OpenCV线拟合算法

时间:2012-07-30 13:27:09

标签: algorithm opencv statistics curve-fitting

我正在尝试了解OpenCV fitLine()算法。

这是来自OpenCV的代码片段: icvFitLine2D函数 - icvFitLine2D

我看到有一些随机函数选择点进行近似,然后计算从点到拟合线的距离(选择点),然后选择其他点并尝试通过选择distType来最小化距离。

如果没有严格的数学并假设没有很好的统计知识,有人可以澄清this moment会发生什么吗? OpenCV代码注释和变量名称无助于我理解这段代码。

1 个答案:

答案 0 :(得分:22)

(这是一个老问题,但主题引起了我的好奇心)

OpenCV FitLine实现了两种不同的机制。

如果参数distType设置为CV_DIST_L2,则使用standard unweighted least squares fit

如果使用其中一个distTypesCV_DIST_L1CV_DIST_L12CV_DIST_FAIRCV_DIST_WELSCHCV_DIST_HUBER),则该过程为某种RANSAC适合:

  • 最多重复20次:
    • 选择10个随机点,仅为它们设置最小二乘法
    • 最多重复30次:
  • 返回找到的最佳线条

以下是伪代码的更详细说明:

repeat at most 20 times:

    RANSAC (line 371)
     - pick 10 random points, 
     - set their weights to 1, 
     - set all other weights to 0

    least squares weighted fit (fitLine2D_wods, line 381)
     - fit only the 10 picked points to the line, using least-squares

    repeat at most 30 times: (line 382)
     - stop if the difference between the found solution and the previous found solution is less than DELTA  (line 390 - 406)
       (the angle difference must be less than adelta, and the distance beween the line centers must be less than rdelta)
     - stop if the sum of squared distances between the found line and the points is less than EPSILON (line 407)
       (The unweighted sum of squared distances is used here ==> the standard L2 norm)

        re-calculate the weights for *all* points (line 412)
         - using the given norm (CV_DIST_L1 / CV_DIST_L12 / CV_DIST_FAIR / ...)
         - normalize the weights so their sum is 1
         - special case, to catch errors: if for some reason all weights are zero, set all weight to 1

        least squares weighted fit (fitLine2D_wods, line 437)
         - fit *all* points to the line, using weighted least squares

    if the last found solution is better than the current best solution (line 440)
        save it as the new best
        (The unweighted sum of squared distances is used here ==> the standard L2 norm)

        if the distance between the found line and the points is less than EPSILON
             break

return the best solution

权重的计算取决于所选的distType,根据the manual,其公式为weight[Point_i] = 1/ p(distance_between_point_i_and_line),其中p为:

  

distType = CV_DIST_L1 enter image description here

     

distType = CV_DIST_L12 enter image description here

     

distType = CV_DIST_FAIR enter image description here

     

distType = CV_DIST_WELSCH enter image description here

     

distType = CV_DIST_HUBER enter image description here

不幸的是,我不知道哪个distType最适合哪种数据,也许其他人可以对此有所了解。


我发现有趣的事情:选择的规范仅用于迭代重新加权,找到的最佳解决方案总是根据L2规范(未加权和最小二乘是最小的)。我不确定这是否正确。