在一定范围内检测矢量数组中的圆弧和圆形状 [英] detecting arc and circle shapes in vector array with a range

问题描述

我有一个二维顶点数组，我想检测数组中是否有任何拱形或圆形.有时值不是那么精确，我需要一个很小的范围.这些是值.第三顶点值保持为0:

  verticle:-0.014848，-13.2684，0角:0.141274垂直:-0.0174556，-4.84519、0角度:90垂直:0，0，0角度:90垂直:-9.53674e-07、14.14、0角:40.7168垂直:-12.1101、14.0709、0 角度:7.94458垂直:-12.0996、10.6442、0角度:0.294751垂直:-12.2305、10.6484、0角:0.24309垂直:-12.325、10.6384、0角:0.349426垂直:-12.4475、10.6125、0角:0.392669垂直:-12.5638、10.564、0角:0.404935垂直:-12.678、10.508、0角:0.34605垂直:-12.7579、10.4453、0 角度:0.391671垂直:-12.8315、10.36、0角:0.390671垂直:-12.9051，10.2747，0角:0.438795垂直:-12.9725、10.1668、0角:0.455425垂直:-13.0377，10.0514，0角:0.300014垂直:-13.0407、9.94252、0角:0.388662垂直:-13.0738、9.83064、0角:0.338041垂直:-13.0725、9.70936、0角:0.254878垂直:-13.0412，9.59645，0角:0.257171垂直:-13.0098，9.48352，0角:0.259443垂直:-12.9785、9.37061、0 角度:0.158259垂直:-12.9192，9.27357，0角:0.0713262垂直:-12.8297、9.18489、0角:0.14537垂直:-12.7724，9.09539，0角:0.0484566垂直:-12.657、9.03012、0角:0.0197823垂直:-12.5738、8.94603、0角:0.125115垂直:-12.4667，8.92887，0角:0.219397垂直:-12.3296、8.90207、0角:0.185575垂直:-12.2288、8.88951、0角:0.299361垂直:-12.1、8.88292、0角:11.3066垂直:-12.1075、5.66474、0角:0.158259垂直:-12.2062、5.65268、0角:0.266879垂直:-12.3329、5.64184、0角:0.312787垂直:-12.4554、5.61594、0角:0.384104垂直:-12.5717，5.56746，0角度:0.322034垂直:-12.6557、5.5198、0角:0.45024垂直:-12.7657、5.44874、0角:0.416371垂直:-12.8415，5.37097，0角:0.464781垂直:-12.913、5.27815、0角:0.514343垂直:-12.9803，5.17027，0角:0.436111垂直:-13.0176，5.07075，0角:0.487788垂直:-13.0506、4.95617、0角:0.439686垂直:-13.0515，4.84242，0角:0.441462垂直:-13.0524、4.72867、0角:0.470222垂直:-13.0511、4.6074、0角:0.399585垂直:-13.0198，4.49448，0角:0.402998垂直:-12.9885、4.38156、0角:0.305828垂直:-12.9291、4.28452、0角:0.237388垂直:-12.8396，4.19585，0角:0.213062垂直:-12.7523，4.1147，0角:0.188712垂直:-12.667、4.04107、0角:0.0625573垂直:-12.5557、3.99086、0角:0.0279765垂直:-12.4466，3.94818，0角:0.0197823垂直:-12.3416、3.