Swift: Group duplicate array elements into new array-CodePudding

I have an Order object as follows:

struct Order {
    let id: Int
    let item: String
    let price: Int
}

These are grouped in an, however sometimes there are duplicate IDs and I need these to be grouped into their own array of Order duplicate objects. So essentially at the moment I have [Order] and I need to convert this into [[Order]] where all duplicates will be grouped together, and where there are no duplicates they will simply be on their own in an array.

So for example, imagine I have the following array of orders:

[Order(id: 123, item: "Test item1", price: 1)
Order(id: 345, item: "Test item2", price: 1)
Order(id: 678, item: "Test item3", price: 1)
Order(id: 123, item: "Test item1", price: 1)]

This needs to be converted to:

[[Order(id: 123, item: "Test item1", price: 1), Order(id: 123, item: "Test item1", price: 1)],
[Order(id: 345, item: "Test item2", price: 1)],
[Order(id: 678, item: "Test item3", price: 1)]]

I have been playing around with a dictionary and have so far come up with the following:

let dictionary = Dictionary(grouping: orders, by: { (element: Order) in
    return element.id
})

This returns the following type:

[Int : [Order]]

Which is close, but I don't really want them in a dictionary like this. I just need to be able to get an [[Order]] array that I can loop through for use in my UI.

CodePudding user response：

All you have to do is to use .values on the dictionary you have created and you have your array,

let values = Dictionary(grouping: orders, by: \.id).values

CodePudding user response：

This somewhat should do the trick, not the best solution but should be enough to get it working.

var orders = [[Order]]()

for order in orders {
    if let index = orders.firstIndex(where: { $0.id == order.id }) {
        // We have this order id already let's append another duplicate
        orders[index].append(order)
    } else {
        // We don't have this order id, create a new array
        orders.append([order])
    }
}