如何检测opencv中的行？

Question

我试图检测停车线,如下所示

我希望得到的是交叉线上清晰的线条和(x,y)位置,但结果并不是很有希望

我想这是由于两个主要原因

1. 有些线条非常破碎或缺失甚至人眼都可以清楚地识别它们.(即使HoughLine可以帮助连接一些丢失的行,因为HoughLine有时会将不必要的行连接在一起,所以我宁愿手动完成它)

2. 有一些重复的线条

工作的一般管道如下所示

1.选择一些特定的颜色(白色或黄色)

import cv2
import numpy as np
import matplotlib
from matplotlib.pyplot import imshow
from matplotlib import pyplot as plt

# white color mask
img = cv2.imread(filein)
#converted = convert_hls(img)
image = cv2.cvtColor(img,cv2.COLOR_BGR2HLS)
lower = np.uint8([0, 200, 0])
upper = np.uint8([255, 255, 255])
white_mask = cv2.inRange(image, lower, upper)
# yellow color mask
lower = np.uint8([10, 0,   100])
upper = np.uint8([40, 255, 255])
yellow_mask = cv2.inRange(image, lower, upper)
# combine the mask
mask = cv2.bitwise_or(white_mask, yellow_mask)
result = img.copy()
cv2.imshow("mask",mask) 

2.重复扩张和侵蚀,直到图像无法更改(参考)

height,width = mask.shape
skel = np.zeros([height,width],dtype=np.uint8)      #[height,width,3]
kernel = cv2.getStructuringElement(cv2.MORPH_CROSS, (3,3))
temp_nonzero = np.count_nonzero(mask)
while(np.count_nonzero(mask) != 0 ):
    eroded = cv2.erode(mask,kernel)
    cv2.imshow("eroded",eroded)   
    temp = cv2.dilate(eroded,kernel)
    cv2.imshow("dilate",temp)
    temp = cv2.subtract(mask,temp)
    skel = cv2.bitwise_or(skel,temp)
    mask = eroded.copy()

cv2.imshow("skel",skel)
#cv2.waitKey(0)

3.应用canny过滤线条并使用HoughLinesP获取线条

edges = cv2.Canny(skel, 50, 150)
cv2.imshow("edges",edges)
lines = cv2.HoughLinesP(edges,1,np.pi/180,40,minLineLength=30,maxLineGap=30)
i = 0
for x1,y1,x2,y2 in lines[0]:
    i+=1
    cv2.line(result,(x1,y1),(x2,y2),(255,0,0),1)
print i

cv2.imshow("res",result)
cv2.waitKey(0)

我想知道在选择某种颜色的第一步之后,线条会有破碎和噪音,我认为在这一步中我们应该做一些事情来使断线完整,不那么嘈杂的线条,然后尝试应用一些东西来做Canny和Hough线,任何想法？

Answer 1

这是我的管道,也许它可以给你一些帮助.

首先,获取灰度图像并处理GaussianBlur.

img = cv2.imread('src.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)

kernel_size = 5
blur_gray = cv2.GaussianBlur(gray,(kernel_size, kernel_size),0)

其次,进程边缘检测使用Canny.

low_threshold = 50
high_threshold = 150
edges = cv2.Canny(blur_gray, low_threshold, high_threshold)

然后,使用HoughLinesP来获取线条.您可以调整参数以获得更好的性能.

rho = 1  # distance resolution in pixels of the Hough grid
theta = np.pi / 180  # angular resolution in radians of the Hough grid
threshold = 15  # minimum number of votes (intersections in Hough grid cell)
min_line_length = 50  # minimum number of pixels making up a line
max_line_gap = 20  # maximum gap in pixels between connectable line segments
line_image = np.copy(img) * 0  # creating a blank to draw lines on

# Run Hough on edge detected image
# Output "lines" is an array containing endpoints of detected line segments
lines = cv2.HoughLinesP(edges, rho, theta, threshold, np.array([]),
                    min_line_length, max_line_gap)

for line in lines:
    for x1,y1,x2,y2 in line:
    cv2.line(line_image,(x1,y1),(x2,y2),(255,0,0),5)

最后,在srcImage上绘制线条.

# Draw the lines on the  image
lines_edges = cv2.addWeighted(img, 0.8, line_image, 1, 0)

这是我最后的表演.

最终图片:

Answer 2

我不确定你究竟在问什么,因为你的帖子毫无疑问.

检测线段的一种不错且强大的技术是LSD(线段检测器),自openCV 3开始在openCV中可用.

