Zhang-Suen细化算法C＃
我正在尝试按照本指南在C＃中编写Zhang-Suen细化算法，而不处理边距。
在功能＆＃39; zhangsuen＆＃39;中，我正在阅读图像＆＃39; imgUndo＆＃39;并写入图像＆＃39; img＆＃39;。 for循环内的指针dataPtrOrigin_aux用于读取3x3窗口内的9个像素，这样dataPtrOrigin_aux5就是该窗口的中心像素，该窗口将沿着整个图像移动，从左到右，从上到下移动。在每个像素中，如果验证语句是否为真，则在指针dataPtrFinal写入的图像中进行相应的更改。
请注意，我将当前像素的邻居存储在8个元素数组中。因此，它们将按照以下顺序存储：
internal static void zhangsuen(Image<Bgr, byte> img, Image<Bgr, byte> imgUndo)
{
unsafe
{
MIplImage m = img.MIplImage; //Image to be written.
MIplImage mUndo = imgUndo.MIplImage; //Image to be read.
byte* dataPtrFinal = (byte*)m.imageData.ToPointer();
byte* dataPtrUndo = (byte*)mUndo.imageData.ToPointer();
int width = img.Width; //Width of the image.
int height = img.Height; //Height of the image.
int nChan = m.nChannels; //3 channels (R, G, B).
int wStep = m.widthStep; //Total width of the image (including padding).
int padding = wStep - nChan * width; //Padding at the end of each line.
int x, y, i;
int[] neighbours = new int[8]; //Store the value of the surrounding neighbours in this array.
int step; //Step 1 or 2.
int[] sequence = { 1, 2, 4, 7, 6, 5, 3, 0, 1 };
int blackn = 0; //Number of black neighbours.
int numtransitions = 0; //Number of transitions from white to black in the sequence specified by the array sequence.
int changed = 1; //Just so it enters the while.
bool isblack = false;
int counter = 0;
while(changed > 0)
{
changed = 0;
if (counter % 2 == 0) //We want to read all the pixels in the image before going to the next step
step = 1;
else
step = 2;
for (y = 0; y < height; y++)
{
for (x = 0; x < width; x++)
{
byte* dataPtrOrigin_aux1 = (byte*)(dataPtrUndo + (y - 1) * m.widthStep + (x - 1) * m.nChannels);
byte* dataPtrOrigin_aux2 = (byte*)(dataPtrUndo + (y - 1) * m.widthStep + (x) * m.nChannels);
byte* dataPtrOrigin_aux3 = (byte*)(dataPtrUndo + (y - 1) * m.widthStep + (x + 1) * m.nChannels);
byte* dataPtrOrigin_aux4 = (byte*)(dataPtrUndo + (y) * m.widthStep + (x - 1) * m.nChannels);
byte* dataPtrOrigin_aux5 = (byte*)(dataPtrUndo + (y) * m.widthStep + (x) * m.nChannels);
byte* dataPtrOrigin_aux6 = (byte*)(dataPtrUndo + (y) * m.widthStep + (x + 1) * m.nChannels);
byte* dataPtrOrigin_aux7 = (byte*)(dataPtrUndo + (y + 1) * m.widthStep + (x - 1) * m.nChannels);
byte* dataPtrOrigin_aux8 = (byte*)(dataPtrUndo + (y + 1) * m.widthStep + (x) * m.nChannels);
byte* dataPtrOrigin_aux9 = (byte*)(dataPtrUndo + (y + 1) * m.widthStep + (x + 1) * m.nChannels);
if (x > 0 && y > 0 && x < width - 1 && y < height - 1)
{
if (dataPtrOrigin_aux5[0] == 0)
isblack = true;
if (isblack)
{
neighbours[0] = dataPtrOrigin_aux1[0];
neighbours[1] = dataPtrOrigin_aux2[0];
neighbours[2] = dataPtrOrigin_aux3[0];
neighbours[3] = dataPtrOrigin_aux4[0];
neighbours[4] = dataPtrOrigin_aux6[0];
neighbours[5] = dataPtrOrigin_aux7[0];
neighbours[6] = dataPtrOrigin_aux8[0];
neighbours[7] = dataPtrOrigin_aux9[0];
for(i = 0; i <= 7; i++)
{
if (neighbours[i] == 0)
blackn++;
if (neighbours[sequence[i]] - neighbours[sequence[i + 1]] == 255) //número de transições de branco para preto, seguindo a ordem do vector sequence
numtransitions++;
}
if ((blackn >= 2 && blackn <= 6) && numtransitions == 1)
{
if (step == 1 && (neighbours[1] == 255 || neighbours[4] == 255 || neighbours[6] == 255) && (neighbours[4] == 255 || neighbours[6] == 255 || neighbours[3] == 255))
{
dataPtrFinal[0] = 255;
dataPtrFinal[1] = 255;
dataPtrFinal[2] = 255;
changed++;
}
if (step == 2 && (neighbours[1] == 255 || neighbours[4] == 255 || neighbours[3] == 255) && (neighbours[1] == 255 || neighbours[6] == 255 || neighbours[3] == 255))
{
dataPtrFinal[0] = 255;
dataPtrFinal[1] = 255;
dataPtrFinal[2] = 255;
changed++;
}
}
}
}
dataPtrFinal += nChan;
isblack = false;
blackn = 0;
numtransitions = 0;
}
dataPtrFinal += padding;
}
dataPtrUndo = (byte*)m.imageData.ToPointer(); //Change the image to be read to the one that has just been written.
