我是opencv的初学者。我正在做一个数独解算器的项目。起初,我的工作是从数独中提取数独网格。我成功地提取了它,并找到了图像中的最大框(很明显,最大的框将是数独网格)。但是当我试图根据需要改变图像的形状(即根据数独网格的大小裁剪图像)时,我得到了错误的结果。pts1,pts2。下面是运行良好的代码。
import numpy as np
def get_sudo_rid(name,size):
img=name
original=img.copy()
img=cv2.cvtColor(img,cv2.COLOR_BGR2RGB)
graymain=cv2.cvtColor(img,cv2.COLOR_RGB2GRAY)
ath=cv2.adaptiveThreshold(graymain,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C,cv2.THRESH_BINARY_INV,39,10)
contours,hierarchy=cv2.findContours(ath,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
area=[]
maxarea=0
cnt=contours[0]
for i in contours:
if cv2.contourArea(i)>maxarea:
cnt=i
maxarea=cv2.contourArea(i)
blank= np.zeros(img.shape,np.uint8)
image=cv2.drawContours(blank,[cnt],-1,(255,255,255),5)
edges=cv2.Canny(image,40,150)
lines=cv2.HoughLines(edges,1,np.pi/180,100)
createhor=[]
createver=[]
created=[]
anglediff=10
rhodiff=10
flag=0
count = 2
for line in lines:
for(rho,theta) in line:
flag=0
for (rho1,theta1) in created:
if(abs(rho-rho1) < rhodiff and abs(theta-theta1)<anglediff):
flag=1
if(flag==0):
a=np.cos(theta)
b=np.sin(theta)
xo=a*rho
yo=b*rho
x1=int( xo + 1000*(-b))
y1=int( yo + 1000*(a))
x2=int( xo + 1000*(-b))
y2=int( yo +1000*(a))
d=np.linalg.norm(np.array((x1,y1,0))-np.array((x2,y2,0)))
cv2.line(img,(x1,y1),(x2,y2),(0,255,0),2)
m=abs(1/np.tan(theta))
if(m<1):
createhor.append((rho,theta))
else:
createver.append((rho,theta))
points=[]
for (rho,theta) in createhor:
for (rho1,theta1) in createver:
if(rho,theta)!=(rho1,theta1):
a=[[np.cos(theta),np.sin(theta)],[np.cos(theta1),np.sin(theta1)]]
b=[rho,rho1]
cos=np.linalg.solve(a,b)
if list(cos) not in points:
points.append(list(cos))
points.sort()
if(points[0][1]>points[1][1]):
points[0],points[1]=points[1],points[0]
if(points[-1][1] < points[-2][1]):
points[-2],points[-1] = points[-1], points[-2]
points[1],points[2] = points[2],points[1]
for i in points:
images=cv2.circle(image,(int(i[0]),int(i[1])),4,(0,0,255),-1)
pts1 = np.float32(points)
pts2 = np.array([[0, 0], [size, 0], [0, size], [size, size]], np.float32)
#pts2 = np.float32([[0, 0], [size, 0], [0, size], [size, size]])
print(pts2)到目前为止,我的代码运行得很好。但当我添加这行时
M= cv2.getPerspectiveTransform(pts1,pts2)
我得到了下面的错误:
error: (-215:Assertion failed) src.checkVector(2, CV_32F) == 4 && dst.checkVector(2, CV_32F) == 4 in function 'cv::getPerspectiveTransform'
实际上,我在这个项目中得到了github的帮助,在他的项目中,他为所有的图像发送了size = 900作为参数,因为我不明白他为什么要这样做,所以我复制了同样的内容。
发布于 2020-08-15 02:17:05
问题是pts1必须是4x2矩阵,而不是28x2矩阵。您需要做的是获得网格的最外面的角,它必须是一个形状为(4,2)的float32数组,然后获得透视变换矩阵,最后将其应用于剩余的点。
dst = np.float32([[0, 0], [size, 0], [size, size], [0, size]])
# perspective transform = matrix transformation
M = cv2.getPerspectiveTransform(outer_grid_corners, dst)
# apply transformation to all the other points
transformed_pts = cv2.warpPerspective(pts1, M)请注意,您的outer_grid_corners必须按顺时针方式排序。
https://stackoverflow.com/questions/63410603
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