Question

我正在学习OpenCv。我有一个斜齿轮图像来寻找牙齿。

到目前为止，我一直试图找到轮廓，然后计算牙齿。我能够找到轮廓以及轮廓的坐标。但是我坚持算数。由于我是OpenCV的新手，这可能是我尝试发现牙齿不正确的方式。

我的代码

import cv2
import numpy as np
import scipy as sp
import imutils
from skimage.morphology import reconstruction

import csv

raw_image = cv2.imread('./Gear Image/new1.jpg')
#cv2.imshow('Original Image', raw_image)
#cv2.waitKey(0)

bilateral_filtered_image = cv2.bilateralFilter(raw_image, 5, 175, 175)
#cv2.imshow('Bilateral', bilateral_filtered_image)
#cv2.waitKey(0)

edge_detected_image = cv2.Canny(bilateral_filtered_image, 75, 200)
#cv2.imshow('Edge', edge_detected_image)
#cv2.waitKey(0)



contours, hierarchy = cv2.findContours(edge_detected_image, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)





contour_list = []
for contour in contours:
    approx = cv2.approxPolyDP(contour,0.01*cv2.arcLength(contour,True),True)
    area = cv2.contourArea(contour)
    if ((len(approx) > 5) & (len(approx) < 25) & (area > 50) ):
        contour_list.append(contour)



cv2.drawContours(raw_image, contour_list,  -1, (255,0,0), 2)


c = max(contours, key = cv2.contourArea)
M = cv2.moments(c)

cX = int(M["m10"] / M["m00"])
cY = int(M["m01"] / M["m00"])

cv2.circle(raw_image, (cX, cY), 5, (142, 152, 100), -1)
cv2.putText(raw_image, "centroid", (cX - 25, cY - 25),cv2.FONT_HERSHEY_SIMPLEX, 0.5, (142, 152, 100), 2)

contour_length = "Number of contours detected: {}".format(len(contours))
cv2.putText(raw_image,contour_length , (20,40),  cv2.FONT_HERSHEY_SIMPLEX, 0.5, (142, 152, 100), 2)

for c in range(len(contours)):
        n_contour = contours[c]
        for d in range(len(n_contour)):
            XY_Coordinates = n_contour[d]


print(len(coordinates))
print(XY_Coordinates)
print(type(XY_Coordinates))
print(XY_Coordinates[0,[0]])
print(XY_Coordinates[0,[1]])



cv2.imshow('Objects Detected',raw_image)
cv2.waitKey(0)

输入图像：

我得到的输出图像：

在此阶段之后，我该如何计算牙齿？我可以使用坐标来计算时间间隔并计算牙齿。

或在此阶段之后还有其他计算牙齿的方法吗？谢谢。

Answer 1

我的解决方案的第一部分类似于@HansHirse发布的答案，但是我使用了另一种方法来计算牙齿。我的完整代码可以在这里找到：link to full code for python3 opencv4。在继续之前，请检查齿轮的外轮廓是否正确检测到。如果未正确检测到齿轮，则其余答案将不起作用。

在数齿之前，我先“解开”齿轮。我通过扫一下齿轮并计算从齿轮中心到牙齿外侧的距离来做到这一点。

这是我用来扫过齿轮并找到从齿轮中心到齿轮外部的距离的代码：

# Start at angle 0, and increment the angle 1/200 rad
angle = 0
increment = 1/200
# Create a list for the distances from the centroid to the edge of the gear tooth
distances = []
# Create an image for display purposes
display_image = raw_image.copy()
# Sweep around the circle (until one full revolution)
while angle < 2*math.pi:
    # Compute a ray from the center of the circle with the current angle
    img_size = max(raw_image.shape)
    ray_end = int(math.sin(angle) * img_size + cX), int(math.cos(angle) * img_size + cY)
    center = cX, cY
    # Create mask
    mask = np.zeros((raw_image.shape[0], raw_image.shape[1]), np.uint8)
    # Draw a line on the mask
    cv2.line(mask, center, ray_end, 255, 2)
    # Mask out the gear slice (this is the portion of the gear the us below the line)
    gear_slice = cv2.bitwise_and(raw_image, raw_image, mask = mask)
    # Threshold the image
    _, thresh = cv2.threshold(cv2.cvtColor(gear_slice, cv2.COLOR_BGR2GRAY), 0 , 255, 0)
    # Find the contours in the edge_slice
    _, edge_slice_contours, _ = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
    # Get the center of the edge slice contours
    M = cv2.moments(max(edge_slice_contours, key = cv2.contourArea))
    edge_location = int(M["m10"] / M["m00"]), int(M["m01"] / M["m00"])
    cv2.circle(display_image, edge_location, 0, (0,255,0), 4)
    # Find the distance from the center of the gear to the edge of the gear...at this specific angle
    edge_center_distance = distance(center, edge_location)
    # Find the xy coordinates for this point on the graph - draw blue circle
    graph_point = int(angle*0.5*raw_image.shape[1]/math.pi), int(edge_center_distance+ 1.5*gear_radius)
    cv2.circle(display_image, graph_point, 0, (0,255,0), 2)    
    # Add this distance to the list of distances
    distances.append(-edge_center_distance)
    # Create a temporary image and draw the ray on it
    temp = display_image.copy()
    cv2.line(temp, ray_end, (cX,cY), (0,0,255), 2)
    # Show the image and wait
    cv2.imshow('raw_image', temp)
    vid_writer.write(temp)
    k = cv2.waitKey(1)
    if k == 27: break
    # Increment the angle
    angle += increment
# Clean up
cv2.destroyAllWindows()

