立方体边缘检测

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英文:

Cube Edge detection

问题

我试图使用Scikit-Image库中的Hough变换获取下面图像中立方体的边缘。

更新
这是我正在使用的代码:

smoothed_image = filters.frangi(gray_image)

# 使用Canny边缘检测器进行边缘检测
低阈值 = 0.1
高阈值 = 3
低阈值 = 低 * smoothed_image.max()
高阈值 = 高 * 低阈值
edges = canny(smoothed_image, sigma, low_threshold=低阈值,
             high_threshold=高阈值)


# 使用Hough变换检测丝状物轮廓
hspace, theta, dist = hough_line(edges)

# 找到垂直线
vertical_peaks = hough_line_peaks(hspace, theta, dist, num_peaks=2)

vertical_lines = []

for _, angle, dist in zip(*vertical_peaks):
    x = dist * np.cos(angle)
    y = dist * np.sin(angle)
    vertical_lines.append((x, y))


# 可视化结果
fig, axes = plt.subplots(1, 3, figsize=(15, 5))
axes[0].imshow(smoothed_image)
axes[0].set_title('平滑后的图像')
#axes[0].axis('off')
axes[1].imshow(edges, cmap='gray')
axes[1].set_title('Canny边缘检测')
axes[1].axis('off')
axes[2].imshow(edges, cmap='gray')
for x, y in vertical_lines:
    axes[2].axvline(x=x, color='red')
axes[2].set_xlim((0, image.shape[1]))
axes[2].set_ylim((image.shape[0], 0))
axes[2].set_title('检测到的线条')
axes[2].axis('off')
plt.tight_layout()
plt.show()

这是生成的输出:
立方体边缘检测

有关如何使Hough变换检测到另一条边的任何想法吗?

英文:

I am trying to get the edges of a cube in the image below using the Hough transform in the Scikit-Image library 立方体边缘检测.

Update
Here is the code I am working with:

smoothed_image = filters.frangi(gray_image)

# Perform edge detection using the Canny edge detector
low_threshold = 0.1  
high_threshold = 3  
low_threshold = low * smoothed_image.max()
high_threshold = high * low_threshold
edges = canny(smoothed_image, sigma, low_threshold=low_threshold,
             high_threshold=high_threshold)


# Hough transform to detect the filament profile
hspace, theta, dist = hough_line(edges)

# Find the vertical lines
vertical_peaks = hough_line_peaks(hspace, theta, dist, num_peaks=2)

vertical_lines = []

for _, angle, dist in zip(*vertical_peaks):
    x = dist * np.cos(angle)
    y = dist * np.sin(angle)
    vertical_lines.append((x, y))


# Visualize the results
fig, axes = plt.subplots(1, 3, figsize=(15, 5))
axes[0].imshow(smoothed_image)
axes[0].set_title('Smoothed Image')
#axes[0].axis('off')
axes[1].imshow(edges, cmap='gray')
axes[1].set_title('Canny Edge Detection')
axes[1].axis('off')
axes[2].imshow(edges, cmap='gray')
for x, y in vertical_lines:
    axes[2].axvline(x=x, color='red')
axes[2].set_xlim((0, image.shape[1]))
axes[2].set_ylim((image.shape[0], 0))
axes[2].set_title('Detected Lines')
axes[2].axis('off')
plt.tight_layout()
plt.show()

And this is the resulting output
立方体边缘检测

Any ideas on how to make the Hough transform detect the other edge?

答案1

得分: 0

如@Pete建议的,"遮住管子",因为它挡住了立方体,并调整以下参数:

  1. Canny 参数
  2. 高斯滤波器 参数
  3. 霍夫直线 检测器参数

这样你就可以检测到另一条边。

P.S:我使用了 OpenCV,但你也可以使用 Scikit Image 重复这些结果。

#!/usr/bin/env python3

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

im_path = "cube.png"

img = cv2.imread(im_path)
  
# 将图像转换为灰度
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

# 高斯滤波
k = 5
sigma = 0

gauss = cv2.GaussianBlur(gray, (k, k), sigma)

