计算轮廓上特定部分/边/线的斜率、长度和角度？
Calculate slope, length and angle of a specific part / side / line on a contour?
我在图像中检测到两个轮廓，需要顶部轮廓的两个垂直边缘之间的直径和下部轮廓的垂直边缘之间的直径。我用这段代码实现了这一点。
import cv2
import numpy as np
import math, os
import imutils
img = cv2.imread("1.jpg")
gray = cv2.cvtColor(img, cv2.COLOR_RGB2GRAY)
gray = cv2.GaussianBlur(gray, (7, 7), 0)
edges = cv2.Canny(gray, 200, 100)
edges = cv2.dilate(edges, None, iterations=1)
edges = cv2.erode(edges, None, iterations=1)
cnts = cv2.findContours(edges.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = imutils.grab_contours(cnts)
# sorting the contours to find the largest and smallest one
c1 = max(cnts, key=cv2.contourArea)
c2 = min(cnts, key=cv2.contourArea)
# determine the most extreme points along the contours
extLeft1 = tuple(c1[c1[:, :, 0].argmin()][0])
extRight1 = tuple(c1[c1[:, :, 0].argmax()][0])
extLeft2 = tuple(c2[c2[:, :, 0].argmin()][0])
extRight2 = tuple(c2[c2[:, :, 0].argmax()][0])
# show contour
cimg = cv2.drawContours(img, cnts, -1, (0,200,0), 2)
# set y of left point to y of right point
lst1 = list(extLeft1)
lst1[1] = extRight1[1]
extLeft1 = tuple(lst1)
lst2 = list(extLeft2)
lst2[1] = extRight2[1]
extLeft2= tuple(lst2)
# compute the distance between the points (x1, y1) and (x2, y2)
dist1 = math.sqrt( ((extLeft1[0]-extRight1[0])**2)+((extLeft1[1]-extRight1[1])**2) )
dist2 = math.sqrt( ((extLeft2[0]-extRight2[0])**2)+((extLeft2[1]-extRight2[1])**2) )
# draw lines
cv2.line(cimg, extLeft1, extRight1, (255,0,0), 1)
cv2.line(cimg, extLeft2, extRight2, (255,0,0), 1)
# draw the distance text
font = cv2.FONT_HERSHEY_SIMPLEX
fontScale = 0.5
fontColor = (255,0,0)
lineType = 1
cv2.putText(cimg,str(dist1),(155,100),font, fontScale, fontColor, lineType)
cv2.putText(cimg,str(dist2),(155,280),font, fontScale, fontColor, lineType)
# show image
cv2.imshow("Image", img)
cv2.waitKey(0)
现在我还需要上等高线底部斜线的角度。
有什么办法可以得到这个吗？是否可以使用等高线？
或者是否有必要使用 HoughLinesP 并以某种方式对相关行进行排序？
然后继续提问：也许它也有可能得到描述该边抛物线斜率的函数？
非常感谢您的帮助！
有几种方法可以只获取斜率。为了知道斜率，我们可以使用 cv2.HoughLines 来检测底部水平线，检测到该线的端点并从中获得斜率。例如，
lines = cv2.HoughLines(edges, rho=1, theta=np.pi/180, threshold=int(dist2*0.66) )
on edges 在你的代码中给出了 4 行，如果我们强制角度是水平的
for line in lines:
rho, theta = line[0]
# here we filter out non-horizontal lines
if abs(theta - np.pi/2) > np.pi/180:
continue
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 1000*(-b))
y1 = int(y0 + 1000*(a))
x2 = int(x0 - 1000*(-b))
y2 = int(y0 - 1000*(a))
cv2.line(img_lines,(x1,y1),(x2,y2),(0,0,255),1)
我们得到：
关于抛物线的扩展题，我们先写一个return左右两点的函数：
def horizontal_scan(gray_img, thresh=50, start=50):
'''
scan horizontally for left and right points until we met an all-background line
@param thresh: threshold for background pixel
@param start: y coordinate to start scanning
'''
ret = []
thickness = 0
for i in range(start,len(gray_img)):
row = gray_img[i]
# scan for left:
left = 0
while left < len(row) and row[left]<thresh:
left += 1
if left==len(row):
break;
# scan for right:
right = left
while right < len(row) and row[right] >= thresh:
right+=1
if thickness == 0:
thickness = right - left
# prevent sudden drop, error/noise
if (right-left) < thickness//5:
continue
else:
thickness = right - left
ret.append((i,left,right))
return ret
# we start scanning from extLeft1 down until we see a blank line
# with some tweaks, we can make horizontal_scan run on edges,
# which would be simpler and faster
horizontal_lines = horizontal_scan(gray, start = extLeft1[1])
# check if horizontal_line[0] are closed to extLeft1 and extRight1
print(horizontal_lines[0], extLeft1, extRight1[0])
