Algorithm 将多维数组转换为一维数组的算法

将二维数组转换为一维数组很容易，但是如何将二维以上的多维数组转换为一维数组呢？例如，假设有int[5][5][5]x和int[125]y，我想将x[3][4][2]处的值放在y的正确位置

希望这是有道理的

m0,m1,.. are dimensions
A(i,j,k,...) -> A0[i + j*m0 + k*m0*m1 + ...]

和有用的C技巧：

double *A;
size_t m;
#define A(i,j) A[(i) + (j)*m];

---

可以使用不同的方法将多维数组映射为线性数组。问题是你必须选择一个惯例。让我们按照下面的惯例来做。第一个索引指定块容器，第二个索引指定前一个容器中的块，最后第三个索引是块内的偏移量。您可以很容易地将其推广到多维，但在本例中，将其保持为3：

#include <cstddef>

std::size_t linear_index
    (std::size_t f,
     std::size_t s, 
     std::size_t t,
     std::size_t f_width,
     std::size_t s_width)
{
    return (f*f_width + s)*s_width + t;
}

#包括
标准：大小与线性指数
（标准：尺寸），
标准：尺寸，
标准：尺寸，
标准：尺寸和宽度，
标准：尺寸（宽度）
{
返回（f*f_宽度+s）*s_宽度+t；
}

这里已经有几个技术上很好的答案，但这里有一个更直观的理解方法

---

好的，你知道如何从一维到二维

一维阵列如下所示：

int [5] :

+-----+-----+-----+-----+-----+
|  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |
+-----+-----+-----+-----+-----+

int [5][5] :

+-----+-----+-----+-----+-----+     
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | 
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     

+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 | 1,0 | 1,1 | 1,2 | 1,3 | 1,4 | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
                             vvv
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
|  0  |  1  |  2  |  3  |  4  |  5  |  6  |  7  |  8  |  9  | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -

int [5][5] :

+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     |  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     |  5  |  6  |  7  |  8  |  9  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | =>  | 10  | 11  | 12  | 13  | 14  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     | 15  | 16  | 17  | 18  | 19  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     | 20  | 21  | 22  | 23  | 24  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+

2-D array index [i][j]          =>  1-D array index [i*5 + j]

二维阵列如下所示：

int [5] :

+-----+-----+-----+-----+-----+
|  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |
+-----+-----+-----+-----+-----+

int [5][5] :

+-----+-----+-----+-----+-----+     
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | 
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     

+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 | 1,0 | 1,1 | 1,2 | 1,3 | 1,4 | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
                             vvv
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
|  0  |  1  |  2  |  3  |  4  |  5  |  6  |  7  |  8  |  9  | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -

int [5][5] :

+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     |  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     |  5  |  6  |  7  |  8  |  9  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | =>  | 10  | 11  | 12  | 13  | 14  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     | 15  | 16  | 17  | 18  | 19  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     | 20  | 21  | 22  | 23  | 24  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+

2-D array index [i][j]          =>  1-D array index [i*5 + j]

您可以将转换为相应的1-D阵列，如下所示：

int [5] :

+-----+-----+-----+-----+-----+
|  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |
+-----+-----+-----+-----+-----+

int [5][5] :

+-----+-----+-----+-----+-----+     
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | 
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     

+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 | 1,0 | 1,1 | 1,2 | 1,3 | 1,4 | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
                             vvv
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
|  0  |  1  |  2  |  3  |  4  |  5  |  6  |  7  |  8  |  9  | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -

int [5][5] :

+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     |  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     |  5  |  6  |  7  |  8  |  9  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | =>  | 10  | 11  | 12  | 13  | 14  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     | 15  | 16  | 17  | 18  | 19  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     | 20  | 21  | 22  | 23  | 24  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+

2-D array index [i][j]          =>  1-D array index [i*5 + j]

但另一种思考方法是描绘原始阵列，但重新标记-如下所示：

int [5] :

+-----+-----+-----+-----+-----+
|  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |
+-----+-----+-----+-----+-----+

int [5][5] :

+-----+-----+-----+-----+-----+     
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | 
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     
|     |     |     |     |     |     
+-----+-----+-----+-----+-----+     

+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 | 1,0 | 1,1 | 1,2 | 1,3 | 1,4 | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
                             vvv
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -
|  0  |  1  |  2  |  3  |  4  |  5  |  6  |  7  |  8  |  9  | etc.
|     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+- - -

int [5][5] :

+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 0,0 | 0,1 | 0,2 | 0,3 | 0,4 |     |  0  |  1  |  2  |  3  |  4  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 1,0 | 1,1 | 1,2 | 1,3 | 1,4 |     |  5  |  6  |  7  |  8  |  9  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 2,0 | 2,1 | 2,2 | 2,3 | 2,4 | =>  | 10  | 11  | 12  | 13  | 14  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 3,0 | 3,1 | 3,2 | 3,3 | 3,4 |     | 15  | 16  | 17  | 18  | 19  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+
| 4,0 | 4,1 | 4,2 | 4,3 | 4,4 |     | 20  | 21  | 22  | 23  | 24  |
|     |     |     |     |     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+     +-----+-----+-----+-----+-----+