92056、0角:0.158259垂直:-12.2107，3.91634，0角:0.111906垂直:-12.1121，3.9113，0角:17.8633垂直:-12.0988，0.00704384，0角:15.2939垂直:-12.0895，-3.29836，0角:0.174713垂直:-12.2204，-3.29415，0角:0.100871垂直:-12.3471，-3.30499、0角:0.034264垂直:-12.4395，-3.32253、0角:0.0395647垂直:-12.5579，-3.36349、0角:0.139882垂直:-12.67，-3.42703、0角:0.170174垂直:-12.7499，-3.8974、0角:0.236563垂直:-12.8557，-3.57586、0角:0.266144垂直:-12.9293，-3.66115，0角:0.363156垂直:-12.9666，-3.76067，0角:0.357727垂直:-13.0339，-3.86855、0角:0.421973垂直:-13.067，-3.98313、0角:0.454565垂直:-13.0678，-4.09688，0角:0.452407垂直:-13.0687，-4.21063，0​​角:0.482545垂直:-13.0675，-4.3319、0角:0.487788垂直:-13.0361，-4.44482，0角:0.463094垂直:-12.9768，-4.54186，0角:0.421973垂直:-12.9496，-4.63972，0角:0.44279垂直:-12.8622，-4.72087，0角:0.402026垂直:-12.8071，-4.80285，0角:0.383084垂直:-12.7239，-4.86895、0角:0.399585垂直:-12.6105，-4.92668、0角:0.29074垂直:-12.5336，-4.97019、0角:0.30901垂直:-12.4266，-5.00535，0 角度:0.245493垂直:-12.3237，-5.02544，0角:0.214891垂直:-12.2229，-5.03801，0角:0.132704垂直:-12.0983，-5.01964，0角:11.875垂直:-12.0995，-8.28741、0角:0.300014垂直:-12.2304，-8.28319，0角:0.199792垂直:-12.327，-8.28568、0角:0.179137垂直:-12.4495，-8.31158、0角:0.121947垂直:-12.5679，-8.35253、0角:0.0395647垂直:-12.6799，-8.41607，0角:0.0279765垂直:-12.7598，-8.47878，0角:0.0442347垂直:-12.8657，-8.56491、0角:0.138476垂直:-12.9372，-8.65773，0角:0.199792垂直:-12.9765，-8.74972、0角:0.214891垂直:-13.0418，-8.86513，0角:0.275536垂直:-13.0749，-8.9797、0角:0.335718垂直:-13.0757，-9.09345，0角:0.359365垂直:-13.0745，-9.21473、0角:0.356083垂直:-13.0733，-9.33601、0角:0.39217垂直:-13.0419，-9.44893、0角:0.428872垂直:-12.9805，-9.55349、0角:0.402512垂直:-12.9211，-9.65052、0角:0.401538垂直:-12.8618，-9.74756，0角:0.417778垂直:-12.7744，-9.82871、0角:0.436559垂直:-12.659，-9.89397，0角:0.370094垂直:-12.5758，-9.96007、0角:0.338041垂直:-12.4687，-9.99522，0角:0.384613垂直:-12.3316，-10.022、0角:0.265408垂直:-12.2308，-10.0346，0角:0.261696垂直:-12.1041，-10.0237，0角:7.8231垂直:-12.1023，-13.1853、0角:42.4836 

