如何在c#中重塑数组

Question

我在c#中有一个3D字节数组,我从位图中读取了这个数组:

byte[w, h, 3]

将此数组重塑为2D(线性)形式的最简单,性能更友好的方法是什么？

byte[w*h, 3]

换句话说,我想保持通道数(特征)但是呈线性形状(而不是方形)

让我试着说明输入和期望的输出:

输入:

|(r1,g1,b1)    (r2,g2,b2)    (r3,g3,b3)|
|(r4,g4,b4)    (r5,g5,b5)    (r6,g6,b6)|
|(r7,g7,b7)    (r8,g8,b8)    (r9,g9,b9)|

注意arr [0,0,0] = r1,arr [0,0,1] = g1,arr [0,0,2] = b1等.

并输出:

|(r1,g1,b1)    (r2,g2,b2)    (r3,g3,b3)    (r4,g4,b4)    (r5,g5,b5)    (r6,g6,b6) ...|

Answer 1

这似乎工作正常,因为数组在内存中已经是正确的形状:

var a = new byte[2,  2, 2] { { { 1, 2 }, { 3, 4 } }, { { 5, 6 }, { 7, 8 } } };
var b = new byte[2 * 2, 2];

//sizeof(byte) is obviously 1 here, but I put it there for documentation
Buffer.BlockCopy(a, 0, b, 0, a.Length * sizeof(byte));

对于那些有兴趣:至于做什么,如果你真的想转一个二维数组到1D:

byte[,] a = {
    {1, 2},
    {3, 4},
    {5, 6},
};
var b = new byte[a.GetLength(1) * a.GetLength(0)]; //Transpose

const int R_STRIDE1 = 8; //Tune this for your CPU
const int C_STRIDE1 = 8; //Tune this for your CPU

//You should hoist the calls to GetLength() out of the loop unlike what I do here
for (int r1 = 0; r1 < a.GetLength(0); r1 += R_STRIDE1)
for (int c1 = 0; c1 < a.GetLength(1); c1 += C_STRIDE1)
    for (int r2 = 0; r2 < R_STRIDE1; r2++)
    for (int c2 = 0; c2 < C_STRIDE1; c2++)
    {
        var r = r1 + r2;
        var c = c1 + c2;
        if (r < a.GetLength(0) && c < a.GetLength(1))
            b[c * a.GetLength(0) + r] = a[r, c];
    }

这应该利用CPU中的缓存.我只对此进行了有限的测试 - 它仍然可能很慢.如果是,请尝试调整它.
你可以(有点非平凡地)将它扩展到3D数组.