c++ - HoughCircles 无法在此图像上检测到圆圈

Question

我试图在我的图像中检测包含圆点的圆圈，但不幸的是我无法这样做。我正在使用 opencv HoughTransform，但找不到使它起作用的参数。

src = imread("encoded.jpg",1);
    /// Convert it to gray
    cvtColor(src, src_gray, CV_BGR2GRAY);

    vector<Vec3f> circles;

    /// Apply the Hough Transform to find the circles
    HoughCircles(src_gray, circles, CV_HOUGH_GRADIENT, 1, 10,
        100, 30, 1, 30 // change the last two parameters
        // (min_radius & max_radius) to detect larger circles
        );

    /// Draw the circles detected
    for (size_t i = 0; i < circles.size(); i++)
    {
        cout << "Positive" << endl;
        Point center(cvRound(circles[i][0]), cvRound(circles[i][1]));
        int radius = cvRound(circles[i][2]);
        // circle center
        circle(src, center, 3, Scalar(0, 255, 0), -1, 8, 0);
        // circle outline
        circle(src, center, radius, Scalar(0, 0, 255), 3, 8, 0);
    }

    /// Show your results
    namedWindow("Hough Circle Transform Demo", CV_WINDOW_AUTOSIZE);
    imshow("Hough Circle Transform Demo", src_gray);
    waitKey(0);

我的图片在这里:

为什么 HoughCircles 无法检测到此图像中的圆圈？它似乎正在处理其他更简单的图像，例如电路板图像。

Answer 1

我遇到了您的确切问题并找到了解决方案

关键在于对 HoughCircles 正在做什么有足够的直觉，这样您就可以构建一个程序，为您想要在其中找到圆圈的所有各种图像自动调整超参数。

核心问题，一些直觉

HoughCircles 并不是独立存在的，尽管它表明它可能具有最小和最大半径参数，但您需要运行数百或数千次迭代才能在正确的设置中自动调整和自动拨号。然后在你完成之后你需要后处理验证步骤来 100% 确定这个圆是你想要的。问题是您正在尝试通过猜测和检查来手动调整 HoughCircles 的输入参数。那根本行不通。让计算机为您自动调整这些参数。

HoughCircles 的手动调整什么时候可以令人满意？

如果您想手动对参数进行硬编码，那么您绝对需要做的一件事就是将圆的精确半径控制在一两个像素以内。您可以猜测 dp 分辨率并设置累加器数组投票阈值，您可能没问题。但是，如果您不知道半径，则 HoughCircles 输出将毫无用处，因为它要么到处都找不到圆，要么找不到任何地方。假设您确实手动找到了一个可接受的调整，您向它展示了几个像素不同的图像，并且您的 HoughCircles 吓坏了并在图像中找到了 200 个圆圈。毫无值(value)。

有希望:

希望来自于 HoughCircles 即使在大图像上也非常快的事实。您可以为 HoughCircles 编写一个程序来完美地自动调整设置。如果您不知道半径并且它可能很小或很大，您可以从一个很大的“最小距离参数”、一个非常好的 dp 分辨率和一个非常高的投票阈值开始。因此，当您开始迭代时，HoughCircles 可以预见地拒绝找到任何圈子，因为设置过于激进并且投票没有清除阈值。但是循环会不断迭代并逐步达到最佳设置，让最佳设置成为表明您已完成的避雷针。您找到的第一个圆圈将是图像中像素完美的最大和最佳圆圈，HoughCircles 会给您留下一个像素完美的圆圈，就在它应该出现的位置。只是您必须运行它 5000 次。

示例 python 代码(抱歉不是 C++):

它的边缘仍然很粗糙，但您应该能够将其清理干净，以便在一秒钟内获得令人满意的像素完美效果。

import numpy as np
import argparse
import cv2
import signal

from functools import wraps
import errno
import os
import copy

# construct the argument parser and parse the arguments
ap = argparse.ArgumentParser()
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"])
orig_image = np.copy(image)
output = image.copy()
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)

cv2.imshow("gray", gray)
cv2.waitKey(0)

circles = None

minimum_circle_size = 100      #this is the range of possible circle in pixels you want to find
maximum_circle_size = 150     #maximum possible circle size you're willing to find in pixels

guess_dp = 1.0

number_of_circles_expected = 1          #we expect to find just one circle
breakout = False

#hand tune this
max_guess_accumulator_array_threshold = 100     #minimum of 1, no maximum, (max 300?) the quantity of votes 
                                                #needed to qualify for a circle to be found.
circleLog = []

guess_accumulator_array_threshold = max_guess_accumulator_array_threshold

while guess_accumulator_array_threshold > 1 and breakout == False:
    #start out with smallest resolution possible, to find the most precise circle, then creep bigger if none found
    guess_dp = 1.0
    print("resetting guess_dp:" + str(guess_dp))
    while guess_dp < 9 and breakout == False:
        guess_radius = maximum_circle_size
        print("setting guess_radius: " + str(guess_radius))
        print(circles is None)
        while True:

            #HoughCircles algorithm isn't strong enough to stand on its own if you don't
            #know EXACTLY what radius the circle in the image is, (accurate to within 3 pixels) 
            #If you don't know radius, you need lots of guess and check and lots of post-processing 
            #verification.  Luckily HoughCircles is pretty quick so we can brute force.

            print("guessing radius: " + str(guess_radius) + 
                    " and dp: " + str(guess_dp) + " vote threshold: " + 
                    str(guess_accumulator_array_threshold))

            circles = cv2.HoughCircles(gray, 
                cv2.HOUGH_GRADIENT, 
                dp=guess_dp,               #resolution of accumulator array.
                minDist=100,                #number of pixels center of circles should be from each other, hardcode
                param1=50,
                param2=guess_accumulator_array_threshold,
                minRadius=(guess_radius-3),    #HoughCircles will look for circles at minimum this size
                maxRadius=(guess_radius+3)     #HoughCircles will look for circles at maximum this size
                )

            if circles is not None:
                if len(circles[0]) == number_of_circles_expected:
                    print("len of circles: " + str(len(circles)))
                    circleLog.append(copy.copy(circles))
                    print("k1")
                break
                circles = None
            guess_radius -= 5 
            if guess_radius < 40:
                break;

        guess_dp += 1.5

    guess_accumulator_array_threshold -= 2

#Return the circleLog with the highest accumulator threshold

# ensure at least some circles were found
for cir in circleLog:
    # convert the (x, y) coordinates and radius of the circles to integers
    output = np.copy(orig_image)

    if (len(cir) > 1):
        print("FAIL before")
        exit()

    print(cir[0, :])

    cir = np.round(cir[0, :]).astype("int")

    # loop over the (x, y) coordinates and radius of the circles
    if (len(cir) > 1):
        print("FAIL after")
        exit()

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

    # show the output image
    cv2.imshow("output", np.hstack([orig_image, output]))
    cv2.waitKey(0)

因此，如果您运行它，它需要 5 秒的时间，但它几乎达到了像素完美(自动调谐器的进一步手动调整使其达到亚像素完美):

上面的代码转换为:

对此:

使这项工作成功的秘诀在于您在开始之前拥有多少信息。如果您知道半径到某个公差(例如 20 像素)，那么您就完成了。但如果你不这样做，你必须聪明地知道你如何通过仔细接近决议和投票阈值来爬上最大选票的半径。如果圆圈形状怪异，则dp分辨率需要更高，投票阈值需要探索更低的范围。