我认为解决此问题的唯一方法是将距离计算与大于某个值的角度相结合.我知道这是一个不好的解决方案.但是我想不出其他任何方法来计算这个.

角度计算是这样的:

  Vec3f :: angle的内联浮点数(Vof3f& vec的常量)const {ofVec3f n1 = this-> normalized();ofVec3f n2 = vec.normalized();return(float)(acos(n1.dot(n2))* RAD_TO_DEG);} 

两点之间的距离:

  Vec3f :: distance(const ofVec3f& pnt)的内联浮点数const {浮点数vx = x-pnt.x;float vy = y-pnt.y;浮点数vz = z-pnt.z;return(float)sqrt(vx * vx + vy * vy + vz * vz);} 

我已经使用OpenFrameworks库实现了这一点:

  float dist =建筑[x].多边形[z] .distance(建筑[x].多边形[z + 1]);浮动角度；如果(z  

这是该库的github

http://en.wikipedia.org/wiki/RANSAC

这些算法适用于嘈杂的数据.

加里·布拉德斯基(Gary Bradski)的《学习OpenCV》一书中有一个名为霍夫圆变换"的章节，该章节运行了几页.尽管您可能会看过OpenCV库本身，但可能会在其他地方找到有关Hough变换和RANSAC的更直接的描述.

