我一直试图在正方形周围绘制 4 条线,以便获得正方形的顶点.由于准确性,我将采用这种方法,而不是直接使用 Harris 或轮廓方法寻找角点.在opencv的内置函数中使用houghlines我无法获得全长线来获得交点,而且我也得到了太多不相关的线.我想知道是否可以微调参数以满足我的要求?如果是,我该怎么做?我的问题与
代码:
#include #include "opencv2highgui.hpp"#include "opencv2imgproc.hpp"#include "opencv2/imgcodecs/imgcodecs.hpp"#include "opencv2/videoio/videoio.hpp"使用命名空间 cv;使用命名空间标准;int main(int argc, const char** argv){垫子图像,src;image = imread("c:/pics/output2_1.bmp");src = image.clone();cvtColor(图像,图像,CV_BGR2GRAY);阈值(图像,图像,0, 255,CV_THRESH_OTSU + CV_THRESH_BINARY_INV);namedWindow("thresh", WINDOW_NORMAL);resizeWindow("thresh", 600, 400);imshow("thresh", image);cv::Mat 边缘;cv::Canny(图像,边缘,0, 255);向量<Vec2f>线条;HoughLines(edges, lines, 1, CV_PI/180, 100, 0, 0);for (size_t i = 0; i
输出图像:
更新:此图像上有一个框架,因此我能够通过删除该框架来减少图像边框中不相关的线条,但是我仍然没有获得覆盖正方形的完整线条.
有很多种方法可以做到这一点,我举一个例子.但是,我在 python
中最快,所以我的代码示例将使用该语言.不过,翻译它应该不难(在您为他人完成后,请随时使用您的 C++ 解决方案编辑您的帖子).
对于预处理,我强烈建议
如果您打算使用 HoughLinesP
,那么您的输出将是线 segments,您所拥有的只是线上的两个点.
由于线条大多是垂直和水平的,您可以根据它们的位置轻松地分割线条.如果一条线的两个 y 坐标彼此靠近,则该线大部分是水平的.如果两个 x 坐标彼此靠近,则该线大部分是垂直的.因此,您可以通过这种方式将线条分割为垂直线和水平线.
def segment_lines(lines, delta):h_lines = []v_lines = []对于线中线:对于 x1, y1, x2, y2 在线:如果 abs(x2-x1)
然后,您可以从端点获取两条线段的交点
然而,这些都在一个向量中,所以你不仅需要平均每个角的点,你还需要将它们组合在一起.您可以使用 k-means 聚类来实现这一点,它在 OpenCV 中实现为
我们有;四个点,在盒子的角落.
这是我用python编写的完整代码,包括生成上图的代码:
导入 cv2将 numpy 导入为 npdef find_intersection(line1, line2):# 提取点x1, y1, x2, y2 = line1[0]x3, y3, x4, y4 = line2[0]# 计算行列式Px = ((x1*y2 - y1*x2)*(x3-x4) - (x1-x2)*(x3*y4 - y3*x4))/((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))Py = ((x1*y2 - y1*x2)*(y3-y4) - (y1-y2)*(x3*y4 - y3*x4))/((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))返回 Px, Pydef segment_lines(线,增量):h_lines = []v_lines = []对于线中线:对于 x1, y1, x2, y2 在线:如果 abs(x2-x1)
<小时>
最后,只有一个问题...为什么不使用
这是使用以下代码生成的:
导入 cv2将 numpy 导入为 npimg = cv2.imread('image.png')# 预处理img = cv2.resize(img, None, fx=.5, fy=.5)灰色 = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)r, gray = cv2.threshold(gray, 120, 255, type=cv2.THRESH_BINARY)灰色 = cv2.GaussianBlur(gray, (3,3), 3)# 运行哈里斯灰色 = np.float32(灰色)dst = cv2.cornerHarris(gray,2,3,0.04)# 扩大角点进行标记dst = cv2.dilate(dst,None)dst = cv2.dilate(dst,None)# 临界点img[dst>0.01*dst.max()]=[0,0,255]cv2.imshow('dst',img)cv2.waitKey(0)cv2.imwrite('harris.png', img)
我认为通过一些细微的调整,Harris 角点检测器可能比外推霍夫线交叉点更准确.
