Python的cv2凸缺陷距离太大？

Question

更新：我自己计算的距离与

cv2

返回的距离之间的比率似乎正好是 256。这并不奇怪，因为查看他们的代码（第 394 行here）显示像素距离成倍增加到 256。我想我只是不明白为什么。

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我正在使用

cv2.convexityDefects

来查找某些形状的凸缺陷。我希望计算缺陷的深度（到凸包的距离，这里解释得很好）。

但是，我得到的距离太大了，没有意义。我还尝试使用 cv2.convexityDefects 的起点和终点输出参数手动计算距离（参见下面的代码），并获得更合理的结果。

cv2.convexityDefects

计算出的距离为，

>> d
array([21315, 26578, 19093, 56472, 35230, 20476, 26825], dtype=int32)

这在大约 500 像素宽的图像上下文中完全没有意义。我做错了什么？

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更多信息：

这是我为本次测试创建的图像（呃，我的意思是艺术品），

这是代码：

from numpy.linalg import norm

# Load an image
img = cv2.imread('buff.png')

# Threshold
ret, img = cv2.threshold(img,127,255,cv2.THRESH_BINARY)

# Detect edges
edges = cv2.Canny(img, 1, 2)

# Find contours
cnts, h = cv2.findContours(edges, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)

# Find convex hull
hull = cv2.convexHull(cnts[0], returnPoints = False)

# Find defects
defects = cv2.convexityDefects(cnts[0], np.sort(np.squeeze(hull)))
# Reshape the output
s = np.reshape(defects,(-1,4))[:,0] # start point idx
e = np.reshape(defects,(-1,4))[:,1] # end point idx
f = np.reshape(defects,(-1,4))[:,2] # defect idx
d = np.reshape(defects,(-1,4))[:,3] # distances as calculated by cv2.convexityDefects

# Draw contours in red, convex hull in blue
img = cv2.drawContours(img, cnts[0], -1, (255,0,0), 2)
img = cv2.polylines(img, [cnts[0][np.squeeze(hull)]], True, (0,0,255), 3)

# Calculate distances manually and put in a list d2
d2 = list()
for i in range(len(f)):
    # Draw the defects in blue, start points in pink:
    img = cv2.circle(img, tuple(cnts[0][f[i]][0]), 5,(0, 0, 255), -1)
    img = cv2.circle(img, tuple(cnts[0][s[i]][0]), 5,(255, 0, 255), -1)

    # Manually calculate the distances as the distances between the defects (blue points)
    # and the line defined between each pair of start and end points (pink here, for demonstration)
    d2.append(norm(np.cross(cnts[0][s[i]][0]-cnts[0][e[i]][0], cnts[0][e[i]][0]-cnts[0][f[i]][0]))/norm(cnts[0][s[i]][0]-cnts[0][e[i]][0]))
    
pp.imshow(img)
pp.show()

这是上述代码应用于测试图像时的输出：

Answer 1

我自己计算的距离与

cv2

返回的距离之间的比率似乎正好是 256。这并不奇怪，因为查看他们的代码（第 394 行here）显示像素距离乘以 256我想我只是不明白为什么。

Answer 2

这是故意的。

fixpt_depth is fixed-point approximation (with 8 fractional bits) of the distance between the farthest contour point and the hull. That is, to get the floating-point value of the depth will be fixpt_depth/256.0.