python - 用python查找不完整圆的交点

我试图找到不完整圆的交点，如下图所示:



参考此链接的解决方案:
Detect semi-circle in opencv

我正在尝试将c++代码转换为python代码，我已经转换了大多数代码，但是我不明白下面的2行c++代码:

为什么半径需要除以25？

// maximal distance of inlier might depend on the size of the circle
float maxInlierDist = radius/25.0f;

我完全不知道如何将此C++行转换为python:

if(dt.at<float>(cY,cX) < maxInlierDist)

希望任何人都能对此有所帮助，谢谢!

我试图用谷歌搜索一些数学公式，但是找不到为什么半径需要除以25。我在c++中也不是很好。

我转换后的代码:

# import the necessary packages
import numpy as np
import argparse
import cv2
import math

# construct the argument parser and parse the arguments
ap = argparse.ArgumentParser()
ap.add_argument("-t", "--thres", required = True, help = "Path to the image")
ap.add_argument("-i", "--image", required = True, help = "Path to the image")
args = vars(ap.parse_args())

# load the image, clone it for output, and then convert it to grayscale
image = cv2.imread(args["image"])

gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)

gray = cv2.Canny(gray, 200,20)

# detect circles in the image
circles = cv2.HoughCircles(gray, cv2.HOUGH_GRADIENT, 1,minDist=300,
                               param1=200, param2=20,
                               minRadius=0, maxRadius=0)

#gray = (255*mask).astype(np.uint8)

dt = cv2.distanceTransform(255-gray, cv2.DIST_L2, 3)

cv2.imshow('Distance Transform', dt/255.0)
# ensure at least some circles were found

if circles is not None:
    # convert the (x, y) coordinates and radius of the circles to integers
    circles = np.round(circles[0, :]).astype("int")

    # loop over the (x, y) coordinates and radius of the circles
    for (x, y, r) in circles:
        # draw the circle in the output image, then draw a rectangle
        # corresponding to the center of the circle
        cv2.circle(image, (x, y), r, (0, 255, 0), 2)
        cv2.rectangle(image, (x - 5, y - 5), (x + 5, y + 5),
                             (0, 128,255),-1)

        minInlierDist = 2.0
        counter =0
        inlier =0
        radius=r
        num_circle = 50

        maxInlierDist=radius/25.0
        if maxInlierDist<minInlierDist:
            maxInlierDist=minInlierDist

        for index in range(num_circle):
            counter +=1
            #angle = t * math.pi / 180
            angle = 2 * math.pi * index / num_circle

            cX = x + math.sin(angle)*radius
            cY = y + math.cos(angle)*radius
            centerxy = cX,cY
            cv2.circle(image,tuple(np.array(centerxy,int)),3,(0,0,255),-1)

#if(dt.at<float>(cY,cX) < maxInlierDist) #c++ ! I'm stuck here!

    cv2.imshow("output", image)#np.hstack([image, gray]))
    cv2.waitKey(0)
else:
    print("no circles found!")
    cv2.waitKey(0)

这里最重要的问题是:如何知道/识别圆上的绿色采样点是离群值，蓝色点是离群值。

最佳答案

1)已选择1/25作为半径的任意分数，以获得最大可接受误差。

2)在opencv python中，矩阵存储为多维numpy数组。要访问(cY，cX)处的点，请使用dt [cY，cX]

关于python - 用python查找不完整圆的交点，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/56967444/