答案 0 :(得分:0)
经过一些搜索,我最终找到了可能的解决方案。关键问题是执行正确的图像预处理以获得实线(与问题中的图像相反:我们需要没有间隙的线)。
此帖子挽救了生命:Gap Filling Contours / Lines。 就我而言,dimKernel = 50,thBin = 160,thDistTrans = 0.07
def preprocessing(imm,dimKernel,thBin,thDistTrans):
grayImage = cv.cvtColor(imm, cv.COLOR_BGR2GRAY)
ret,binImage=cv.threshold(grayImage,thBin,255,cv.THRESH_BINARY_INV)
structVerticale = kernelVerticale(dimKernel,1)
im1 = cv.morphologyEx(binImage, cv.MORPH_OPEN, structVerticale)
structOrizzontale = kernelOrizzontale(dimKernel,3)
im2 = cv.morphologyEx(binImage, cv.MORPH_OPEN, structOrizzontale)
result = overlaps(im1,im2)
out = ndi.distance_transform_edt(np.invert(result))
out = out < thDistTrans * out.max()
out = morphology.skeletonize(out)
out = (out.astype(int)*255).astype("uint8")
kernel = np.ones((3,3),np.uint8)
out = cv.dilate(out,kernel)
return out
然后,我需要使用cv.findContours确定正确的矩形;根据经验证据,我知道可以使用区域(从原始区域图像的1/6到1/3)来识别我要查找的直肠。最后,轮廓已使用cv.boundingRect近似裁剪为矩形,然后进行裁剪:
contours, hierarchy = cv.findContours(ris,cv.RETR_CCOMP , cv.CHAIN_APPROX_SIMPLE)
hierarchy = hierarchy[0]
aree = []
for i,figura in enumerate(contours):
area = cv.contourArea(figura)
aree.append([area,i])
aree.sort(reverse=True)
areaMax = (ris.shape[0]*ris.shape[1])/3
areaMin = (ris.shape[0]*ris.shape[1])/6
i = 0
while i<len(aree) and (aree[i][0]<areaMin or aree[i][0]>areaMax):
i+=1
cnt = contours[aree[i][1]]
x,y,w,h = cv.boundingRect(cnt)
immOrg = immOrg.crop((x, y, x+w, y+h))
我确定这个解决方案远非最佳解决方案,因为我是一名业余程序员,我以前从未使用过cv,但希望能对某人有所帮助