如何将数组中的每个Int值乘以Swift中的常量?

时间:2022-02-05 13:42:41

Say, I have an array [20, 2, 3]
How can I multiply each Int value of this array in Swift?
So 2 x array becomes [40, 4, 6], 3 x array becomes [60, 6, 9] and so on?

说,我有一个数组[20,2,3]如何在Swift中乘以这个数组的每个Int值?那么2 x数组变为[40,4,6],3 x数组变为[60,6,9]等等?

3 个解决方案

#1


11  

You can use .map():

你可以使用.map():

let values = [20, 2, 3]
let doubles = values.map { $0 * 2 }
let triples = values.map { $0 * 3 }

If you want to do the update in-place:

如果要进行就地更新:

var values = [20, 2, 3]

values.enumerate().forEach { index, value in
  values[index] = value * 2
}
// values is now [40, 4, 6]

#2


4  

map is definitely a good way to go – although if you want to get the syntax that you asked for in your question "2 x array becomes ...", you could overload the * operator.

map肯定是一个很好的方法 - 虽然如果你想在你的问题“2 x array变成......”中得到你要求的语法,你可以重载*运算符。

This will allow you to multiply an array by a scalar inline, if the array's elements are multipliable and of the same type as the scalar.

如果数组的元素是可乘的且与标量相同的类型,这将允许您将数组乘以标量内联。

protocol Multipliable {
    func *(lhs:Self, rhs:Self) -> Self
}

// add any additional types that you want to be multipliable here
extension Int:Multipliable {}
extension Double:Multipliable {}
extension Float:Multipliable {}

// map through the array elements, multiplying each by the rhs argument
func *<T:Multipliable>(lhs:[T], rhs:T) -> [T] {
    return lhs.map {$0 * rhs}
}

// use an inout argument in order to multiply an array in-place
func *=<T:Multipliable>(inout lhs:[T], rhs:T) {
    lhs = lhs*rhs
}

You can then use it like this:

然后你可以像这样使用它:

let yourArray = [2, 3, 4]

var arr = yourArray * 4 // arr = [8, 12, 16]

arr *= 3 // arr = [24, 36, 48]

#3


1  

You can use a for loop.

你可以使用for循环。

    var array = [20, 2, 3]
    let multiplier = 2

    for i in 0...array.count - 1{
        array[i] = array[i] * multiplier
    }

#1


11  

You can use .map():

你可以使用.map():

let values = [20, 2, 3]
let doubles = values.map { $0 * 2 }
let triples = values.map { $0 * 3 }

If you want to do the update in-place:

如果要进行就地更新:

var values = [20, 2, 3]

values.enumerate().forEach { index, value in
  values[index] = value * 2
}
// values is now [40, 4, 6]

#2


4  

map is definitely a good way to go – although if you want to get the syntax that you asked for in your question "2 x array becomes ...", you could overload the * operator.

map肯定是一个很好的方法 - 虽然如果你想在你的问题“2 x array变成......”中得到你要求的语法,你可以重载*运算符。

This will allow you to multiply an array by a scalar inline, if the array's elements are multipliable and of the same type as the scalar.

如果数组的元素是可乘的且与标量相同的类型,这将允许您将数组乘以标量内联。

protocol Multipliable {
    func *(lhs:Self, rhs:Self) -> Self
}

// add any additional types that you want to be multipliable here
extension Int:Multipliable {}
extension Double:Multipliable {}
extension Float:Multipliable {}

// map through the array elements, multiplying each by the rhs argument
func *<T:Multipliable>(lhs:[T], rhs:T) -> [T] {
    return lhs.map {$0 * rhs}
}

// use an inout argument in order to multiply an array in-place
func *=<T:Multipliable>(inout lhs:[T], rhs:T) {
    lhs = lhs*rhs
}

You can then use it like this:

然后你可以像这样使用它:

let yourArray = [2, 3, 4]

var arr = yourArray * 4 // arr = [8, 12, 16]

arr *= 3 // arr = [24, 36, 48]

#3


1  

You can use a for loop.

你可以使用for循环。

    var array = [20, 2, 3]
    let multiplier = 2

    for i in 0...array.count - 1{
        array[i] = array[i] * multiplier
    }