我用C#编写了一个快速的Hough算法，它可以很好地处理您的数据.您可以添加随机噪声，但仍然可以使用.核心算法是大约100行带有注释和空格的代码，这是一种草率算法.

该算法的部分输出图像如下所示，数据点为白色，霍夫拟合圆为红色.

我没有执行的一个标准算法步骤是过滤数据，以消除当存在多个具有大致相同的中心和大致相同的半径的合理的圆拟合时，除最佳圆拟合之外的所有圆.这就是为什么您看到一个双圆穿过点.

尽管Hough算法似乎有些过分，但很好的是它可以一次又一次地工作，可重复使用，易于参数化，无论数据干净还是嘈杂，都会产生良好的结果.

注:为了简化实现，我将(x，y)的float值转换为整数，并对值进行了归一化，以便所有(x，y)均为正.像这样:

  int x =(int)(100 * floatX + 1600);//floatX =数据中的原始X值int y =(int)(100 * floatY + 3200);//floatY =数据中的原始Y值 

I have an array of 2d vertices and I want to detect if there are any arch or circle shapes in a array. Sometimes the values are not that precise and I need a small range. Here are the values. The 3rd verticle value value remains 0:

verticle:  -0.014848, -13.2684, 0 angle : 0.141274
verticle:  -0.0174556, -4.84519, 0 angle : 90
verticle:  0, 0, 0 angle : 90
verticle:  -9.53674e-07, 14.14, 0 angle : 40.7168
verticle:  -12.1101, 14.0709, 0 angle : 7.94458
verticle:  -12.0996, 10.6442, 0 angle : 0.294751
verticle:  -12.2305, 10.6484, 0 angle : 0.24309
verticle:  -12.325, 10.6384, 0 angle : 0.349426
verticle:  -12.4475, 10.6125, 0 angle : 0.392669
verticle:  -12.5638, 10.564, 0 angle : 0.404935
verticle:  -12.678, 10.508, 0 angle : 0.34605
verticle:  -12.7579, 10.4453, 0 angle : 0.391671
verticle:  -12.8315, 10.36, 0 angle : 0.390671
verticle:  -12.9051, 10.2747, 0 angle : 0.438795
verticle:  -12.9725, 10.1668, 0 angle : 0.455425
verticle:  -13.0377, 10.0514, 0 angle : 0.300014
verticle:  -13.0407, 9.94522, 0 angle : 0.388662
verticle:  -13.0738, 9.83064, 0 angle : 0.338041
verticle:  -13.0725, 9.70936, 0 angle : 0.254878
verticle:  -13.0412, 9.59645, 0 angle : 0.257171
verticle:  -13.0098, 9.48352, 0 angle : 0.259443
verticle:  -12.9785, 9.37061, 0 angle : 0.158259
verticle:  -12.9192, 9.27357, 0 angle : 0.0713262
verticle:  -12.8297, 9.18489, 0 angle : 0.14537
verticle:  -12.7724, 9.09539, 0 angle : 0.0484566
verticle:  -12.657, 9.03012, 0 angle : 0.0197823
verticle:  -12.5738, 8.96403, 0 angle : 0.125115
verticle:  -12.4667, 8.92887, 0 angle : 0.219397
verticle:  -12.3296, 8.90207, 0 angle : 0.185575
verticle:  -12.2288, 8.88951, 0 angle : 0.299361
verticle:  -12.1, 8.89282, 0 angle : 11.3066
verticle:  -12.1075, 5.64764, 0 angle : 0.158259
verticle:  -12.2062, 5.65268, 0 angle : 0.266879
verticle:  -12.3329, 5.64184, 0 angle : 0.312787
verticle:  -12.4554, 5.61594, 0 angle : 0.384104
verticle:  -12.5717, 5.56746, 0 angle : 0.322034
verticle:  -12.6557, 5.5198, 0 angle : 0.45024
verticle:  -12.7657, 5.44874, 0 angle : 0.416371
verticle:  -12.8415, 5.37097, 0 angle : 0.464781
verticle:  -12.913, 5.27815, 0 angle : 0.514343
verticle:  -12.9803, 5.17027, 0 angle : 0.436111
verticle:  -13.0176, 5.07075, 0 angle : 0.487788
verticle:  -13.0506, 4.95617, 0 angle : 0.439686
verticle:  -13.0515, 4.84242, 0 angle : 0.441462
verticle:  -13.0524, 4.72867, 0 angle : 0.470222
verticle:  -13.0511, 4.6074, 0 angle : 0.399585
verticle:  -13.0198, 4.49448, 0 angle : 0.402998
verticle:  -12.9885, 4.38156, 0 angle : 0.305828
verticle:  -12.9291, 4.28452, 0 angle : 0.237388
verticle:  -12.8396, 4.19585, 0 angle : 0.213062