其结果是齿距齿轮中心的距离与角度成函数关系。

import matplotlib.pyplot as plt
plt.plot(distances)
plt.show()

现在计数牙齿要容易得多，因为它们是函数中的峰值（或者在这种情况下为山谷-稍后会详细介绍）。为了计算峰数，我把 Fourier transform的牙齿距离功能。

import scipy.fftpack
# Calculate the Fourier transform
yf = scipy.fftpack.fft(distances)
fig, ax = plt.subplots()
# Plot the relevant part of the Fourier transform (a gear will have between 2 and 200 teeth)
ax.plot(yf[2:200])
plt.show()

傅立叶变换的峰值出现在37。因此，有37个谷和38个齿轮齿。

num_teeth = list(yf).index(max(yf[2:200])) - 1
print('Number of teeth in this gear: ' + str(num_teeth))

Answer 2

也许以下解决方案适合您。

我在双边滤波后添加了一些轻微的中值模糊，以改善后续边缘检测（较小的边缘）。
在findContours中，我从RETR_TREE切换到RETR_EXTERNAL，仅获得了最外面的轮廓。
为此，我确定轮廓的凸包，并确保每个齿上只有一个凸包点。
这些“稀疏”凸包点的最终数量就是齿的数量。

（为了使答案简短，我删除了一些不必要的代码。）

import cv2
import numpy as np

raw_image = cv2.imread('images/vChAL.jpg')

bilateral_filtered_image = cv2.bilateralFilter(raw_image, 5, 175, 175)

# Added median blurring to improve edge detection
median_blurred_images = cv2.medianBlur(bilateral_filtered_image, 5)

edge_detected_image = cv2.Canny(median_blurred_images, 75, 200)

# Switched from RETR_TREE to RETR_EXTERNAL to only extract most outer contours
contours, _ = cv2.findContours(edge_detected_image, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)

contour_list = []
for contour in contours:
    approx = cv2.approxPolyDP(contour,0.01*cv2.arcLength(contour,True),True)
    area = cv2.contourArea(contour)
    if ((len(approx) > 5) & (len(approx) < 25) & (area > 50) ):
        contour_list.append(contour)

cv2.drawContours(raw_image, contour_list, -1, (255, 0, 0), 2)

c = max(contours, key = cv2.contourArea)

contour_length = "Number of contours detected: {}".format(len(contours))
cv2.putText(raw_image,contour_length , (20, 40),  cv2.FONT_HERSHEY_SIMPLEX, 0.5, (142, 152, 100), 2)

# Determine convex hull of largest contour
hull = cv2.convexHull(c, clockwise = True, returnPoints = False)

# Debug: Draw "raw" convex hull points (green)
cv2.drawContours(raw_image, c[hull], -1, (0, 255, 0), 3)

# Determine convex hull, such that nearby convex hull points are "grouped"
sparsehull = []
for idx in hull:
    if (len(sparsehull) == 0):
        sparsehull.append(idx)
    else:
        last = sparsehull[-1]
        diff = c[idx] - c[last]
        if (cv2.norm(diff) > 40):
            sparsehull.append(idx)
sparsehull = np.asarray(sparsehull)

# Debug: Draw "sparse2 convex hull points (red)
cv2.drawContours(raw_image, c[sparsehull], -1, (0, 0, 255), 3)

# Additional output on image
teeth_length = "Number of teeth detected: {}".format(len(sparsehull))
cv2.putText(raw_image, teeth_length , (20, 60), cv2.FONT_HERSHEY_SIMPLEX, 0.5, (142, 152, 100), 2)

cv2.imshow('Objects Detected', raw_image)
cv2.waitKey(0)

免责声明：我是Python的新手，尤其是OpenCV（胜利的C ++）的Python API。非常欢迎评论，改进，强调Python的执行！！

通过python opencv查找齿轮齿

2 个答案: