java - 如何将2D阵列以小于90°的角度旋转到最佳近似值？

假设我有一个以0°旋转存储的数组:

而且我希望如果我通过，例如以30°作为参数，则返回的近似值非常好:

45°将是

I am aware of the solutions posted for 90° rotations.但是我认为这不会对我有帮助吗？

我没有任何示例代码，因为我目前甚至都不知道从哪里开始寻找。如果有任何我可以用谷歌搜索的关键词，可以指出我可以适应的某些公式的方向，那也很好。

Spectre在C#中的代码解决方案:

    class Rotation
{

    public Rotation() {
        A = new int[xs,ys]{
{0,0,0,9,0,0,0},
{0,0,0,9,0,0,0},
{0,0,0,9,0,0,0},
{9,9,9,9,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
};
        B = new int[xs, ys];

        deg = (float)(Math.PI / 180.0);
    }

    public const int xs = 7; // matrix size
    public const int ys = 7;
    const int x0 = 3; // rotation center cell
    const int y0 = 3;
    readonly float deg;
    public int[,] A;
    public int[,] B;

    //---------------------------------------------------------------------------

    public void rotcv(float ang) {
        rotcw(Rotation.x0, Rotation.y0, ang);
    }
    private void rotcw(int x0, int y0, float ang) // rotate A -> B by angle ang around (x0,y0) CW if ang>0
    {
        int x, y, ix0, iy0, ix1, iy1, q;
        double xx, yy, fx, fy, c, s;
        // circle kernel
        c = Math.Cos(-ang); s = Math.Sin(-ang);
        // rotate
        for (y = 0; y < ys; y++)
            for (x = 0; x < xs; x++)
            {
                // offset so (0,0) is center of rotation
                xx = x - x0;
                yy = y - y0;
                // rotate (fx,fy) by ang
                fx = ((xx * c) - (yy * s));
                fy = ((xx * s) + (yy * c));
                // offset back and convert to ints and weights
                fx += x0; ix0 = (int)Math.Floor(fx); fx -= ix0; ix1 = ix0 + 1; if (ix1 >= xs) ix1 = ix0;
                fy += y0; iy0 = (int)Math.Floor (fy); fy -= iy0; iy1 = iy0 + 1; if (iy1 >= ys) iy1 = iy0;
                // bilinear interpolation A[fx][fy] -> B[x][y]
                if ((ix0 >= 0) && (ix0 < xs) && (iy0 >= 0) && (iy0 < ys))
                {
                    xx = (A[ix0,iy0]) + ((A[ix1,iy0] - A[ix0,iy0]) * fx);
                    yy = (A[ix0,iy0]) + ((A[ix1,iy0] - A[ix0,iy0]) * fx);
                    xx = xx + ((yy - xx) * fy); q =(int) xx;
                }
                else q = 0;
                B[x,y] = q;
            }
    }
}

测试:

 static void Main(string[] args)
    {
        Rotation rot = new Rotation();

        for (int x = 0; x < Rotation.xs; x++) {
            for (int y = 0; y < Rotation.xs; y++) {
                Console.Write(rot.A[x,y] + " ");
            }
            Console.WriteLine();
        }
        Console.WriteLine();
        float rotAngle = 0;
        while (true)
        {
            rotAngle += (float)(Math.PI/180f)*90;
            rot.rotcv(rotAngle);
            for (int x = 0; x < Rotation.xs; x++)
            {
                for (int y = 0; y < Rotation.xs; y++)
                {
                    Console.Write(rot.B[x, y] + " ");
                }
                Console.WriteLine();
            }
            Console.WriteLine();
            Console.ReadLine();
        }

    }

最佳答案

好的，这是 promise 的。第一个 C++ 代码:

//---------------------------------------------------------------------------
#include <math.h>
//---------------------------------------------------------------------------
const int xs=7; // matrix size
const int ys=7;
const int x0=3; // rotation center cell
const int y0=3;
const float deg=M_PI/180.0;
int A[xs][ys]=
    {
    {0,0,0,9,0,0,0},
    {0,0,0,9,0,0,0},
    {0,0,0,9,0,0,0},
    {9,9,9,9,0,0,0},
    {0,0,0,0,0,0,0},
    {0,0,0,0,0,0,0},
    {0,0,0,0,0,0,0},
    };
int B[xs][ys];
//---------------------------------------------------------------------------
void rotcw(int x0,int y0,float ang) // rotate A -> B by angle ang around (x0,y0) CW if ang>0
    {
    int x,y,ix0,iy0,ix1,iy1,q;
    float xx,yy,fx,fy,c,s;
    // circle kernel
    c=cos(-ang); s=sin(-ang);
    // rotate
    for (y=0;y<ys;y++)
     for (x=0;x<xs;x++)
        {
        // offset so (0,0) is center of rotation
        xx=x-x0;
        yy=y-y0;
        // rotate (fx,fy) by ang
        fx=float((xx*c)-(yy*s));
        fy=float((xx*s)+(yy*c));
        // offset back and convert to ints and weights
        fx+=x0; ix0=floor(fx); fx-=ix0; ix1=ix0+1; if (ix1>=xs) ix1=ix0;
        fy+=y0; iy0=floor(fy); fy-=iy0; iy1=iy0+1; if (iy1>=ys) iy1=iy0;
        // bilinear interpolation A[ix0+fx][iy0+fy] -> B[x][y]
        if ((ix0>=0)&&(ix0<xs)&&(iy0>=0)&&(iy0<ys))
            {
            xx=float(A[ix0][iy0])+(float(A[ix1][iy0]-A[ix0][iy0])*fx);
            yy=float(A[ix0][iy0])+(float(A[ix1][iy0]-A[ix0][iy0])*fx);
            xx=xx+((yy-xx)*fy); q=xx;
            } else q=0;
        B[x][y]=q;
        }
    }
//---------------------------------------------------------------------------

这里7x7预览15度步长:

可能需要通过一半的细胞或其他东西来稍微调整中心(中心流得太多了，我很喜欢)

矩阵A是源，而B是目标...

您还可以添加tresholding ...，例如:

if (q>=5) q=9; else q=0;