arrays - N维数组swift

Question

有什么方法可以在 swift 中创建一个 n 维数组吗？我希望能够创建一个创建 n 维数组的函数，但我不知道该怎么做。

基本上是这样的:

func ndarray <T> (dimensions: Int...) -> [[T]] { // What do I tell it I return?
    var out
    for d in dimensions {
        out = Array<T>(repeating: out, count: d)
    }
    return out
} 

上面的代码由于明显的原因不起作用，但我认为它指出了我遇到的主要问题:

• 如何定义返回类型
• 我如何实际创建数组
• 创建后如何遍历和填充数组

Answer 1

这是 N 维数组的实现。其内部使用普通数组存储，并将多维索引转换为内部数组的单个索引。

struct NDimArray<T> {
    let dimensions: [Int]
    var data: [T]

    init(dimensions: Int..., initialValue: T) {
        self.dimensions = dimensions
        data = Array(repeating: initialValue, count: dimensions.reduce(1, *))
    }

    init(dimensions: Int..., initUsing initializer: () -> T) {
        self.dimensions = dimensions
        data = (0 ..< dimensions.reduce(1, *)).map { _ in initializer() }
    }

    // Compute index into data from indices
    private func computeIndex(_ indices: [Int]) -> Int {
        guard indices.count == dimensions.count else { fatalError("Wrong number of indices: got \(indices.count), expected \(dimensions.count)") }
        zip(dimensions, indices).forEach { dim, idx in
            guard (0 ..< dim) ~= idx else { fatalError("Index out of range") }
        }

        var idx = indices
        var dims = dimensions
        var product = 1
        var total = idx.removeLast()
        while !idx.isEmpty {
            product *= dims.removeLast()
            total += (idx.removeLast() * product)
        }

        return total
    }

    subscript(_ indices: Int...) -> T {
        get {
            return data[computeIndex(indices)]
        }
        set {
            data[computeIndex(indices)] = newValue
        }
    }
}

示例:

// Create a 3 x 4 x 5 array of String with initial value ""

var arr = NDimArray<String>(dimensions: 3, 4, 5, initialValue: "")
for x in 0 ..< 3 {
    for y in 0 ..< 4 {
        for z in 0 ..< 5 {
            // Encode indices in the string
            arr[x, y, z] = "(\(x),\(y),\(z))"
        }
    }
}


// Show internal storage of data
print(arr.data)

["(0,0,0)", "(0,0,1)", "(0,0,2)", "(0,0,3)", "(0,0,4)", "(0,1,0)", "(0,1,1)", "(0,1,2)", "(0,1,3)", "(0,1,4)", "(0,2,0)", "(0,2,1)", "(0,2,2)", "(0,2,3)", "(0,2,4)", "(0,3,0)", "(0,3,1)", "(0,3,2)", "(0,3,3)", "(0,3,4)", "(1,0,0)", "(1,0,1)", "(1,0,2)", "(1,0,3)", "(1,0,4)", "(1,1,0)", "(1,1,1)", "(1,1,2)", "(1,1,3)", "(1,1,4)", "(1,2,0)", "(1,2,1)", "(1,2,2)", "(1,2,3)", "(1,2,4)", "(1,3,0)", "(1,3,1)", "(1,3,2)", "(1,3,3)", "(1,3,4)", "(2,0,0)", "(2,0,1)", "(2,0,2)", "(2,0,3)", "(2,0,4)", "(2,1,0)", "(2,1,1)", "(2,1,2)", "(2,1,3)", "(2,1,4)", "(2,2,0)", "(2,2,1)", "(2,2,2)", "(2,2,3)", "(2,2,4)", "(2,3,0)", "(2,3,1)", "(2,3,2)", "(2,3,3)", "(2,3,4)"]

print(arr[2, 2, 2])  // "(2,2,2)"
print(arr[3, 0, 0])  // Fatal error: Index out of range
print(arr[0, 4, 0])  // Fatal error: Index out of range
print(arr[2])        // Fatal error: Wrong number of indices: got 1, expected 3

---

使用引用类型初始化数组

正如@DuncanC 在评论中指出的那样，在使用引用类型的值初始化数组时必须小心，因为数组将填充对对象的引用，并且在任何索引处修改对象都会修改所有

为了解决这个问题，我添加了第二个初始化程序:

init(dimensions: Int..., initUsing initializer: () -> T)

它有一个闭包 () -> T 可以用来为数组的每个元素创建一个新对象。

例如:

class Person {
    var name = ""
}

// Pass a closure which creates a `Person` instance to fill the array
// with 25 person objects   
let arr = NDimArray(dimensions: 5, 5, initUsing: { Person() })
arr[3, 3].name = "Fred"
arr[2, 2].name = "Wilma"

print(arr[3, 3].name, arr[2, 2].name)

Fred Wilma