python - OpenCV轮廓检测

Question

我一直在尝试使用 OpenCV 检测轮廓。我正在尝试检测白细胞的细胞核。我在我的其他图像上测试了它，结果证明它没问题，除了原子核彼此距离太远的图像。这是我制作的程序的结果:

在底部，核没有被检测为一个，而是被检测为两个，因为它们没有连接或粘在一起。如何让程序将其检测为只有一个细胞？

这是我的代码:

import cv2
import numpy as np

limit_area = 1000   
x = 0   
y = 0   
w = 0   
h = 0   
nuclei = []   
count = 0   
number_name = 1   

img1 = cv2.imread('7.bmp')
img = cv2.add(img1, 0.70)
img_3 = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
mask1 = cv2.inRange(img_3, (90,140,0), (255,255,255))   
mask2 = cv2.inRange(img_3, (90,90,0), (255,255,255))   
mask1 = cv2.equalizeHist(mask1)
mask2 = cv2.equalizeHist(mask2)
mask = mask1 + mask2   
kernel = np.ones((1,4),np.uint8)   
mask = cv2.dilate(mask,kernel,iterations = 1)   
kernel_close = np.ones((3,3),np.uint8)
mask = cv2.morphologyEx(mask, cv2.MORPH_OPEN, kernel_close)   
blur2 = cv2.medianBlur(mask,7)   
canny = cv2.Canny(blur2, 100,200)   
im2, contours, hierarchy = cv2.findContours(canny,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)   

for cnt in contours:   
    if cv2.contourArea(cnt) >= limit_area:   
        nuclei.append(cnt)   
        print(cv2.contourArea(cnt))
        x, y, w, h = cv2.boundingRect(cnt)   
        roi = blur2[y:y+h, x:x+w]
        outfile = '%d.jpg' % number_name
        image_roi = cv2.resize(roi, (128,128), interpolation=cv2.INTER_AREA)
        image_roi = cv2.medianBlur(image_roi, 5)
        (T, thresh) = cv2.threshold(image_roi, 10, 255, cv2.THRESH_BINARY)
        _, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
        contours = [i for i in contours if cv2.contourArea(i) <= 5000]
        cv2.fillPoly(thresh, contours, color=(0,0,0))
        image_roi = thresh
        cv2.imshow(outfile, image_roi)
        cv2.rectangle(img, (x, y), (x+w, y+h), (0,255,0), 7)  
        number_name += 1   

    count += 1   

cv2.imshow('img', img)
cv2.waitKey(0)
cv2.destroyAllWindows()

原图如下:

Answer 1

一种简单的方法是合并紧密检测到的区域。 Image Localization 中有一个概念叫做 Intersection over Union，如果两个边界框的 IoU 分数大于某个阈值，则将它们合并。一个伪 cade 就像

xi1 = max(box1[0], box2[0])
yi1 = max(box1[1], box2[1])
xi2 = min(box1[2], box2[2])
yi2 = min(box1[3], box2[3])
inter_area = max((xi2 - xi1), 0) * max((yi2 - yi1), 0)
box1_area = (box1[2] - box1[0]) * (box1[3] - box1[1])
box2_area = (box2[2] - box2[0]) * (box2[3] - box2[1])
union_area = box1_area + box2_area - inter_area
iou = inter_area/union_area

您可以尝试的另一种方法是 Flood Fill Algorithm ，我认为它应该可以正常工作，因为核毕竟是一个实体(完全连接)。在拨号之前尝试此操作，这可能是您的轮廓分成两半的原因。