How to group an array with multiple parameters

You can adopt to this example by conforming InventoryItem to Equatable and Hashable as below,

class InventoryItem: Equatable, Hashable {

    var hashValue: Int { return a.hashValue ^ b.hashValue ^ c.hashValue }

    static func == (lhs: InventoryItem, rhs: InventoryItem) -> Bool {
        return  lhs.a == rhs.a &&
                lhs.b == rhs.b &&
                lhs.c == rhs.c
    }

    var a: String
    var b: String
    var c: String

    init(_ a: String, b: String, c: String) {
        self.a = a
        self.b = b
        self.c = c
    }
}

Lets say we have this data,

let inv1 = InventoryItem("a", b: "b", c: "c")
let inv2 = InventoryItem("a", b: "b", c: "c")
let inv3 = InventoryItem("p", b: "q", c: "r")
let inv4 = InventoryItem("x", b: "y", c: "z")
let list = [inv1, inv2, inv3, inv4]

var groupedItems: [String: [InventoryItem]] = [:]

let unique = Set(list)
unique.forEach { (item) in
   groupedItems["\(item.hashValue)"] = list.filter({ $0 == item })
}

groupedItems.keys.forEach { key in
   let items = groupedItems[key]!
   print(items.count)
   items.forEach({ (item) in
       print("\(item.a)\(item.b)\(item.c)")
   })
}

Output:

1
xyz
1
pqr
2
abc
abc

Note If your InventoryItem is a struct or enum then you don't need to explicitly conform to Equatable, only conforming to Hashable is enough and you can just remove hashValue and equator method(static func == (lhs: InventoryItem, rhs: InventoryItem) -> Bool) from InventoryItem.