Swift 3.0
i had the same problem, i found on the documentation Apple this solution.
let numbers = [1, 2, 3, 4]
let addTwo: (Int, Int) -> Int = { x, y in x + y }
let numberSum = numbers.reduce(0, addTwo)
// 'numberSum' == 10
But, in my case i had a list of object, then i needed transform my value of my list:
let numberSum = self.list.map({$0.number_here}).reduce(0, { x, y in x + y })
this work for me.
For me, it was like this using property
let blueKills = match.blueTeam.participants.reduce(0, { (result, participant) -> Int in
result + participant.kills
})
Swift3 has changed to :
let multiples = [...]
sum = multiples.reduce(0, +)
Swift 4 example
class Employee {
var salary: Int = 0
init (_ salary: Int){
self.salary = salary
}
}
let employees = [Employee(100),Employee(300),Employee(600)]
var sumSalary = employees.reduce(0, {$0 + $1.salary}) //1000
@IBOutlet var valueSource: [MultipleIntBoundSource]!
private var allFieldsCount: Int {
var sum = 0
valueSource.forEach { sum += $0.count }
return sum
}
used this one for nested parameters
For sum of elements in array of Objects
self.rankDataModelArray.flatMap{$0.count}.reduce(0, +)
Swift 3+ one liner to sum properties of objects
var totalSum = scaleData.map({$0.points}).reduce(0, +)
Where points is the property in my custom object scaleData that I am trying to reduce
This also works:
let arr = [1,2,3,4,5,6,7,8,9,10]
var sumedArr = arr.reduce(0, combine: {$0 + $1})
print(sumedArr)
The result will be: 55
Swift 3
If you have an array of generic objects and you want to sum some object property then:
class A: NSObject {
var value = 0
init(value: Int) {
self.value = value
}
}
let array = [A(value: 2), A(value: 4)]
let sum = array.reduce(0, { $0 + $1.value })
// ^ ^
// $0=result $1=next A object
print(sum) // 6
Despite of the shorter form, many times you may prefer the classic for-cycle:
let array = [A(value: 2), A(value: 4)]
var sum = 0
array.forEach({ sum += $0.value})
// or
for element in array {
sum += element.value
}
In Swift 4 You can also constrain the sequence elements to Numeric protocol to return the sum of all elements in the sequence as follow
extension Sequence where Element: Numeric {
/// Returns the sum of all elements in the collection
func sum() -> Element { return reduce(0, +) }
}
edit/update:
Xcode 10.2 • Swift 5 or later
We can simply constrain the sequence elements to the new AdditiveArithmetic protocol to return the sum of all elements in the collection
extension Sequence where Element: AdditiveArithmetic {
func sum() -> Element {
return reduce(.zero, +)
}
}
Xcode 11 • Swift 5.1 or later
extension Sequence where Element: AdditiveArithmetic {
func sum() -> Element { reduce(.zero, +) }
}
let numbers = [1,2,3]
numbers.sum() // 6
let doubles = [1.5, 2.7, 3.0]
doubles.sum() // 7.2
To sum a property of a custom object we can extend Sequence to take a predicate to return a value that conforms to AdditiveArithmetic:
extension Sequence {
func sum<T: AdditiveArithmetic>(_ predicate: (Element) -> T) -> T { reduce(.zero) { $0 + predicate($1) } }
}
Usage:
struct Product {
let id: String
let price: Decimal
}
let products: [Product] = [.init(id: "abc", price: 21.9),
.init(id: "xyz", price: 19.7),
.init(id: "jkl", price: 2.9)
]
products.sum(\.price) // 44.5
this is my approach about this. however I believe that the best solution is the answer from the user username tbd
var i = 0
var sum = 0
let example = 0
for elements in multiples{
i = i + 1
sum = multiples[ (i- 1)]
example = sum + example
}
Swift 3
From all the options displayed here, this is the one that worked for me.
let arr = [6,1,2,3,4,10,11]
var sumedArr = arr.reduce(0, { ($0 + $1)})
print(sumedArr)
How about the simple way of
for (var i = 0; i < n; i++) {
sum = sum + Int(multiples[i])!
}
//where n = number of elements in the array
A possible solution: define a prefix operator for it. Like the reduce "+/" operator as in APL (e.g. GNU APL)
A bit of a different approach here.
Using a protocol en generic type allows us to to use this operator on Double, Float and Int array types
protocol Number
{
func +(l: Self, r: Self) -> Self
func -(l: Self, r: Self) -> Self
func >(l: Self, r: Self) -> Bool
func <(l: Self, r: Self) -> Bool
}
extension Double : Number {}
extension Float : Number {}
extension Int : Number {}
infix operator += {}
func += <T:Number> (inout left: T, right: T)
{
left = left + right
}
prefix operator +/ {}
prefix func +/ <T:Number>(ar:[T]?) -> T?
{
switch true
{
case ar == nil:
return nil
case ar!.isEmpty:
return nil
default:
var result = ar![0]
for n in 1..<ar!.count
{
result += ar![n]
}
return result
}
}
use like so:
let nmbrs = [ 12.4, 35.6, 456.65, 43.45 ]
let intarr = [1, 34, 23, 54, 56, -67, 0, 44]
+/nmbrs // 548.1
+/intarr // 145
(updated for Swift 2.2, tested in Xcode Version 7.3)
Keep it simple...
var array = [1, 2, 3, 4, 5, 6, 7, 9, 0]
var n = 0
for i in array {
n += i
}
print("My sum of elements is: \(n)")
Output:
My sum of elements is: 37
Source: Stackoverflow.com