使用霍夫线变换的最长线检测
use*_*312 1 c# image-processing hough-transform
输入图像
预期产出
目前的输出
我们可以看到它检测到错误的行.
在下面的代码中,我应该查找错误吗？
虽然有一个问题.如果我将阈值从50增加到150,源代码似乎会产生正确的输出.但是,对我来说,这没有任何意义,因为增加的阈值意味着排除低投票线.
.
源代码
HoughLineTransform.cs
public class Line
{
public Point Start { get; set; }
public Point End { get; set; }
public int Length
{
get
{
return (int)Math.Sqrt(Math.Pow(End.X - Start.X, 2) + Math.Pow(End.Y - Start.Y, 2)); ;
}
}
public Line()
{
}
public Line(Point start, Point end)
{
Start = start;
End = end;
}
}
public class HoughLineTransform
{
public HoughMap Accumulator { get; set; }
public HoughLineTransform() {}
public Line GetLongestLine()
{
List<Line> lines = GetLines(50);
int maxIndex = 0;
double maxLength = -1.0;
for (int i = 0; i < lines.Count; i++)
{
if (maxLength < lines[i].Length)
{
maxIndex = i;
maxLength = lines[i].Length;
}
}
return lines[maxIndex];
}
public List<Line> GetLines(int threshold)
{
if (Accumulator == null)
{
throw new Exception("HoughMap is null");
}
int houghWidth = Accumulator.Width;
int houghHeight = Accumulator.Height;
int imageWidth = Accumulator.Image.GetLength(0);
int imageHeight = Accumulator.Image.GetLength(1);
List<Line> lines = new List<Line>();
if (Accumulator == null)
return lines;
for (int rho = 0; rho < houghWidth; rho++)
{
for (int theta = 0; theta < houghHeight; theta++)
{
if ((int)Accumulator[rho, theta] > threshold)
{
//Is this point a local maxima (9x9)
int peak = Accumulator[rho, theta];
for (int ly = -4; ly <= 4; ly++)
{
for (int lx = -4; lx <= 4; lx++)
{
if ((ly + rho >= 0 && ly + rho < houghWidth) && (lx + theta >= 0 && lx + theta < houghHeight))
{
if ((int)Accumulator[rho + ly, theta + lx] > peak)
{
peak = Accumulator[rho + ly, theta + lx];
ly = lx = 5;
}
}
}
}
if (peak > (int)Accumulator[rho, theta])
continue;
int x1, y1, x2, y2;
x1 = y1 = x2 = y2 = 0;
double rad = theta * Math.PI / 180;
if (theta >= 45 && theta <= 135)
{
//y = (r - x Math.Cos(t)) / Math.Sin(t)
x1 = 0;
y1 = (int)(((double)(rho - (houghWidth / 2)) - ((x1 - (imageWidth / 2)) * Math.Cos(rad))) / Math.Sin(rad) + (imageHeight / 2));
x2 = imageWidth - 0;
y2 = (int)(((double)(rho - (houghWidth / 2)) - ((x2 - (imageWidth / 2)) * Math.Cos(rad))) / Math.Sin(rad) + (imageHeight / 2));
}
else
{
//x = (r - y Math.Sin(t)) / Math.Cos(t);
y1 = 0;
x1 = (int)(((double)(rho - (houghWidth / 2)) - ((y1 - (imageHeight / 2)) * Math.Sin(rad))) / Math.Cos(rad) + (imageWidth / 2));
y2 = imageHeight - 0;
x2 = (int)(((double)(rho - (houghWidth / 2)) - ((y2 - (imageHeight / 2)) * Math.Sin(rad))) / Math.Cos(rad) + (imageWidth / 2));
}
lines.Add(new Line(new Point(x1, y1), new Point(x2, y2)));
}
}
}
return lines;
}
}
该算法实际上很容易理解,即使乍一看似乎相反.它基于以下行公式:
? = cos(?) * x + sin(?) * y
其中 ρ 是从原点到直线的垂直距离,θ是由该垂直线和水平轴形成的角度.