# 在图像上应用边缘检测方法
edges = cv2.Canny(gauss, 50, 150, apertureSize=3)

# 移除不需要的管子
# https://stackoverflow.com/a/64175996/8618242
sum_ = np.sum(edges, axis=1)
mean_ = np.mean(sum_)
edges[sum_ > mean_, :] = 0

# 这将返回一个r和theta值的数组
lines = cv2.HoughLines(edges, 1, np.pi/180, 150)
  
# 以下for循环在r和theta值在2D数组的范围内运行
for r_theta in lines:
    arr = np.array(r_theta[0], dtype=np.float64)
    r, theta = arr
    # 在a中存储cos(theta)的值
    a = np.cos(theta)
  
    # 在b中存储sin(theta)的值
    b = np.sin(theta)
  
    # x0存储rcos(theta)的值
    x0 = a*r
  
    # y0存储rsin(theta)的值
    y0 = b*r
  
    # x1存储(rcos(theta)-1000sin(theta))的四舍五入值
    x1 = int(x0 + 1000*(-b))
  
    # y1存储(rsin(theta)+1000cos(theta))的四舍五入值
    y1 = int(y0 + 1000*(a))
  
    # x2存储(rcos(theta)+1000sin(theta))的四舍五入值
    x2 = int(x0 - 1000*(-b))
  
    # y2存储(rsin(theta)-1000cos(theta))的四舍五入值
    y2 = int(y0 - 1000*(a))
  
    # cv2.line在图像中从点(x1, y1)到(x2, y2)绘制一条线。
    # (0,0,255)表示要绘制的线的颜色。在这种情况下,是红色。
    cv2.line(img, (x1, y1), (x2, y2), (0, 0, 255), 2)

cv2.namedWindow("output", cv2.WINDOW_NORMAL)
cv2.imshow("output", img)
cv2.waitKey(0)

立方体边缘检测

英文:

As suggested by @Pete "Mask out the tube" as it is occluding the cube, and adjust the parameters:

  1. Canny parameters
  2. Gaussian Filter parameters
  3. Hough Line detector parameters

so you can detect the other edge.

> P.S: I used OpenCV, but you might repeat the results using Scikit Image

#!/usr/bin/env python3
import cv2 
import numpy as np 
import matplotlib.pyplot as plt 
im_path = "cube.png"
img = cv2.imread(im_path)
# Convert the img to grayscale
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# Gaussian Filter
k = 5
sigma = 0
gauss = cv2.GaussianBlur(gray,(k,k),sigma)
# Apply edge detection method on the image
edges = cv2.Canny(gauss, 50, 150, apertureSize=3)
# Remove unwanted tube
# https://stackoverflow.com/a/64175996/8618242
sum_ = np.sum(edges, axis=1)
mean_ = np.mean(sum_)
edges[sum_> mean_, :] = 0
# This returns an array of r and theta values
lines = cv2.HoughLines(edges, 1, np.pi/180, 150)
# The below for loop runs till r and theta values
# are in the range of the 2d array
for r_theta in lines:
arr = np.array(r_theta[0], dtype=np.float64)
r, theta = arr
# Stores the value of cos(theta) in a
a = np.cos(theta)
# Stores the value of sin(theta) in b
b = np.sin(theta)
# x0 stores the value rcos(theta)
x0 = a*r
# y0 stores the value rsin(theta)
y0 = b*r
# x1 stores the rounded off value of (rcos(theta)-1000sin(theta))
x1 = int(x0 + 1000*(-b))
# y1 stores the rounded off value of (rsin(theta)+1000cos(theta))
y1 = int(y0 + 1000*(a))
# x2 stores the rounded off value of (rcos(theta)+1000sin(theta))
x2 = int(x0 - 1000*(-b))
# y2 stores the rounded off value of (rsin(theta)-1000cos(theta))
y2 = int(y0 - 1000*(a))
# cv2.line draws a line in img from the point(x1,y1) to (x2,y2).
# (0,0,255) denotes the colour of the line to be
# drawn. In this case, it is red.
cv2.line(img, (x1, y1), (x2, y2), (0, 0, 255), 2)
cv2.namedWindow("output", cv2.WINDOW_NORMAL)
cv2.imshow("output", img)
cv2.waitKey(0)

立方体边缘检测

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  • 本文由 发表于 2023年7月17日 23:52:03
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