0..<dimensions-1 ~= count($0.1) {
                return ($0.0, $0.1 + [$1])
            }
            else {
                return ($0.0 + [$0.1 + [$1]], [])
            }
        }.0
}

answered Sep 8, 2015 at 19:36

Vladimir Kofman

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