counter++;
}
}
}
当我结束阅读第一张图片并将更改写入图片时，我会立即阅读图片。 （一旦（y = 0; y＆lt; height; y ++）的循环结束，我希望我刚写的图像是我将在下一个循环中读取的图像，以便进一步细化。我试图完成这与行
dataPtrUndo = (byte*)m.imageData.ToPointer();
虽然计数器的某个值大于0（取决于读取的图像）但是我收到的错误表明已经尝试写入受保护的内存，这表示我已尝试在图像限制之外写入，但我不明白为什么。它是我错误地做的最后一次归因于dataPtrUndo吗？
3 个答案:
答案 0 :(得分：6)
这是我对Zhang-Suen细化算法的C＃实现
public static bool[][] ZhangSuenThinning(bool[][] s)
{
bool[][] temp = s;
bool even = true;
for (int a = 1; a < s.Length-1; a++)
{
for (int b = 1; b < s[0].Length-1; b++)
{
if (SuenThinningAlg(a, b, temp, even))
{
temp[a][b] = false;
}
even = !even;
}
}
return temp;
}
static bool SuenThinningAlg(int x, int y, bool[][] s, bool even)
{
bool p2 = s[x][y - 1];
bool p3 = s[x + 1][y - 1];
bool p4 = s[x + 1][y];
bool p5 = s[x + 1][y + 1];
bool p6 = s[x][y + 1];
bool p7 = s[x - 1][y + 1];
bool p8 = s[x - 1][y];
bool p9 = s[x - 1][y - 1];
int bp1 = NumberOfNonZeroNeighbors(x, y, s);
if (bp1 >= 2 && bp1 <= 6)//2nd condition
{
if (NumberOfZeroToOneTransitionFromP9(x, y, s) == 1)
{
if (even)
{
if (!((p2 && p4) && p8))
{
if (!((p2 && p6) && p8))
{
return true;
}
}
}
else
{
if (!((p2 && p4) && p6))
{
if (!((p4 && p6) && p8))
{
return true;
}
}
}
}
}
return false;
}
static int NumberOfZeroToOneTransitionFromP9(int x, int y, bool[][]s)
{
bool p2 = s[x][y - 1];
bool p3 = s[x + 1][y - 1];
bool p4 = s[x + 1][y];
bool p5 = s[x + 1][y + 1];
bool p6 = s[x][y + 1];
bool p7 = s[x - 1][y + 1];
bool p8 = s[x - 1][y];
bool p9 = s[x - 1][y - 1];
int A = Convert.ToInt32((p2 == false && p3 == true)) + Convert.ToInt32((p3 == false && p4 == true)) +
Convert.ToInt32((p4 == false && p5 == true)) + Convert.ToInt32((p5 == false && p6 == true)) +
Convert.ToInt32((p6 == false && p7 == true)) + Convert.ToInt32((p7 == false && p8 == true)) +
Convert.ToInt32((p8 == false && p9 == true)) + Convert.ToInt32((p9 == false && p2 == true));
return A;
}
static int NumberOfNonZeroNeighbors(int x, int y, bool[][]s)
{
int count = 0;
if (s[x-1][y])
count++;
if (s[x-1][y+1])
count++;
if (s[x-1][y-1])
count++;
if (s[x][y+1])
count++;
if (s[x][y-1])
count++;
if (s[x+1][y])
count++;
if (s[x+1][y+1])
count++;
if (s[x+1][y-1])
count++;
return count;
}
答案 1 :(得分：6)
bwang22的答案有效。有点。但有两个问题：它不会进行迭代，直到不再发生更改。它做了一个浅层的阵列..两个问题合作可以说，相互抵消，导致变薄，但不是最好看的。
这是更正后的代码，它可以提供更好看的结果：
从Image转换为bool [] []并返回的前两种方法;功能没有针对速度进行优化;如果你需要那个去lockbits / unsafe ..：
public static bool[][] Image2Bool(Image img)
{
Bitmap bmp = new Bitmap(img);
bool[][] s = new bool[bmp.Height][];
for (int y = 0; y < bmp.Height; y++ )
{
s[y] = new bool[bmp.Width];
for (int x = 0; x < bmp.Width; x++)
s[y][x] = bmp.GetPixel(x, y).GetBrightness() < 0.3;
}
return s;
}
public static Image Bool2Image(bool[][] s)
{
Bitmap bmp = new Bitmap(s[0].Length, s.Length);
using (Graphics g = Graphics.FromImage(bmp)) g.Clear(Color.White);
for (int y = 0; y < bmp.Height; y++)
for (int x = 0; x < bmp.Width; x++)
if (s[y][x]) bmp.SetPixel(x, y, Color.Black);
return (Bitmap)bmp;
}
现在已修正的细化代码，其中大部分或多或少与bwang22的答案不变：
public static bool[][] ZhangSuenThinning(bool[][] s)
{
bool[][] temp = ArrayClone(s); // make a deep copy to start..