I've been trying to get 4 lines around the square so that I can obtain the vertices of the square. I'm going with this approach rather than finding corners directly using Harris or contours method due to accuracy. Using houghlines in built function in opencv I'm unable to get full length lines to get intersection points and I'm also getting too many irrelevant lines. I'd like to know if the parameters can be fine tuned to obtain my requirements? If yes how do I go about it? My question is exactly the same as this one here. However I'm not getting those lines itself even after changing those parameters. I've attached the original image along with the code and output:
Original Image:
Code:
#include <Windows.h>
#include "opencv2highgui.hpp"
#include "opencv2imgproc.hpp"
#include "opencv2/imgcodecs/imgcodecs.hpp"
#include "opencv2/videoio/videoio.hpp"
using namespace cv;
using namespace std;
int main(int argc, const char** argv)
{
Mat image,src;
image = imread("c:/pics/output2_1.bmp");
src = image.clone();
cvtColor(image, image, CV_BGR2GRAY);
threshold(image, image, 0, 255, CV_THRESH_OTSU + CV_THRESH_BINARY_INV);
namedWindow("thresh", WINDOW_NORMAL);
resizeWindow("thresh", 600, 400);
imshow("thresh", image);
cv::Mat edges;
cv::Canny(image, edges, 0, 255);
vector<Vec2f> lines;
HoughLines(edges, lines, 1, CV_PI / 180, 100, 0, 0);
for (size_t i = 0; i < lines.size(); i++)
{
float rho = lines[i][0], theta = lines[i][1];
Point pt1, pt2;
double a = cos(theta), b = sin(theta);
double x0 = a*rho, y0 = b*rho;
pt1.x = cvRound(x0 + 1000 * (-b));
pt1.y = cvRound(y0 + 1000 * (a));
pt2.x = cvRound(x0 - 1000 * (-b));
pt2.y = cvRound(y0 - 1000 * (a));
line(src, pt1, pt2, Scalar(0, 0, 255), 3, CV_AA);
}
namedWindow("Edges Structure", WINDOW_NORMAL);
resizeWindow("Edges Structure", 600, 400);
imshow("Edges Structure", src);
waitKey(0);
return(0);
}
Output Image:
Update: There is a frame on this image, so I was able to reduce the irrelevant lines in the border of the image by removing that frame, however I'm still not getting complete lines covering the square.
There are many ways to do this, I will give an example of just one. However, I'm quickest in python
, so my code example will be in that language. Should not be hard to translate it, though (please feel free to edit your post with your C++ solution after you've finished it for others).
For preprocessing, I highly suggest dilate()
ing your edge image. This will make the lines thicker which will help fit the Hough lines better. What the Hough lines function does in the abstract is basically make a grid of lines passing through a ton of angles and distances, and if the lines go over any white pixels from Canny, then it gives that line a score for each point it goes through. However, the lines from Canny won't be perfectly straight, so you'll get a few different lines scoring. Making those Canny lines thicker will mean each line that is really close to fitting well will have better chances of scoring higher.
If you're going to use HoughLinesP
, then your output will be line segments, where all you have is two points on the line.
Since the lines are mostly vertical and horizontal, you can easily split the lines based on their position. If the two y-coordinates of one line are near each other, then the line is mostly horizontal. If the two x-coordinates are near each other, then the line is mostly vertical. So you can segment your lines into vertical lines and horizontal lines that way.
def segment_lines(lines, delta):
h_lines = []
v_lines = []
for line in lines:
for x1, y1, x2, y2 in line:
if abs(x2-x1) < delta: # x-values are near; line is vertical
v_lines.append(line)
elif abs(y2-y1) < delta: # y-values are near; line is horizontal
h_lines.append(line)
return h_lines, v_lines
Then, you can obtain intersection points of two line segments from their endpoints using determinants.
def find_intersection(line1, line2):
# extract points
x1, y1, x2, y2 = line1[0]
x3, y3, x4, y4 = line2[0]
# compute determinant
Px = ((x1*y2 - y1*x2)*(x3-x4) - (x1-x2)*(x3*y4 - y3*x4))/
((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
Py = ((x1*y2 - y1*x2)*(y3-y4) - (y1-y2)*(x3*y4 - y3*x4))/
((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
return Px, Py
So now if you loop through all your lines, you'll have intersection points from all your horizontal and vertical lines, but you have many lines, so you'll have many intersection points for the same corner of the box.
However, these are all in one vector, so not only do you need to average the points in each corner, you need to actually group them together, too. You can achieve this with k-means clustering, which is implemented in OpenCV as kmeans()
.
def cluster_points(points, nclusters):
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
_, _, centers = cv2.kmeans(points, nclusters, None, criteria, 10, cv2.KMEANS_PP_CENTERS)
return centers
Finally, we can simply plot those centers (making sure we round first---since so far everything is a float) onto the image with circle()
to make sure we've done it right.
And we have it; four points, at the corners of the box.
