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使用 opencv 和 python 的 HoughCircles 圓檢測-

[英]HoughCircles circle detection using opencv and python-

我正在嘗試使用 OpenCV 的 (Hough)Circle 檢測來檢測圓圈。 我在黑色背景上創建了一個實心圓圈,嘗試使用參數,使用模糊和所有內容,但我無法讓它找到任何東西。

任何想法,建議等都會很棒,謝謝!

我目前的代碼是這樣的:

import cv2
import numpy as np

"""
params = dict(dp=1,
              minDist=1,
              circles=None,
              param1=300,
              param2=290,
              minRadius=1,
              maxRadius=100)
"""

img = np.ones((200,250,3), dtype=np.uint8)
for i in range(50, 80, 1):
    for j in range(40, 70, 1):
        img[i][j]*=200

cv2.circle(img, (120,120), 20, (100,200,80), -1)


gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
canny = cv2.Canny(gray, 200, 300)

cv2.imshow('shjkgdh', canny)
gray = cv2.medianBlur(gray, 5)
circles = cv2.HoughCircles(gray, cv2.cv.CV_HOUGH_GRADIENT, 1, 20,
              param1=100,
              param2=30,
              minRadius=0,
              maxRadius=0)

print circles
circles = np.uint16(np.around(circles))
for i in circles[0,:]:
    cv2.circle(img,(i[0],i[1]),i[2],(0,255,0),2)
    cv2.circle(img,(i[0],i[1]),2,(0,0,255),3)

cv2.imshow('circles', img)
k = cv2.waitKey(0)
if k == 27:
    cv2.destroyAllWindows()

你的代碼工作得很好。 問題在於您的HoughCircles閾值參數。

讓我們嘗試從OpenCV Docs了解您使用的參數:

param1 – 第一個特定於方法的參數。 在 CV_HOUGH_GRADIENT 的情況下,它是傳遞給 Canny() 邊緣檢測器的兩個閾值中較高的閾值(較低的閾值小兩倍)。

param2 – 第二個特定於方法的參數。 在 CV_HOUGH_GRADIENT 的情況下,它是檢測階段圓心的累加器閾值。 它越小,檢測到的假圓就越多。 與較大累加器值相對應的圓圈將首先返回。

因此,如您所見,HoughCircles 函數在內部調用 Canny 邊緣檢測器,這意味着您可以在函數中使用灰度圖像,而不是它們的輪廓。

現在將param1減少到 30,將param2減少到 15,並在下面的代碼中查看結果:

import cv2
import numpy as np

img = np.ones((200,250,3), dtype=np.uint8)
for i in range(50, 80, 1):
    for j in range(40, 70, 1):
        img[i][j]*=200

cv2.circle(img, (120,120), 20, (100,200,80), -1)

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

circles = cv2.HoughCircles(gray, cv2.cv.CV_HOUGH_GRADIENT, 1, 20,
              param1=30,
              param2=15,
              minRadius=0,
              maxRadius=0)

print circles
circles = np.uint16(np.around(circles))
for i in circles[0,:]:
    cv2.circle(img,(i[0],i[1]),i[2],(0,255,0),2)
    cv2.circle(img,(i[0],i[1]),2,(0,0,255),3)

cv2.imshow('circles', img)

k = cv2.waitKey(0)
if k == 27:
    cv2.destroyAllWindows()

霍夫圈子

如果您沒有讓 HoughCircles 為您提供針對明顯圓圈的像素完美解決方案,那么您就沒有正確使用它

你的錯誤是你試圖自己手動調整你的超參數。 那是行不通的。 讓計算機為您自動調整參數:

import numpy as np
import argparse
import cv2
import signal

from functools import wraps
import errno
import os
import copy

ap = argparse.ArgumentParser()
ap.add_argument("-i", "--image", required = True, help = "Path to the image")
args = vars(ap.parse_args())

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

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.cv.CV_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")

    for (x, y, r) in cir:
        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)

    cv2.imshow("output", np.hstack([orig_image, output]))
    cv2.waitKey(0)

上面的代碼對此進行了轉換: 原來的

對此:

霍夫圈子

有關此操作的更多信息,請參閱: https : //stackoverflow.com/a/46500223/445131

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