从RGB转换为LAB色彩空间 - 对L*A*B*值范围的任何了解？

Question

在OpenCV(Python)中将图像从RGB转换为LAB时,我无法找到有关L*A*B*值范围的文档.寻找一些确认我的洞察力是正确的,因为数字是相当奇特的.我的亮度结果是0-255,但对于a和b我分别得到42-226和20-223.我知道这些值不需要有预定的范围,但任何人都可以了解为什么选择这些范围？

我试图在LAB空间中创建颜色直方图,并且需要知道以空间有效的方式存储bin值的值范围.

import cv2
import numpy as np
import sys
import urllib

print cv2.__version__ # 2.4.7
print sys.version # 2.7.5+ (default, Sep 19 2013, 13:48:49) \n[GCC 4.8.1]

# Load an image that contains all possible colors.
request = urllib.urlopen('http://www.brucelindbloom.com/downloads/RGB16Million.png')
image_array = np.asarray(bytearray(request.read()), dtype=np.uint8)
image = cv2.imdecode(image_array, cv2.CV_LOAD_IMAGE_COLOR)

# I was uncertain if it was BGR or RGB but in this case it doesn't matter because
# of my input image.
lab_image = cv2.cvtColor(image, cv2.COLOR_BGR2LAB)
l_channel,a_channel,b_channel = cv2.split(lab_image)

# Print the minimum and maximum of lightness.
print np.min(l_channel) # 0
print np.max(l_channel) # 255

# Print the minimum and maximum of a.
print np.min(a_channel) # 42
print np.max(a_channel) # 226

# Print the minimum and maximum of b.
print np.min(b_channel) # 20
print np.max(b_channel) # 223

谢谢!

Answer 1

查看OpenCV 文档(向下滚动到RGB↔CIE L*a*b*的转换定义),我们可以看到这些值重新调整为0-255范围:

L←L*255/100; a←a + 128; b←b + 128

此外:LAB色彩空间跨越整个可感知的色彩范围,RGB则不然.因此,从RGB转换时,您将看不到整个值范围.