Here's my full code in python, including the code to generate the figures above:
import cv2
import numpy as np
def find_intersection(line1, line2):
# extract points
x1, y1, x2, y2 = line1[0]
x3, y3, x4, y4 = line2[0]
# compute determinant
Px = ((x1*y2 - y1*x2)*(x3-x4) - (x1-x2)*(x3*y4 - y3*x4))/
((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
Py = ((x1*y2 - y1*x2)*(y3-y4) - (y1-y2)*(x3*y4 - y3*x4))/
((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
return Px, Py
def segment_lines(lines, delta):
h_lines = []
v_lines = []
for line in lines:
for x1, y1, x2, y2 in line:
if abs(x2-x1) < delta: # x-values are near; line is vertical
v_lines.append(line)
elif abs(y2-y1) < delta: # y-values are near; line is horizontal
h_lines.append(line)
return h_lines, v_lines
def cluster_points(points, nclusters):
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
_, _, centers = cv2.kmeans(points, nclusters, None, criteria, 10, cv2.KMEANS_PP_CENTERS)
return centers
img = cv2.imread('image.png')
# preprocessing
img = cv2.resize(img, None, fx=.5, fy=.5)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray, 50, 150)
dilated = cv2.dilate(edges, np.ones((3,3), dtype=np.uint8))
cv2.imshow("Dilated", dilated)
cv2.waitKey(0)
cv2.imwrite('dilated.png', dilated)
# run the Hough transform
lines = cv2.HoughLinesP(dilated, rho=1, theta=np.pi/180, threshold=100, maxLineGap=20, minLineLength=50)
# segment the lines
delta = 10
h_lines, v_lines = segment_lines(lines, delta)
# draw the segmented lines
houghimg = img.copy()
for line in h_lines:
for x1, y1, x2, y2 in line:
color = [0,0,255] # color hoz lines red
cv2.line(houghimg, (x1, y1), (x2, y2), color=color, thickness=1)
for line in v_lines:
for x1, y1, x2, y2 in line:
color = [255,0,0] # color vert lines blue
cv2.line(houghimg, (x1, y1), (x2, y2), color=color, thickness=1)
cv2.imshow("Segmented Hough Lines", houghimg)
cv2.waitKey(0)
cv2.imwrite('hough.png', houghimg)
# find the line intersection points
Px = []
Py = []
for h_line in h_lines:
for v_line in v_lines:
px, py = find_intersection(h_line, v_line)
Px.append(px)
Py.append(py)
# draw the intersection points
intersectsimg = img.copy()
for cx, cy in zip(Px, Py):
cx = np.round(cx).astype(int)
cy = np.round(cy).astype(int)
color = np.random.randint(0,255,3).tolist() # random colors
cv2.circle(intersectsimg, (cx, cy), radius=2, color=color, thickness=-1) # -1: filled circle
cv2.imshow("Intersections", intersectsimg)
cv2.waitKey(0)
cv2.imwrite('intersections.png', intersectsimg)
# use clustering to find the centers of the data clusters
P = np.float32(np.column_stack((Px, Py)))
nclusters = 4
centers = cluster_points(P, nclusters)
print(centers)
# draw the center of the clusters
for cx, cy in centers:
cx = np.round(cx).astype(int)
cy = np.round(cy).astype(int)
cv2.circle(img, (cx, cy), radius=4, color=[0,0,255], thickness=-1) # -1: filled circle
cv2.imshow("Center of intersection clusters", img)
cv2.waitKey(0)
cv2.imwrite('corners.png', img)
Finally, just one question...why not use the Harris corner detector implemented in OpenCV as cornerHarris()
? Because it works really well with very minimal code. I thresholded the grayscale image, and then gave a little blur to remove spurious corners, and, well...
This was produced with the following code:
import cv2
import numpy as np
img = cv2.imread('image.png')
# preprocessing
img = cv2.resize(img, None, fx=.5, fy=.5)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
r, gray = cv2.threshold(gray, 120, 255, type=cv2.THRESH_BINARY)
gray = cv2.GaussianBlur(gray, (3,3), 3)
# run harris
gray = np.float32(gray)
dst = cv2.cornerHarris(gray,2,3,0.04)
# dilate the corner points for marking
dst = cv2.dilate(dst,None)
dst = cv2.dilate(dst,None)
# threshold
img[dst>0.01*dst.max()]=[0,0,255]
cv2.imshow('dst',img)
cv2.waitKey(0)
cv2.imwrite('harris.png', img)
I think with some minor adjustments the Harris corner detector can probably be much more accurate than extrapolating Hough line intersections.
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