我想检测一条线上的圆形侵蚀和扩张。对于扩张,我试图递归地侵蚀图像,在每次递归时,我都会检查宽高宽高比。如果比率小于4,我假设它的轮廓是圆形的,对于每个这样的轮廓,我根据矩和面积计算圆的圆心和半径。这是检测圆形扩张的函数:
def detect_circular_dilations(img, contours):
contours_current, hierarchy = cv2.findContours(img, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
if len(contours_current) == 0:
return get_circles_from_contours(contours)
for c in contours_current:
x, y, w, h = cv2.boundingRect(c)
if w > h:
aspect_ratio = float(w) / h
else:
aspect_ratio = float(h) / w
if aspect_ratio < 4 and w < 20 and h < 20 and w > 5 and h > 5:
contours.append(c)
return detect_circular_dilations(cv2.erode(img, None, iterations=1), contours)下面是我想要检测的圆形膨胀的一个例子:

我还没有解决的另一个问题是圆形腐蚀的检测。圆周侵蚀的示例如下:

在这里,我用红色矩形标记了要检测的圆形侵蚀。可能会有一些较小的圆形图案(在左侧),不应该被视为实际的圆形侵蚀。
有人知道检测这种圆形的最好方法是什么吗?对于循环扩张,我希望有任何评论/建议,以便潜在地使检测更健壮。
谢谢!
发布于 2019-05-05 00:01:13
我会尝试用cv2.Canny()找到这条线的两条边,然后搜索轮廓。如果您按边框的宽度对轮廓进行排序,则前两个轮廓将是您的线条边缘。之后,您可以计算一条边中的每个点到另一条边的最小距离。然后,你可以计算距离的中位数,如果一个点的距离大于或小于中位数(+-公差),那么该点就是直线的扩张或侵蚀,并将其附加到列表中。如果需要,您可以通过遍历列表来筛选噪声,如果它们不连续(在x轴上),则删除这些点。
下面是一个简单的例子:
import cv2
import numpy as np
from scipy import spatial
def detect_dilation(median, mindist, tolerance):
count = 0
for i in mindist:
if i > median + tolerance:
dilate.append((reshape_e1[count][0], reshape_e1[count][1]))
elif i < median - tolerance:
erode.append((reshape_e1[count][0], reshape_e1[count][1]))
else:
pass
count+=1
def other_axis(dilate, cnt):
temp = []
for i in dilate:
temp.append(i[0])
for i in cnt:
if i[0] in temp:
dilate.append((i[0],i[1]))
img = cv2.imread('1.jpg')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray,100,200)
_, contours, hierarchy = cv2.findContours(edges,cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
contours.sort(key= lambda cnt :cv2.boundingRect(cnt)[3])
edge_1 = contours[0]
edge_2 = contours[1]
reshape_e1 = np.reshape(edge_1, (-1,2))
reshape_e2 =np.reshape(edge_2, (-1,2))
tree = spatial.cKDTree(reshape_e2)
mindist, minid = tree.query(reshape_e1)
median = np.median(mindist)
dilate = []
erode = []
detect_dilation(median,mindist,5)
other_axis(dilate, reshape_e2)
other_axis(erode, reshape_e2)
dilate = np.array(dilate).reshape((-1,1,2)).astype(np.int32)
erode = np.array(erode).reshape((-1,1,2)).astype(np.int32)
x,y,w,h = cv2.boundingRect(dilate)
cv2.rectangle(img,(x,y),(x+w,y+h),(255,0,0),2)
x,y,w,h = cv2.boundingRect(erode)
cv2.rectangle(img,(x,y),(x+w,y+h),(0,0,255),2)
cv2.imshow('img', img)
cv2.waitKey(0)
cv2.destroyAllWindows()结果:

