我发现在图像中的大多数线的交点 [英] I am struck at finding the point of intersection of most lines in an image

问题描述

我使用标准的hough变换来获得使用OpenCV的图像中的直线。现在，我需要找到图像中大多数线的交叉点的x和y坐标。我的想法是将图像分成几个8 * 8像素段（我还没有决定大小），然后搜索该区域的线。但我不知道如何计算线段中的线。我在这一点上打了。所以我无法做的事情（使用openCV）是
1.将图像分为8 * 8像素段（我不知道如果有一个函数，如果不是我做什么）
2.计算每个段中的线数。

I used standard hough transforms to obtain the straight lines in an image using OpenCV. Now, I need to find the x and y coordinates of the point of intersection of most of the lines in the image. My idea was to divide the image into several 8*8 pixel segments( I haven't decided on the size yet) and then search the region for lines. But I am not sure as to how I count the lines in the segments. I am struck at this point. So the things I am unable to do( using openCV) are 1.Divide the image into 8*8 pixel segments( i don't know if there is a function for it and if not what do I do) 2.Count the number of lines in each segment. Any reading material or hints for the code will be of really helpful.

推荐答案

您检测交叉点的方法是完全错了。有一个简单的数学公式。我给你一个例子代码在纯C：

Your way of detecting intersection points is totally wrong. There is a simple mathematical formula for that. I am giving you an example code in plain C. :

// This is a point
typedef struct{
   int x,y;
} MYintPOINT;

// This is line
typedef struct {
    MYintPOINT  pStart;
    MYintPOINT  pEnd;
} MyLine;

#define PointMinusPoint(P,Q,R)      {(P).x = (Q).x - (R).x; (P).y = (Q).y - (R).y;}
#define PointCross(P,Q)             (((P).x*(Q).y)-((P).y*(Q).x))
#define SIGN(X)             (((X)>=0)? 1:-1 )
#define ABS(a)              ((a) >= 0 ? (a) : (-(a)))
#define ROUND(a)            ((SIGN(a)) * ( ( int )( ABS(a) + 0.5 ) ) ) 

// Given 2 line segments, find their intersection point
// rerurns [Px,Py] point in 'res' or FALSE if parallel. Uses vector cross product technique.
int findLinesIntersectionPoint(const MyLine*l1, const MyLine*l2, MYintPOINT *res){
    MYintPOINT  p  = l1->pStart;
    MYintPOINT  dp;
    MYintPOINT  q  = l2->pStart;
    MYintPOINT  dq;
    MYintPOINT  qmp;            // q-p
    int         dpdq_cross;     // 2 cross products
    int         qpdq_cross;     // dp with dq,  q-p with dq
    float       a;

    PointMinusPoint(dp,l1->pEnd,l1->pStart);
    PointMinusPoint(dq,l2->pEnd,l2->pStart);
    PointMinusPoint(qmp,q,p);

    dpdq_cross = PointCross(dp,dq);
    if (!dpdq_cross){
        // Perpendicular Lines
        return 0;
    }

    qpdq_cross = PointCross(qmp,dq);
    a = (qpdq_cross*1.0f/dpdq_cross);

    res->x = ROUND(p.x+a*dp.x);
    res->y = ROUND(p.y+a*dp.y);
    return 1;
}

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