int count = 0;
do // the missing iteration
{
count = step(1, temp, s);
temp = ArrayClone(s); // ..and on each..
count += step(2, temp, s);
temp = ArrayClone(s); // ..call!
}
while (count > 0);
return s;
}
static int step(int stepNo, bool[][] temp, bool[][] s)
{
int count = 0;
for (int a = 1; a < temp.Length - 1; a++)
{
for (int b = 1; b < temp[0].Length - 1; b++)
{
if (SuenThinningAlg(a, b, temp, stepNo == 2))
{
// still changes happening?
if (s[a][b]) count++;
s[a][b] = false;
}
}
}
return count;
}
static bool SuenThinningAlg(int x, int y, bool[][] s, bool even)
{
bool p2 = s[x][y - 1];
bool p3 = s[x + 1][y - 1];
bool p4 = s[x + 1][y];
bool p5 = s[x + 1][y + 1];
bool p6 = s[x][y + 1];
bool p7 = s[x - 1][y + 1];
bool p8 = s[x - 1][y];
bool p9 = s[x - 1][y - 1];
int bp1 = NumberOfNonZeroNeighbors(x, y, s);
if (bp1 >= 2 && bp1 <= 6) //2nd condition
{
if (NumberOfZeroToOneTransitionFromP9(x, y, s) == 1)
{
if (even)
{
if (!((p2 && p4) && p8))
{
if (!((p2 && p6) && p8))
{
return true;
}
}
}
else
{
if (!((p2 && p4) && p6))
{
if (!((p4 && p6) && p8))
{
return true;
}
}
}
}
}
return false;
}
static int NumberOfZeroToOneTransitionFromP9(int x, int y, bool[][] s)
{
bool p2 = s[x][y - 1];
bool p3 = s[x + 1][y - 1];
bool p4 = s[x + 1][y];
bool p5 = s[x + 1][y + 1];
bool p6 = s[x][y + 1];
bool p7 = s[x - 1][y + 1];
bool p8 = s[x - 1][y];
bool p9 = s[x - 1][y - 1];
int A = Convert.ToInt32((!p2 && p3 )) + Convert.ToInt32((!p3 && p4 )) +
Convert.ToInt32((!p4 && p5 )) + Convert.ToInt32((!p5 && p6 )) +
Convert.ToInt32((!p6 && p7 )) + Convert.ToInt32((!p7 && p8 )) +
Convert.ToInt32((!p8 && p9 )) + Convert.ToInt32((!p9 && p2 ));
return A;
}
static int NumberOfNonZeroNeighbors(int x, int y, bool[][] s)
{
int count = 0;
if (s[x - 1][y]) count++;
if (s[x - 1][y + 1]) count++;
if (s[x - 1][y - 1]) count++;
if (s[x][y + 1]) count++;
if (s[x][y - 1]) count++;
if (s[x + 1][y]) count++;
if (s[x + 1][y + 1]) count++;
if (s[x + 1][y - 1]) count++;
return count;
}
我保留了原始的偶数标记，但是通过比较步骤编号来调用它。我直接使用bools保存了几个字符..