2-D array index [i][j]          =>  1-D array index [i*5 + j]

…如果你这样想的话，三维的情况也遵循同样的原则（对于更高的维度，也是如此——它变得越来越难以可视化！）：

---

您可以在C#中执行以下操作

公共类数组索引器
{
私有只读int[]\u边界；
私有只读int[]\u和；
公共ArrayIndex（参数int[]界限）
{
_界限=界限；
//预先计算速度的界限和。
_总和=新整数[bounds.Length-1]；
对于（int i=1，sum=_-bounds[i-1]；i<_-bounds.Length；++i，sum*=_-bounds[i-1]）
_总和[i-1]=总和；
}
公共T索引（T[]数组，参数int[]索引）
{
if（index.Length！=\u bounds.Length）
抛出新的ArgumentException（“应该有和边界一样多的索引”，“索引”）；
var指数=指数[0]；
对于（inti=1，sum=_界[i-1]；i

或者转换成n维数组

public static class ArrayExtensions
{
    public static Array CreateArray<T>(this T[] array1d, params int[] bounds)
    {
        var arrayNd = Array.CreateInstance(typeof(T), bounds);

        var indices = new int[bounds.Length];
        for (var i = 0; i < array1d.Length; ++i)
        {
            arrayNd.SetValue(array1d[i], indices);

            for (int j = 0; j < bounds.Length; ++j)
            {
                if (++indices[j] < bounds[j])
                    break;
                indices[j] = 0;
            }
        }

        return arrayNd;
    }
}

公共静态类ArrayExtensions
{
公共静态数组CreateArray（此T[]array1d，参数int[]bounds）
{
var arrayNd=Array.CreateInstance（typeof（T），bounds）；
var指数=新整数[bounds.Length]；
对于（变量i=0；i

并进行测试

int[] array3d =
    new[]
    {
        0, 1, 2, 3,
        4, 5, 6, 7,
        8, 9, 10, 11,

        12, 13, 14, 15,
        16, 17, 18, 19,
        20, 21, 22, 23
    };

var copied3d = (int[, ,])array3d.CreateArray(4, 3, 2);
var indexer3d = new ArrayIndexer(4, 3, 2);

for (int i = 0; i < 4; ++i)
{
    for (int j = 0; j < 3; ++j)
    {
        for (int k = 0; k < 2; ++k)
        {
            var x = indexer3d.FastIndex(array3d, i, j, k);
            var y = copied3d[i, j, k];
            Debug.Print("Array[{0},{1},{2}] = {3} and {4} match = {5}", i, j, k, x, y, x == y);
        }
    }
}

int[]数组3d=
新[]
{
0, 1, 2, 3,
4, 5, 6, 7,
8, 9, 10, 11,
12, 13, 14, 15,
16, 17, 18, 19,
20, 21, 22, 23
};
var copied3d=（int[，]）array3d.CreateArray（4,3,2）；
var indexer3d=新的ArrayIndexer（4,3,2）；
对于（int i=0；i<4；++i）
{
对于（int j=0；j<3；++j）
{
对于（int k=0；k<2；++k）
{
var x=索引器3d.FastIndex（数组3d，i，j，k）；
var y=复制的3D[i，j，k]；
Print（“数组[{0}，{1}，{2}]={3}和{4}匹配={5}”，i，j，k，x，y，x==y）；
}
}
}

实际上有一种非常酷的方式来思考这个问题，现在还没有人在这里发布过

在最简单的情况下，您可以将X、Y、Z坐标想象为您创建的虚数系统中的数字。这些数字写为XYZ，因此您的示例[3][4][2]将写为：342

我们中那些习惯于八进制和十六进制的人习惯于认为这并不意味着三百、四个十和两个一，而是相反

三个64、四个8和两个1

或

三个256，四个16和两个1

这确实是虚数系统需要做的，但每个数字都是数组中相应边的长度的基数，乘以下一个较低的基数（除非没有，否则在这种情况下，1.最后一个数组长度不用于此计算，而是用于限制循环。此计算中的排序取决于您希望如何将边长转换为线性元素

对于5x5x5阵列，这很简单：

25s | 5s | 1s * 3 | 4 | 2 ----+----+--- 75 + 20 + 2 = 97 25s | 5s | 1s * 3 | 4 | 2 ----+----+--- 75 + 20 + 2 = 97 其他基础可能更复杂，尤其是尺寸不均匀的基础，但这只是思考问题的另一种方式

下面是一个非统一的565示例：

30s | 5s | 1s * 3 | 4 | 2 ----+----+--- 90 + 20 + 2 = 102 30秒| 5秒| 1秒 * 3 | 4 | 2 ----+----+--- 90 + 20 + 2 = 102

---

谢谢！很好的说明方式！有关int[dimX][dimY][dimZ]：一维数组索引[i*dimY*dimZ+j*dimZ+k]的更多详细信息此示例启发我理解多项式方程和数组（向量）之间的关系。