这里有一些简单的基本C++代码,可以很容易地转换为python:

int main(int argc, char* argv[])
{
    cv::Mat input = cv::imread("C:/StackOverflow/Input/parking.png");
    cv::Mat gray;
    cv::cvtColor(input, gray, CV_BGR2GRAY);


    cv::Ptr<cv::LineSegmentDetector> det;
    det = cv::createLineSegmentDetector();



    cv::Mat lines;
    det->detect(gray, lines);

    det->drawSegments(input, lines);

    cv::imshow("input", input);
    cv::waitKey(0);
    return 0;
}

给出这个结果:

对于进一步处理而言,这比图像看起来更好(没有行重复等)

Answer 3

这里有一些很好的答案,你的问题的第一部分,但至于第二部分(找到线交叉点),我没有看到很多.

我建议你看看Bentley-Ottmann算法.

这里和这里有一些算法的python实现.

编辑:使用VeraPoseidon的Houghlines实现和此处链接的第二个库,我设法获得以下交叉检测结果.感谢Vera和图书馆作者的出色工作.绿色方块表示检测到的交叉点.有一些错误,但这对我来说似乎是一个非常好的起点.似乎您实际想要检测交叉点的大多数位置都检测到多个交叉点,因此您可以在图像上运行适当大小的窗口,该窗口查找多个交叉点,并将该交叉点视为该窗口激活的交叉点.

这是我用来产生结果的代码:

import cv2
import numpy as np
import isect_segments_bentley_ottmann.poly_point_isect as bot


img = cv2.imread('parking.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)

kernel_size = 5
blur_gray = cv2.GaussianBlur(gray,(kernel_size, kernel_size),0)

low_threshold = 50
high_threshold = 150
edges = cv2.Canny(blur_gray, low_threshold, high_threshold)

rho = 1  # distance resolution in pixels of the Hough grid
theta = np.pi / 180  # angular resolution in radians of the Hough grid
threshold = 15  # minimum number of votes (intersections in Hough grid cell)
min_line_length = 50  # minimum number of pixels making up a line
max_line_gap = 20  # maximum gap in pixels between connectable line segments
line_image = np.copy(img) * 0  # creating a blank to draw lines on

# Run Hough on edge detected image
# Output "lines" is an array containing endpoints of detected line segments
lines = cv2.HoughLinesP(edges, rho, theta, threshold, np.array([]),
                    min_line_length, max_line_gap)
print(lines)
points = []
for line in lines:
    for x1, y1, x2, y2 in line:
        points.append(((x1 + 0.0, y1 + 0.0), (x2 + 0.0, y2 + 0.0)))
        cv2.line(line_image, (x1, y1), (x2, y2), (255, 0, 0), 5)

lines_edges = cv2.addWeighted(img, 0.8, line_image, 1, 0)
print(lines_edges.shape)
#cv2.imwrite('line_parking.png', lines_edges)

print points
intersections = bot.isect_segments(points)
print intersections

for inter in intersections:
    a, b = inter
    for i in range(3):
        for j in range(3):
            lines_edges[int(b) + i, int(a) + j] = [0, 255, 0]

cv2.imwrite('line_parking.png', lines_edges)

您可以使用类似此代码块的方法来删除小区域中的多个交叉点:

for idx, inter in enumerate(intersections):
    a, b = inter
    match = 0
    for other_inter in intersections[idx:]:
        c, d = other_inter
        if abs(c-a) < 15 and abs(d-b) < 15:
            match = 1
            intersections[idx] = ((c+a)/2, (d+b)/2)
            intersections.remove(other_inter)

    if match == 0:
        intersections.remove(inter)

输出图像:

你必须使用窗口功能.

Answer 4

如果调整 maxLineGap 或侵蚀内核的大小会发生什么。或者，您可以找到线之间的距离。您必须遍历成对的线，例如 ax1,ay1 到 ax2,ay2 cf bx1,by1 到 bx2,by2 您可以找到与线 b 成直角（-1 超过线的梯度）的梯度与线 b 相交的点。基础学校几何和联立方程，例如：

x = (ay1 - by1) / ((by2 - by1) / (bx2 - bx1) + (ax2 - ax1) / (ay2 - ay1))
# then
y = by1 + x * (by2 - by1) / (bx2 - bx1)

并将 x,y 与 ax1,ay1 进行比较

PS，您可能需要添加对 ax1,ay1 和 bx1,by1 之间距离的检查，因为您的某些线看起来是其他线的延续，并且这些线可能会被最近点技术消除。