如果你知道ρ和θ就知道了这条线.如果你采用ρ和θ的所有可能的对(在给定精度内),你实际上得到了图像中可能存在的所有可能的线.这就是
Map[?, ? ]商店.如果您希望角度精度为1度,则需要180列.对于ρ,可能的最大距离是图像的对角线长度.因此,取一个像素精度,行数可以是图像的对角线长度.但不是你的形象,正方形(在HoughMap.cs中):
int maxTheta = 180;
int houghHeight = (int)( Math.Sqrt( 2 ) * Math.Max( imgWidth, imgHeight ) ) / 2;
int doubleHoughHeight = houghHeight * 2;
doubleHoughHeight是广场的对角线,这就是它需要的原因Math.Max!
图像的每个点都映射到Map数组:
? ? number of points in that line( pair (?, ?) )
Map 0 0 num0
0 1 num1
0 2 num2
. . .
. . .
doubleHoughHeight – 1 179 numN
阈值过滤掉少于50个点的线.以下代码过滤掉了行:
//Is this point a local maxima (9x9)
int peak = Accumulator[ rho, theta ];
for( int ly = -4; ly <= 4; ly++ ) {
for( int lx = -4; lx <= 4; lx++ ) {
if( ( ly + rho >= 0 && ly + rho < houghWidth ) && ( lx + theta >= 0 && lx + theta < houghHeight ) ) {
if( (int)Accumulator[ rho + ly, theta + lx ] > peak ) {
peak = Accumulator[ rho + ly, theta + lx ];
ly = lx = 5;
}
}
}
}
if( peak > (int)Accumulator[ rho, theta ] )
continue;
您可以在以下图片中看到您遇到的实际问题:
您获得的结束点和起点实际上是线和两个轴的交点:
int x1, y1, x2, y2;
x1 = y1 = x2 = y2 = 0;
double rad = theta * Math.PI / 180;
if( theta >= 45 && theta <= 135 ) {
//y = (r - x Math.Cos(t)) / Math.Sin(t)
x1 = 0;
y1 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( x1 - ( imageWidth / 2 ) ) * Math.Cos( rad ) ) ) / Math.Sin( rad ) + ( imageHeight / 2 ) );
x2 = imageWidth - 0;
y2 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( x2 - ( imageWidth / 2 ) ) * Math.Cos( rad ) ) ) / Math.Sin( rad ) + ( imageHeight / 2 ) );
}
else {
//x = (r - y Math.Sin(t)) / Math.Cos(t);
y1 = 0;
x1 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( y1 - ( imageHeight / 2 ) ) * Math.Sin( rad ) ) ) / Math.Cos( rad ) + ( imageWidth / 2 ) );
y2 = imageHeight - 0;
x2 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( y2 - ( imageHeight / 2 ) ) * Math.Sin( rad ) ) ) / Math.Cos( rad ) + ( imageWidth / 2 ) );
}
lines.Add( new Line( new Point( x1, y1 ), new Point( x2, y2 ) ) );
编辑
您的代码工作正常,但它不计算实际长度.我找到的解决方案是存储2D Map数组的每个位置的所有点.在你的HoughMap.cs中:
public List<Point>[] lstPnts { get; set; }
public void Compute() {
if( Image != null ) {
...
...
...
Map = new int[ doubleHoughHeight, maxTheta ];
//Add this code////////////////////////////////////////////////
//lstPnts is an doubleHoughHeight * maxTheta size array of list Points
lstPnts = new List<Point>[ doubleHoughHeight * maxTheta ];
for(int i = 0; i < doubleHoughHeight * maxTheta; i++ ) {
lstPnts[ i ] = new List<Point>();
}
///////////////////////////////////////////////////////////////
....
....