verticle:  -12.7523, 4.1147, 0 angle : 0.188712
verticle:  -12.667, 4.04107, 0 angle : 0.0625573
verticle:  -12.5557, 3.99086, 0 angle : 0.0279765
verticle:  -12.4466, 3.94818, 0 angle : 0.0197823
verticle:  -12.3416, 3.92056, 0 angle : 0.158259
verticle:  -12.2107, 3.91634, 0 angle : 0.111906
verticle:  -12.1121, 3.9113, 0 angle : 17.8633
verticle:  -12.0988, 0.00704384, 0 angle : 15.2939
verticle:  -12.0895, -3.29836, 0 angle : 0.174713
verticle:  -12.2204, -3.29415, 0 angle : 0.100871
verticle:  -12.3471, -3.30499, 0 angle : 0.034264
verticle:  -12.4395, -3.32253, 0 angle : 0.0395647
verticle:  -12.5579, -3.36349, 0 angle : 0.139882
verticle:  -12.67, -3.42703, 0 angle : 0.170174
verticle:  -12.7499, -3.48974, 0 angle : 0.236563
verticle:  -12.8557, -3.57586, 0 angle : 0.266144
verticle:  -12.9293, -3.66115, 0 angle : 0.363156
verticle:  -12.9666, -3.76067, 0 angle : 0.357727
verticle:  -13.0339, -3.86855, 0 angle : 0.421973
verticle:  -13.067, -3.98313, 0 angle : 0.454565
verticle:  -13.0678, -4.09688, 0 angle : 0.452407
verticle:  -13.0687, -4.21063, 0 angle : 0.482545
verticle:  -13.0675, -4.3319, 0 angle : 0.487788
verticle:  -13.0361, -4.44482, 0 angle : 0.463094
verticle:  -12.9768, -4.54186, 0 angle : 0.421973
verticle:  -12.9496, -4.63972, 0 angle : 0.44279
verticle:  -12.8622, -4.72087, 0 angle : 0.402026
verticle:  -12.8071, -4.80285, 0 angle : 0.383084
verticle:  -12.7239, -4.86895, 0 angle : 0.399585
verticle:  -12.6105, -4.92668, 0 angle : 0.29074
verticle:  -12.5336, -4.97019, 0 angle : 0.30901
verticle:  -12.4266, -5.00535, 0 angle : 0.245493
verticle:  -12.3237, -5.02544, 0 angle : 0.214891
verticle:  -12.2229, -5.03801, 0 angle : 0.132704
verticle:  -12.0983, -5.01964, 0 angle : 11.875
verticle:  -12.0995, -8.28741, 0 angle : 0.300014
verticle:  -12.2304, -8.28319, 0 angle : 0.199792
verticle:  -12.327, -8.28568, 0 angle : 0.179137
verticle:  -12.4495, -8.31158, 0 angle : 0.121947
verticle:  -12.5679, -8.35253, 0 angle : 0.0395647
verticle:  -12.6799, -8.41607, 0 angle : 0.0279765
verticle:  -12.7598, -8.47878, 0 angle : 0.0442347
verticle:  -12.8657, -8.56491, 0 angle : 0.138476
verticle:  -12.9372, -8.65773, 0 angle : 0.199792
verticle:  -12.9765, -8.74972, 0 angle : 0.214891
verticle:  -13.0418, -8.86513, 0 angle : 0.275536
verticle:  -13.0749, -8.9797, 0 angle : 0.335718
verticle:  -13.0757, -9.09345, 0 angle : 0.359365
verticle:  -13.0745, -9.21473, 0 angle : 0.356083
verticle:  -13.0733, -9.33601, 0 angle : 0.39217
verticle:  -13.0419, -9.44893, 0 angle : 0.428872
verticle:  -12.9805, -9.55349, 0 angle : 0.402512
verticle:  -12.9211, -9.65052, 0 angle : 0.401538
verticle:  -12.8618, -9.74756, 0 angle : 0.417778
verticle:  -12.7744, -9.82871, 0 angle : 0.436559
verticle:  -12.659, -9.89397, 0 angle : 0.370094
verticle:  -12.5758, -9.96007, 0 angle : 0.338041
verticle:  -12.4687, -9.99522, 0 angle : 0.384613
verticle:  -12.3316, -10.022, 0 angle : 0.265408
verticle:  -12.2308, -10.0346, 0 angle : 0.261696
verticle:  -12.1041, -10.0237, 0 angle : 7.8231
verticle:  -12.1023, -13.1853, 0 angle : 42.4836

