Swift: Getting sorted()

Sticking with the spreadsheet analogy from previous posts, to sort a column of figures low to high or high to low using sort() or sorted() is straightforward (note: sorted() has in Swift 2 changed to sortInPlace()), but usually in a spreadsheet the reordering of one column means the reordering of other columns to keep rows together. And here we need a bit more mutability, because simply put things are mutating.

Rows

Let's suppose there are three rows in the spreadsheet, one with objects, another with prices and a third with colours.

var objects = ["car","van","balloon","flower"]
var prices = [100,123.33,207.66,36.5]
var colours = ["blue","yellow","green","red"]

And that one of those rows is highlighted and determined to be the one to be in charge of reordering all the rows in our spreadsheet table.

Mapping

In this instance we're going to suppose that the object names are going to be in charge of ordering and so they are going to be at the centre of the mapping. In the process of mapping, an array of tuples is created of type [(object:String, price:Double, colour:String)]

var rows = objects.map{(object:$0, price:removeAtIndex(&prices, 0), colour: removeAtIndex(&colours, 0))}

And this array of tuples, which represents the rows displayed can now be sorted however we wish.

Sorting

For now I'm going to stick with sorting by object names:

rows.sort{$0.object < $1.object}

If you wish to sort by other means simply replace 'object' with 'price' or 'colour'.

Time for a refill

With the sorting complete our original arrays can be replaced with those in the new and correct order:

objects = rows.map{$0.object}
prices = rows.map{$0.price}
colours = rows.map{$0.colour}

objects // ["balloon", "car", "flower", "van"]
prices // [207.66, 100, 36.5, 123.33]
colours // ["green", "blue", "red", "yellow"]

Full code

And for completenes, here's the code all in one place:

var objects = ["car","van","balloon","flower"]
var prices = [100,123.33,207.66,36.5]
var colours = ["blue","yellow","green","red"]

var rows = objects.reverse().map{(object:$0, price:prices.removeLast(), colour: colours.removeLast())}
rows.sort{$0.object < $1.object}

objects = rows.map{$0.object}
prices = rows.map{$0.price}
colours = rows.map{$0.colour}

objects // ["balloon", "car", "flower", "van"]
prices // [207.66, 100, 36.5, 123.33]
colours // ["green", "blue", "red", "yellow"]

For the sake of performance, I've tweaked the code by reversing the objects array so that I can work backwards through the prices and colours arrays (using removeLast() lowers the complexity over removeAtIndex()).

Conclusion

It might be that storing separate arrays is considered fragile, mutability undesirable or that the need to identify the names of items unnecessary, but from here experiments can be made to make things more functional and generic. For example:

var objects = ["car","van","balloon","flower"]
var prices = [100,123.33,207.66,36.5]
var colours = ["blue","yellow","green","red"]


func secondarySort<A:Comparable, B>(arr1:[A], inout arr2:[B]) {
    arr2 = arr1.map{(object1:$0, object2:arr2.removeFirst())}.sort({$0.object1 < $1.object1}).map{$0.object2}
}

secondarySort(objects, &colours) // ["green", "blue", "red", "yellow"]
secondarySort(objects, &prices) // [207.66, 100, 36.5, 123.33]
// the array around which the sort is based must be sorted last
objects.sort(<) // ["balloon", "car", "flower", "van"]

Here two arrays are passed to a function and the second array is sorted based on the first, it is a function that can be called time and again for any number of "columns". And it overcomes the hurdle of trying to process multiple "columns" (or arrays, of potentially different types) all at once.

To remain in line with sort() and sorted() being mutating and non-mutating functions, I'll also throw in a secondarySorted() function:

func secondarySorted()<A:Comparable, B>(arr1:[A], var arr2:[B]) -> [B] {

    var rows = arr1.reverse().map{(object1:$0, object2:arr2.removeLast())}
    rows.sort({$0.object1 < $1.object1})
    return rows.map{$0.object2}

}

Follow @sketchytech