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

For example, when I run this:

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)

I get the following output:

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, ...]

I see a couple possibilities. First is that BGR -> HSV conversion isn't strictly invertible. Other is that I'm using OpenCV incorrectly. Which is it?

Answer 1

You should get pretty much the same image, but not, except in edge cases, identical bits. BGR to HSV (and back) involve a transformation between spaces that don't have identical representations for all values in the other space. Small rounding errors will creep in.

See the note in https://docs.opencv.org/3.4/df/d9d/tutorial_py_colorspaces.html about the Hue range being [0, 179]. If you think about it, that implies information loss when converting from BGR (or RGB). You can't get those bits back reliably.