OpenCv反转：将BGR转换为HSV，然后再返回BGR-我是否应该获得相同的图像？

Question

例如，当我运行此命令时：

import cv2
import numpy as np
from collections import Counter

# load image, calculate some basic stats
path = 'path/to/my/image.png'
bgr_img = cv2.imread(path)
h, w, c = bgr_img.shape
print("There are %d entries in this %d channel image" % (h*w*c, c))

# convert and then invert, we should get the same image
bgr_img2 = cv2.cvtColor(cv2.cvtColor(bgr_img, cv2.COLOR_BGR2HSV), cv2.COLOR_HSV2BGR)
print("all entries match: %s" % np.all(bgr_img == bgr_img2))

# hmm, we don't. let's see where the differences are
mismatches = np.where(bgr_img != bgr_img2)
print("there are %d mismatched entries" % len(mismatches[0]))
print("mismatches are along the following channels: %s" % Counter(mismatches[2]))

# ok, clearly lots of them. maybe they're all just off by 1 or 2?
mm = zip(mismatches[0], mismatches[1], mismatches[2])
differences = []
for x, y, c in mm:
    diff = bgr_img[x, y, c] - bgr_img2[x, y, c]
    differences.append(diff)

print(differences)

我得到以下输出：

There are 1228800 entries in this 3 channel image
all entries match: False
there are 524511 mismatched entries
mismatches are along the following channels: Counter({1: 270572, 0: 253939})
[1, 1, 1, 254, 1, 254, 1, 3, 1, 3, 1, 3, 254, 254, 254, 254, 254, 254, 254, 254, 254, 254, 254, 254, 254, 254, 1, 3, 1, 3, 1, 3, 1, 254, 1, 254, 254, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, ...]

我看到了几种可能性。 首先是BGR-> HSV转换不是严格可逆的。 其他是我使用的OpenCV不正确。 哪有

Answer 1

您应该获得几乎相同的图像，但除了边缘情况外，不会获得相同的位。 从BGR到HSV（以及向后）涉及到空间之间的转换，这些空间对于其他空间中的所有值没有相同的表示形式。 小数舍入误差将蔓延。

请参阅https://docs.opencv.org/3.4/df/d9d/tutorial_py_colorspaces.html中有关色调范围[0，179]的注释。 如果您考虑一下，这意味着从BGR（或RGB）转换时会丢失信息。 您无法可靠地收回这些位。