编辑:
如果图片中有一条折线(这意味着更多的轮廓),则必须将每个轮廓视为一条单独的线。您可以通过在cv2.boundingRect()的帮助下创建一个感兴趣的区域来实现这一点。但是当我尝试新上传的图片时,这个过程并不是很健壮,因为你必须改变容差才能得到想要的结果。由于我不知道其他图像是什么样子的,你可能需要一种更好的方法来获得平均距离和容差因子。这里的任何方式都是我描述的一个示例( 15表示容差):
import cv2
import numpy as np
from scipy import spatial
def detect_dilation(median, mindist, tolerance):
count = 0
for i in mindist:
if i > median + tolerance:
dilate.append((reshape_e1[count][0], reshape_e1[count][1]))
elif i < median - tolerance:
erode.append((reshape_e1[count][0], reshape_e1[count][1]))
else:
pass
count+=1
def other_axis(dilate, cnt):
temp = []
for i in dilate:
temp.append(i[0])
for i in cnt:
if i[0] in temp:
dilate.append((i[0],i[1]))
img = cv2.imread('2.jpg')
gray_original = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_, thresh_original = cv2.threshold(gray_original, 0, 255, cv2.THRESH_BINARY+cv2.THRESH_OTSU)
# Filling holes
_, contours, hierarchy = cv2.findContours(thresh_original,cv2.RETR_CCOMP,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contours:
cv2.drawContours(thresh_original,[cnt],0,255,-1)
_, contours, hierarchy = cv2.findContours(thresh_original,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_NONE)
for cnt in contours:
x2,y,w2,h = cv2.boundingRect(cnt)
thresh = thresh_original[0:img.shape[:2][1], x2+20:x2+w2-20] # Region of interest for every "line"
edges = cv2.Canny(thresh,100,200)
_, contours, hierarchy = cv2.findContours(edges,cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
contours.sort(key= lambda cnt: cv2.boundingRect(cnt)[3])
edge_1 = contours[0]
edge_2 = contours[1]
reshape_e1 = np.reshape(edge_1, (-1,2))
reshape_e2 =np.reshape(edge_2, (-1,2))
tree = spatial.cKDTree(reshape_e2)
mindist, minid = tree.query(reshape_e1)
median = np.median(mindist)
dilate = []
erode = []
detect_dilation(median,mindist,15)
other_axis(dilate, reshape_e2)
other_axis(erode, reshape_e2)
dilate = np.array(dilate).reshape((-1,1,2)).astype(np.int32)
erode = np.array(erode).reshape((-1,1,2)).astype(np.int32)
x,y,w,h = cv2.boundingRect(dilate)
if len(dilate) > 0:
cv2.rectangle(img[0:img.shape[:2][1], x2+20:x2+w2-20],(x,y),(x+w,y+h),(255,0,0),2)
x,y,w,h = cv2.boundingRect(erode)
if len(erode) > 0:
cv2.rectangle(img[0:img.shape[:2][1], x2+20:x2+w2-20],(x,y),(x+w,y+h),(0,0,255),2)
cv2.imshow('img', img)
cv2.waitKey(0)
cv2.destroyAllWindows()结果:

发布于 2019-05-05 13:12:56
这类问题通常使用距离变换和中轴变换来解决。这些在某种程度上是相关的,因为中轴沿着距离变换的脊线运行。一般的想法是:
,
例如,我使用下面的MATLAB代码获得了以下输出。

下面是我使用的代码。它使用MATLAB和DIPimage 3,只是作为原理的快速证明。使用您喜欢使用的任何图像处理库,这应该可以直接转换为Python。
% Read in image and remove the red markup:
img = readim('https://i.stack.imgur.com/bNOTn.jpg');
img = img{3}>100;
img = closing(img,5);
% This is the algorithm described above:
img = fillholes(img); % Get rid of holes
radius = dt(img); % Distance transform
m = bskeleton(img); % Medial axis
radius(~m) = 0; % Ignore all pixels outside the medial axis
detection = dilation(radius,25)==radius & radius>25; % Local maxima with radius > 25
pos = findcoord(detection); % Coordinates of detections
radius = double(radius(detection)); % Radii of detections
% This is just to make the markup:
detection = newim(img,'bin');
for ii=1:numel(radius)
detection = drawshape(detection,2*radius(ii),pos(ii,:),'disk');
end
overlay(img,detection)https://stackoverflow.com/questions/55980638
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