在Swift数组中查找元素的和

What is the easiest (best) way to find the sum of an array of integers in swift? I have an array called multiples and I would like to know the sum of the multiples.

用swift查找整数数组和最简单(最好)的方法是什么?我有一个被称为倍数的数组，我想知道倍数的和。

15 个解决方案

#1

This is the easiest/shortest method I can find.

这是我能找到的最简单/最短的方法。

Swift 3:

斯威夫特3:

let multiples = [...]
sum = multiples.reduce(0, +)

Swift 2:

斯威夫特2:

let multiples = [...]
sum = multiples.reduce(0, combine: +)

---

Some more info:

更多信息:

This uses Array's reduce method (documentation here), which allows you to "reduce a collection of elements down to a single value by recursively applying the provided closure". We give it 0 as the initial value, and then, essentially, the closure { $0 + $1 }. Of course, we can simplify that to a single plus sign, because that's how Swift rolls.

这使用了Array的reduce方法(这里的文档)，它允许您“通过递归地应用提供的闭包，将元素集合减少到单个值”。我们给它0作为初始值，然后，本质上，闭包{$0 + $1}。当然，我们可以把它简化成一个加号，因为这就是快速滚动的方式。

#2

Swift 3 / Swift 4 one liner to sum properties of objects

Swift 3 / Swift 4一个行，用于对对象属性进行求和

var totalSum = scaleData.map({$0.points}).reduce(0, +)

Where points is the property in my custom object scaleData that I am trying to reduce

我要减少的自定义对象scaleData中的属性在哪里

#3

Swift3 has changed to :

Swift3已改为:

let multiples = [...]
sum = multiples.reduce(0, +)

#4

This also works:

这同样适用:

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

结果将是:55

#5

How about the simple way of

那么简单的方法呢?

for (var i = 0; i < n; i++) {
 sum = sum + Int(multiples[i])!
}

//where n = number of elements in the array

//其中n =数组中的元素个数

#6

Swift 3

斯威夫特3

If you have an array of generic objects and you want to sum some object property then:

如果你有一个泛型对象数组，你想要求和一些对象属性，那么:

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:

尽管形式较短，但很多时候你可能更喜欢经典的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
}

#7

Swift 4 example

斯威夫特4例

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

#8

A possible solution: define a prefix operator for it. Like the reduce "+/" operator as in APL (e.g. GNU APL)

一种可能的解决方案:为它定义一个前缀运算符。类似于APL中的reduce“+/”运算符(例如GNU APL)

A bit of a different approach here.

这里有一点不同的方法。

Using a protocol en generic type allows us to to use this operator on Double, Float and Int array types

使用通用类型的协议允许我们在双、浮点和Int数组类型上使用该操作符

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:

使用像这样:

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)

(更新为Swift 2.2，在Xcode 7.3版本中测试)

#9

Swift 3.0

斯威夫特3.0

i had the same problem, i found on the documentation Apple this solution.

我也有同样的问题，我在文档中找到了这个解决方案。

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:

但是，在我的例子中，我有一个对象列表，然后我需要转换我列表的值:

let numberSum = self.list.map({$0.number_here}).reduce(0, { x, y in x + y })

this work for me.

这项工作对我来说。

#10

@IBOutlet var valueSource: [MultipleIntBoundSource]!

private var allFieldsCount: Int {
    var sum = 0
    valueSource.forEach { sum += $0.count }
    return sum
}

used this one for nested parameters

将此参数用于嵌套参数

#11

In Swift 4 You can also extend the numeric type to return the sum of all elements in the collection as follow

在Swift 4中，还可以扩展数字类型，以返回集合中所有元素的总和，如下所示

extension Collection where Element: Numeric {
    /// Returns the sum of all elements in the collection
    func sum() -> Element { return reduce(0, +) }
}

---

let numbers = [1,2,3]
numbers.sum()    // 6

let doubles = [1.5, 2.7, 3.0]
doubles.sum()    // 7.2

#12

For sum of elements in array of Objects

用于对象数组中元素的和

self.rankDataModelArray.flatMap{$0.count}.reduce(0, +)

#13

Swift 3

斯威夫特3

From all the options displayed here, this is the one that worked for me.

从这里显示的所有选项中，这是对我有用的。

let arr = [6,1,2,3,4,10,11]


var sumedArr = arr.reduce(0, { ($0 + $1)})
print(sumedArr)

#14

this is my approach about this. however I believe that the best solution is the answer from the user username tbd

这是我的方法。然而，我认为最好的解决方案是用户名tbd的答案。

var i = 0 
var sum = 0
let example = 0
for elements in multiples{
    i = i + 1
    sum = multiples[ (i- 1)]
    example = sum + example
}

#15

Keep it simple...

保持简单…

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:

输出:

My sum of elements is: 37

我的元素和是:37。