python - 打开CV分水岭无法正确分割椭圆形对象

试图创建一种处理图像以计数不同类型平板电脑的方式。以下代码对于圆形对象一直运行良好，但是椭圆形会产生一些无法解决的问题。

kernel = np.ones((5,5),np.uint8)


image = cv2.imread('sample.jpg')
shifted = cv2.GaussianBlur(image, (15, 15), 1)
shifted = cv2.pyrMeanShiftFiltering(shifted, 21, 51)
shifted = cv2.erode(shifted,kernel,iterations=1)
shifted = cv2.dilate(shifted,kernel,iterations=1)
cv2.imwrite("step1.jpg", shifted)
gray = cv2.cvtColor(shifted, cv2.COLOR_BGR2GRAY)
thresh = cv2.threshold(gray, 0, 255,
    cv2.THRESH_BINARY | cv2.THRESH_OTSU)[1]
cv2.imwrite("step2.jpg", thresh)
thresh = cv2.morphologyEx(thresh,cv2.MORPH_OPEN,kernel, iterations = 2)
cv2.imwrite("step3.jpg", thresh)
thresh = cv2.bitwise_not(thresh)
thresh = cv2.erode(thresh,kernel,iterations=1)
cv2.imwrite("step4.jpg", thresh)
D = ndimage.distance_transform_edt(thresh)
localMax = peak_local_max(D, indices=False, min_distance=10,
    labels=thresh)
markers = ndimage.label(localMax, structure=np.ones((3, 3)))[0]

labels = watershed(-D, markers, mask=thresh)
print("[INFO] {} unique segments found".format(len(np.unique(labels)) - 1))
for label in np.unique(labels):
    if label == 0:
        continue
    mask = np.zeros(gray.shape, dtype="uint8")
    mask[labels == label] = 255
    cnts = cv2.findContours(mask.copy(), cv2.RETR_EXTERNAL,
        cv2.CHAIN_APPROX_SIMPLE)
    cnts = imutils.grab_contours(cnts)
    c = max(cnts, key=cv2.contourArea)
    ((x, y), r) = cv2.minEnclosingCircle(c)
    cv2.circle(image, (int(x), int(y)), int(r), (0, 255, 0), 2)
    cv2.putText(image, "#{}".format(label), (int(x) - 10, int(y)),
        cv2.FONT_HERSHEY_SIMPLEX, 0.6, (0, 0, 255), 2)
cv2.imwrite("step5.jpg", image)
cv2.waitKey(0)

正在使用的图像是:

https://imgur.com/a/1U49DeT

过滤后输出:

https://imgur.com/a/vXwrWlG

任何有关如何解决此问题的教学要点将不胜感激。

最佳答案

我认为有一种更好的方法来使用分水岭运算符。

它依赖于良好的渐变，但是如果图像与此图像相似，则应该能够有效地做到这一点。而且，今天有非常强大的边缘检测器，比我在本演示中使用的检测器要好得多。

import cv2
import numpy as np
import higra as hg
from skimage.segmentation import relabel_sequential
import matplotlib.pyplot as plt

def main():
    img_path = "pills.jpg"
    img = cv2.imread(img_path)
    img = cv2.resize(img, (256, 256))
    img = cv2.GaussianBlur(img, (9, 9), 0)

    edges = cv2.Canny(img, 100, 100)

    size = img.shape[:2]
    graph = hg.get_4_adjacency_graph(size)
    edge_weights = hg.weight_graph(graph, edges, hg.WeightFunction.mean)

    tree, altitudes = hg.watershed_hierarchy_by_area(graph, edge_weights)
    segments = hg.labelisation_horizontal_cut_from_threshold(tree, altitudes, 500)
    segments, _, _ = relabel_sequential(segments)
    print('The number of pills is ', segments.max() - 1)
    plt.imshow(segments)
    plt.show()

if __name__ == "__main__":
    main()

最初，我调整图像大小以加快计算速度，并应用模糊处理以减少背景渐变。我检测它的边缘(渐变)并创建一个带有边缘权重的图形。然后我计算按面积排序的分水岭层次结构，并对其进行阈值处理，以获取该级别的连接组件，由此您可以计算分段的数量。

关于python - 打开CV分水岭无法正确分割椭圆形对象，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/61318473/