请注意，我们可以使用此函数来查找由 HoughLines return编辑的水平线的端点。
# last line of horizontal_lines would be the points we need:
upper_lowest_y, upper_lowest_left, upper_lowest_right = horizontal_lines[-1]
img_lines = img.copy()
cv2.line(img_lines, (upper_lowest_left, upper_lowest_y), extLeft1, (0,0,255), 1)
cv2.line(img_lines, (upper_lowest_right, upper_lowest_y), extRight1, (0,0,255),1)
这给出了：
让我们return回答扩展问题，我们有左右两点：
left_points = [(x,y) for y,x,_ in horizontal_lines]
right_points = [(x,y) for y,_,x in horizontal_lines]
显然，它们不能完全符合抛物线，因此我们需要某种 approximation/fitting。为此，我们可以构建一个 LinearRegression 模型：
from sklearn.linear_model import LinearRegression
class BestParabola:
def __init__(self, points):
x_x2 = np.array([(x**2,x) for x,_ in points])
ys = np.array([y for _,y in points])
self.lr = LinearRegression()
self.lr.fit(x_x2,ys)
self.a, self.b = self.lr.coef_
self.c = self.lr.intercept_
self.coef_ = (self.c,self.b,self.a)
def transform(self,points):
x_x2 = np.array([(x**2,x) for x,_ in points])
ys = self.lr.predict(x_x2)
return np.array([(x,y) for (_,x),y in zip(x_x2,ys)])
然后，我们可以拟合给定的 left_points, right_points 以获得所需的抛物线：
# construct the approximate parabola
# the parabollas' coefficients are accessible by BestParabola.coef_
left_parabola = BestParabola(left_points)
right_parabola = BestParabola(right_points)
# get points for rendering
left_parabola_points = left_parabola.transform(left_points)
right_parabola_points = right_parabola.transform(right_points)
# render with matplotlib, cv2.drawContours would work
plt.figure(figsize=(8,8))
plt.imshow(cv2.cvtColor(img,cv2.COLOR_BGR2RGB))
plt.plot(left_parabola_points[:,0], left_parabola_points[:,1], linewidth=3)
plt.plot(right_parabola_points[:,0], right_parabola_points[:,1], linewidth=3, color='r')
plt.show()
给出：
左边的抛物线并不完美，但如果需要的话你应该算出来:-)
我在图像中检测到两个轮廓，需要顶部轮廓的两个垂直边缘之间的直径和下部轮廓的垂直边缘之间的直径。我用这段代码实现了这一点。
import cv2
import numpy as np
import math, os
import imutils
img = cv2.imread("1.jpg")
gray = cv2.cvtColor(img, cv2.COLOR_RGB2GRAY)
gray = cv2.GaussianBlur(gray, (7, 7), 0)
edges = cv2.Canny(gray, 200, 100)
edges = cv2.dilate(edges, None, iterations=1)
edges = cv2.erode(edges, None, iterations=1)
cnts = cv2.findContours(edges.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = imutils.grab_contours(cnts)
# sorting the contours to find the largest and smallest one
c1 = max(cnts, key=cv2.contourArea)
c2 = min(cnts, key=cv2.contourArea)
# determine the most extreme points along the contours
extLeft1 = tuple(c1[c1[:, :, 0].argmin()][0])
extRight1 = tuple(c1[c1[:, :, 0].argmax()][0])
extLeft2 = tuple(c2[c2[:, :, 0].argmin()][0])
extRight2 = tuple(c2[c2[:, :, 0].argmax()][0])
# show contour
cimg = cv2.drawContours(img, cnts, -1, (0,200,0), 2)
# set y of left point to y of right point
lst1 = list(extLeft1)
lst1[1] = extRight1[1]
extLeft1 = tuple(lst1)
lst2 = list(extLeft2)
lst2[1] = extRight2[1]
extLeft2= tuple(lst2)
# compute the distance between the points (x1, y1) and (x2, y2)
dist1 = math.sqrt( ((extLeft1[0]-extRight1[0])**2)+((extLeft1[1]-extRight1[1])**2) )
dist2 = math.sqrt( ((extLeft2[0]-extRight2[0])**2)+((extLeft2[1]-extRight2[1])**2) )
# draw lines
cv2.line(cimg, extLeft1, extRight1, (255,0,0), 1)
cv2.line(cimg, extLeft2, extRight2, (255,0,0), 1)
# draw the distance text
font = cv2.FONT_HERSHEY_SIMPLEX
fontScale = 0.5
fontColor = (255,0,0)
lineType = 1
cv2.putText(cimg,str(dist1),(155,100),font, fontScale, fontColor, lineType)
cv2.putText(cimg,str(dist2),(155,280),font, fontScale, fontColor, lineType)