The only way I thought would fix this problem is a distance calculation combined with the angles that are bigger than a certain value. I know that is a bad solution. But I cannot think of any other way to calculate this.

This is how the angle calculation goes:

 inline float ofVec3f::angle( const ofVec3f& vec ) const {
ofVec3f n1 = this->normalized();
ofVec3f n2 = vec.normalized();
return (float)(acos( n1.dot(n2) )*RAD_TO_DEG);
 }

The distance between two points:

 inline float ofVec3f::distance( const ofVec3f& pnt) const {
float vx = x-pnt.x;
float vy = y-pnt.y;
float vz = z-pnt.z;
return (float)sqrt(vx*vx + vy*vy + vz*vz);
  }

I've used the OpenFrameworks library to realize this:

float dist = buildings[x].polygon[z].distance(buildings[x].polygon[z+1]);
        float angle ;
        if ( z < buildings[x].polygon.size()-1){
            angle  =  buildings[x].polygon[z].angle(buildings[x].polygon[z+1]);
        }
         if ( ( dist >  0.100)&& ( dist <  0.150)) {
          //Is part of ellipse
       }

Here's a github of the library

https://github.com/openframeworks/openFrameworks/blob/master/libs/openFrameworks/math/ofVec3f.h

Here are two screen shots of the points and the line

Here's one vector array of verts x,y x,y

-0.11878395,-106.14753 -0.13964462,-38.761494 0.0,0.0 -7.6293945E-6,113.11968 -96.88052,112.56717 -96.79668,85.153725 -97.843834,85.18742 -98.599945,85.107315 -99.58024,84.900116 -100.51039,84.51225 -101.42383,84.06417 -102.06295,83.562485 -102.65193,82.88013 -103.240906,82.197784 -103.77975,81.33476 -104.30187,80.41151 -104.325485,79.56175 -104.59001,78.64513 -104.5802,77.67491 -104.32949,76.77156 -104.07879,75.868195 -103.82809,74.96484 -103.35321,74.18856 -102.63742,73.47914 -102.17926,72.763084 -101.25601,72.24097 -100.59037,71.71221 -99.73398,71.43098 -98.63669,71.2166 -97.83043,71.11605 -96.8,71.14256 -96.859665,45.18112 -97.6492,45.22146 -98.66293,45.134716 -99.64322,44.92752 -100.57338,44.539658 -101.245926,44.158424 -102.12593,43.58989 -102.73163,42.96776 -103.303894,42.22518 -103.84273,41.36216 -104.14067,40.56599 -104.40518,39.649376 -104.412094,38.739384 -104.41901,37.8294 -104.409195,36.859184 -104.15849,35.955826 -103.90778,35.052475 -103.43291,34.27619 -102.71712,33.566765 -102.01806,32.917564 -101.33571,32.328587 -100.445885,31.92691 -99.57278,31.585457 -98.73309,31.364452 -97.68595,31.330748 -96.89641,31.290417 -96.790054,0.056350708 -96.71601,-26.386883 -97.76316,-26.353178 -98.77688,-26.439924 -99.51628,-26.580261 -100.46315,-26.907902 -101.35987,-27.416212 -101.99899,-27.917892 -102.84558,-28.606876 -103.434555,-29.289228 -103.732506,-30.0854 -104.27134,-30.948421 -104.53585,-31.86504 -104.542755,-32.77503 -104.54967,-33.685017 -104.53986,-34.655228 -104.28916,-35.558586 -103.81427,-36.33486 -103.59699,-37.117775 -102.897934,-37.766975 -102.456474,-38.422806 -101.79083,-38.95156 -100.8843,-39.41346 -100.2688,-39.761543 -99.41241,-40.04277 -98.58944,-40.203552 -97.78319,-40.30411 -96.78618,-40.157143 -96.79571,-66.299255 -97.84286,-66.26555 -98.615685,-66.285446 -99.59598,-66.49263 -100.54285,-66.820274 -101.439575,-67.32858 -102.07869,-67.83027 -102.92528,-68.51925 -103.497536,-69.261826 -103.8122,-69.99777 -104.33433,-70.92102 -104.59884,-71.83764 -104.60574,-72.74762 -104.59593,-73.717834 -104.58613,-74.68805 -104.33543,-75.5914 -103.84383,-76.42791 -103.36895,-77.204185 -102.89406,-77.98047 -102.195,-78.62967 -101.27175,-79.151794 -100.6061,-79.68055 -99.74972,-79.96178 -98.65242,-80.17615 -97.84617,-80.27671 -96.83245,-80.189964 -96.81836,-105.482315 -0.11878395,-106.14753

Other vector array

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解决方案

A Hough circle fit or RANSAC circle fit would probably work for you; in image processing we use these algorithms to find circles, arcs, ellipses, and other shapes.

http://en.wikipedia.org/wiki/Hough_transform

http://en.wikipedia.org/wiki/RANSAC

These algorithms work well with noisy data.

Gary Bradski's book Learning OpenCV has a section entitled "Hough Circle Transform" that runs a few pages. Although you might look at the OpenCV library itself, you'll probably find more straightforward descriptions of the Hough transform and RANSAC elsewhere.

[EDIT]

I wrote a quick Hough algorithm in C# and it worked well with your data. You could add random noise and it would still work. The core algorithm is about 100 lines of code with comments and spaces, and that's for a sloppy algorithm.

A portion of the output image from this algorithm is shown below with data points in white and Hough fit circles in red.

One standard algorithmic step I didn't implement is to filter data to eliminate all but the best circle fit when there are several reasonable circle fits with approximately the same center and approximately the same radius. That's why you see a double circle passing through the points.

Although a Hough algorithm might seem like overkill, what's nice is that it will work time and time again, it's reusable, it's easy to parameterize, and it will yield good results whether the data is clean or noisy.

NOTE: to simplify the implementation, I converted the float values for (x,y) to integers and normalized the values so that all (x,y) were positive. Something like this:

int x = (int)(100 * floatX + 1600);  //floatX = original X value in data
int y = (int)(100 * floatY + 3200);  //floatY = original Y value in data

这篇关于在一定范围内检测矢量数组中的圆弧和圆形状的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！