最后一个函数来获取嵌套的2d数组的深层副本：
public static T[][] ArrayClone<T>(T [][] A)
{ return A.Select(a => a.ToArray()).ToArray(); }
这是使用两个PictureBoxes来调用它的方法：
pictureBox1.Image = Image.FromFile("D:\\RCdemo.png");
bool[][] t = Image2Bool(pictureBox1.Image);
t = ZhangSuenThinning(t);
pictureBox2.Image = Bool2Image(t);
我附上一张测试图片。
答案 2 :(得分：0)
bwang22的答案很慢。尝试以下方法：
public readonly struct ConnectivityData
{
public readonly int[] N;
public readonly int NumNeighbors;
public readonly int NumChanges;
public ConnectivityData(in int[] n, in int numNeighbors, in int numChanges)
{
N = n;
NumNeighbors = numNeighbors;
NumChanges = numChanges;
}
}
public static void ZhangSuen(in HashSet<Pixel> pixels)
{
while (true)
{
// Pass #1:
List<Pixel> mark1 = new List<Pixel>();
foreach (Pixel p in pixels)
{
ConnectivityData conn = ComputeConnectivity(p, pixels);
if (conn.NumNeighbors > 1 &&
conn.NumNeighbors < 7 &&
conn.NumChanges == 1 &&
conn.N[0] * conn.N[2] * conn.N[4] == 0 &&
conn.N[2] * conn.N[4] * conn.N[6] == 0)
{
mark1.Add(p);
}
}
//delete all marked:
foreach (Pixel p in mark1)
{
pixels.Remove(p);
}
// PASS #2:
List<Pixel> mark2 = new List<Pixel>();
foreach (Pixel p in pixels)
{
ConnectivityData conn = ComputeConnectivity(p, pixels);
if (conn.NumNeighbors > 1 &&
conn.NumNeighbors < 7 &&
conn.NumChanges == 1 &&
conn.N[0] * conn.N[2] * conn.N[6] == 0 &&
conn.N[0] * conn.N[4] * conn.N[6] == 0)
{
mark2.Add(p);
}
}
//delete all marked:
foreach (Pixel p in mark2)
{
pixels.Remove(p);
}
if (mark1.Count == 0 && mark2.Count == 0)
{
break;
}
}
}
private static ConnectivityData ComputeConnectivity(
in Pixel p,
in HashSet<Pixel> pixels)
{
// calculate #neighbors and number of changes:
int[] n = new int[8];
if (pixels.Contains(new Pixel(p.X, p.Y - 1)))
{
n[0] = 1;
}
if (pixels.Contains(new Pixel(p.X + 1, p.Y - 1)))
{
n[1] = 1;
}
if (pixels.Contains(new Pixel(p.X + 1, p.Y)))
{
n[2] = 1;
}
if (pixels.Contains(new Pixel(p.X + 1, p.Y + 1)))
{
n[3] = 1;
}
if (pixels.Contains(new Pixel(p.X, p.Y + 1)))
{
n[4] = 1;
}
if (pixels.Contains(new Pixel(p.X - 1, p.Y + 1)))
{
n[5] = 1;
}
if (pixels.Contains(new Pixel(p.X - 1, p.Y)))
{
n[6] = 1;
}
if (pixels.Contains(new Pixel(p.X - 1, p.Y - 1)))
{
n[7] = 1;
}
return new ConnectivityData(
n,
n[0] + n[1] + n[2] + n[3] + n[4] + n[5] + n[6] + n[7],
ComputeNumberOfChanges(n));
}
private static int ComputeNumberOfChanges(in int[] n)
{
int numberOfChanges = 0;
// Iterate over each location and see if it is has changed from 0 to 1:
int current = n[0];
for (int i = 1; i < 8; i++)
{
if (n[i] == 1 && current == 0)
{
numberOfChanges++;
}
current = n[i];
}
// Also consider the change over the discontinuity between n[7] and n[0]:
if (n[0] == 1 && n[7] == 0)
{
numberOfChanges++;
}
return numberOfChanges;
}
要使用：
在您的位图等中，创建一个类型为Pixel的哈希集（包含您要稀疏的所有黑色像素），例如：
public class Pixel
{
public int X;
public int Y;
public Pixel(in int x, in int y)
{
X = x;
Y = y;
}
public override bool Equals(object pixel)
{
Pixel b = pixel as Pixel;
return X == b.X && Y == b.Y;
}
public override int GetHashCode()
{
//return (a.X << 2) ^ a.Y; // this is also commonly used as a pixel hash code
return X * 100000 + Y; // a bit hacky [will fail if bitmap width is > 100000]
}
}
...然后调用ZhangSuen（pixels）。这将从集合中删除适当的像素。
请注意，此方法不适用于所有图像。它会使部分图像消失。具体来说，我遇到的问题是右下角的对角线的粗细约为11像素宽。
我目前正在研究一种改进方法，但是在我测试过的大多数文件（CAD文件）中，它的性能都优于类似的斯坦尼福德算法。
相关问题
最新问题
- 我写了这段代码，但我无法理解我的错误
- 我无法从一个代码实例的列表中删除 None 值，但我可以在另一个实例中。为什么它适用于一个细分市场而不适用于另一个细分市场？
- 是否有可能使 loadstring 不可能等于打印？卢阿
- java中的random.expovariate()
- Appscript 通过会议在 Google 日历中发送电子邮件和创建活动
- 为什么我的 Onclick 箭头功能在 React 中不起作用？
- 在此代码中是否有使用“this”的替代方法？
- 在 SQL Server 和 PostgreSQL 上查询，我如何从第一个表获得第二个表的可视化
- 每千个数字得到
- 更新了城市边界 KML 文件的来源？