# show image
cv2.imshow("Image", img)
cv2.waitKey(0)
现在我还需要上等高线底部斜线的角度。
有什么办法可以得到这个吗？是否可以使用等高线？
或者是否有必要使用 HoughLinesP 并以某种方式对相关行进行排序？
然后继续提问：也许它也有可能得到描述该边抛物线斜率的函数？
非常感谢您的帮助！
有几种方法可以只获取斜率。为了知道斜率，我们可以使用 cv2.HoughLines 来检测底部水平线，检测到该线的端点并从中获得斜率。例如，
lines = cv2.HoughLines(edges, rho=1, theta=np.pi/180, threshold=int(dist2*0.66) )
on edges 在你的代码中给出了 4 行，如果我们强制角度是水平的
for line in lines:
rho, theta = line[0]
# here we filter out non-horizontal lines
if abs(theta - np.pi/2) > np.pi/180:
continue
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 1000*(-b))
y1 = int(y0 + 1000*(a))
x2 = int(x0 - 1000*(-b))
y2 = int(y0 - 1000*(a))
cv2.line(img_lines,(x1,y1),(x2,y2),(0,0,255),1)
我们得到：
关于抛物线的扩展题，我们先写一个return左右两点的函数：
def horizontal_scan(gray_img, thresh=50, start=50):
'''
scan horizontally for left and right points until we met an all-background line
@param thresh: threshold for background pixel
@param start: y coordinate to start scanning
'''
ret = []
thickness = 0
for i in range(start,len(gray_img)):
row = gray_img[i]
# scan for left:
left = 0
while left < len(row) and row[left]<thresh:
left += 1
if left==len(row):
break;
# scan for right:
right = left
while right < len(row) and row[right] >= thresh:
right+=1
if thickness == 0:
thickness = right - left
# prevent sudden drop, error/noise
if (right-left) < thickness//5:
continue
else:
thickness = right - left
ret.append((i,left,right))
return ret
# we start scanning from extLeft1 down until we see a blank line
# with some tweaks, we can make horizontal_scan run on edges,
# which would be simpler and faster
horizontal_lines = horizontal_scan(gray, start = extLeft1[1])
# check if horizontal_line[0] are closed to extLeft1 and extRight1
print(horizontal_lines[0], extLeft1, extRight1[0])
请注意，我们可以使用此函数来查找由 HoughLines return编辑的水平线的端点。
# last line of horizontal_lines would be the points we need:
upper_lowest_y, upper_lowest_left, upper_lowest_right = horizontal_lines[-1]
img_lines = img.copy()
cv2.line(img_lines, (upper_lowest_left, upper_lowest_y), extLeft1, (0,0,255), 1)
cv2.line(img_lines, (upper_lowest_right, upper_lowest_y), extRight1, (0,0,255),1)
这给出了：
让我们return回答扩展问题，我们有左右两点：
left_points = [(x,y) for y,x,_ in horizontal_lines]
right_points = [(x,y) for y,_,x in horizontal_lines]
显然，它们不能完全符合抛物线，因此我们需要某种 approximation/fitting。为此，我们可以构建一个 LinearRegression 模型：
from sklearn.linear_model import LinearRegression
class BestParabola:
def __init__(self, points):
x_x2 = np.array([(x**2,x) for x,_ in points])
ys = np.array([y for _,y in points])
self.lr = LinearRegression()
self.lr.fit(x_x2,ys)
self.a, self.b = self.lr.coef_
self.c = self.lr.intercept_
self.coef_ = (self.c,self.b,self.a)
def transform(self,points):
x_x2 = np.array([(x**2,x) for x,_ in points])
ys = self.lr.predict(x_x2)
return np.array([(x,y) for (_,x),y in zip(x_x2,ys)])
然后，我们可以拟合给定的 left_points, right_points 以获得所需的抛物线：
# construct the approximate parabola
# the parabollas' coefficients are accessible by BestParabola.coef_
left_parabola = BestParabola(left_points)
right_parabola = BestParabola(right_points)
# get points for rendering
left_parabola_points = left_parabola.transform(left_points)
right_parabola_points = right_parabola.transform(right_points)
# render with matplotlib, cv2.drawContours would work
plt.figure(figsize=(8,8))
plt.imshow(cv2.cvtColor(img,cv2.COLOR_BGR2RGB))
plt.plot(left_parabola_points[:,0], left_parabola_points[:,1], linewidth=3)
plt.plot(right_parabola_points[:,0], right_parabola_points[:,1], linewidth=3, color='r')
plt.show()
给出：
左边的抛物线并不完美，但如果需要的话你应该算出来:-)