....
if( ( rho > 0 ) && ( rho <= Map.GetLength( 0 ) ) ) {
Map[ rho, theta ]++;
//Add this line of code////////////////////////////////////////
lstPnts[ rho * maxTheta + theta ].Add( new Point( x, y ) );
///////////////////////////////////////////////////////////////
PointsCount++;
}
....
}
}
在HoughLineTransform.cs中
public List<Line> GetLines( int threshold ) {
if( Accumulator == null ) {
throw new Exception( "HoughMap is null" );
}
int houghWidth = Accumulator.Width;
int houghHeight = Accumulator.Height;
int imageWidth = Accumulator.Image.GetLength( 0 );
int imageHeight = Accumulator.Image.GetLength( 1 );
List<Line> lines = new List<Line>();
if( Accumulator == null )
return lines;
for( int rho = 0; rho < houghWidth; rho++ ) {
for( int theta = 0; theta < houghHeight; theta++ ) {
if( (int)Accumulator[ rho, theta ] > threshold ) {
//Is this point a local maxima (9x9)
int peak = Accumulator[ rho, theta ];
int dd = 10;
for( int ly = -dd; ly <= dd; ly++ ) {
for( int lx = -dd; lx <= dd; lx++ ) {
if( ( ly + rho >= 0 && ly + rho < houghWidth ) && ( lx + theta >= 0 && lx + theta < houghHeight ) ) {
if( (int)Accumulator[ rho + ly, theta + lx ] > peak ) {
peak = Accumulator[ rho + ly, theta + lx ];
ly = lx = dd + 1;
}
}
}
}
if( peak > (int)Accumulator[ rho, theta ] )
continue;
//Map[ rho, theta ] contains these points -> lstPnts[ rho * houghHeight + theta ].
//The points in that list with min and max X coordinate are the Start and End ones
int x1 = houghWidth, y1 = 0, x2 = -1, y2 = 0;
for(int i = 0; i < Accumulator.lstPnts[ rho * houghHeight + theta ].Count; i++ ) {
if( Accumulator.lstPnts[ rho * houghHeight + theta ][ i ].X > x2 ) {
x2 = Accumulator.lstPnts[ rho * houghHeight + theta ][ i ].X;
y2 = Accumulator.lstPnts[ rho * houghHeight + theta ][ i ].Y;
}
if( Accumulator.lstPnts[ rho * houghHeight + theta ][ i ].X < x1 ) {
x1 = Accumulator.lstPnts[ rho * houghHeight + theta ][ i ].X;
y1 = Accumulator.lstPnts[ rho * houghHeight + theta ][ i ].Y;
}
}
//Remove this code
/*int x1, y1, x2, y2;
x1 = y1 = x2 = y2 = 0;
double rad = theta * Math.PI / 180;
if( theta >= 45 && theta <= 135 ) {
//y = (r - x Math.Cos(t)) / Math.Sin(t)
x1 = 0;
y1 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( x1 - ( imageWidth / 2 ) ) * Math.Cos( rad ) ) ) / Math.Sin( rad ) + ( imageHeight / 2 ) );
x2 = imageWidth - 0;
y2 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( x2 - ( imageWidth / 2 ) ) * Math.Cos( rad ) ) ) / Math.Sin( rad ) + ( imageHeight / 2 ) );
}
else {
//x = (r - y Math.Sin(t)) / Math.Cos(t);
y1 = 0;
x1 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( y1 - ( imageHeight / 2 ) ) * Math.Sin( rad ) ) ) / Math.Cos( rad ) + ( imageWidth / 2 ) );
y2 = imageHeight - 0;
x2 = (int)( ( (double)( rho - ( houghWidth / 2 ) ) - ( ( y2 - ( imageHeight / 2 ) ) * Math.Sin( rad ) ) ) / Math.Cos( rad ) + ( imageWidth / 2 ) );
}*/
lines.Add( new Line( new Point( x1, y1 ), new Point( x2, y2 ) ) );
}
}
}
return lines;
}