N维数组快速 [英] N-Dimensional array swift
问题描述
有什么办法可以快速建立n维数组?我希望能够创建一个可以创建具有n个维度的数组的函数，但是我不知道该怎么做.
Is there any way to have an n dimensional array in swift? I would like to be able to make a function that creates an array with n dimensions but I cannot figure out how.
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
}
以上代码出于种种原因无法正常运行，但是，我认为它指出了我遇到的主要问题:
The above code does not work for obvios reasons but, I think it points out the main problems I am having:
- 如何定义返回类型
- 如何实际创建数组
- 创建后如何遍历并填充数组
推荐答案
这里是N维数组的实现.它在内部使用普通数组进行存储，并将多维索引转换为内部数组的单个索引.
Here is the implementation of an N-Dimensional Array. It uses a normal array internally for storage and converts the multi-dimensional indices into a single index for the internal array.
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)]
["(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在注释中指出的那样，在初始化带有引用类型值的数组时必须要小心，因为该数组将填充对对象的引用，并且在任何索引处修改该对象都会修改所有对象.其中
As @DuncanC noted in the comments, you have to be careful when initializing an array with a value which is a reference type, because the array will be filled with references to the object and modifying the object at any index will modify all of them.
为解决此问题，我添加了第二个初始化程序:
To solve this, I added a second initializer:
init(dimensions: Int..., initUsing initializer: () -> T)
带一个闭包() -> T，可用于为数组的每个元素创建一个新对象.
which takes a closure () -> T which can be used to create a new object for each element of the array.
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)
弗雷德·威尔玛
Fred Wilma
这篇关于N维数组快速的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
