I came across two code sample snippets of a function that claim to decompose an image and return an image with just hue, just saturation, and just value, depending on what color was passed as a parameter.
def break_down(image, channel):
hsv = color.rgb2hsv(image)
if (channel == "H"):
out = hsv[:, :, 0].copy()
if (channel == "S"):
out = hsv[:, :, 1].copy()
if (channel == "V"):
out = hsv[:, :, 2].copy()
return out
def break_down(image, channel):
hsv = color.rgb2hsv(image)
out = hsv.copy()
if (channel == "H"):
out[:, :, 1] = 0
out[:, :, 2] = 0
if (channel == "S"):
out[:, :, 0] = 0
out[:, :, 2] = 0
if (channel == "V"):
out[:, :, 0] = 0
out[:, :, 1] = 0
return out
The first one gives output as colored images (red/green/blue). However, the second one gives outputs in black/white/grey shades. I am unable to understand which is the correct code. It's possible that none of them are accurate.
Logically, what should happen when for example we show just saturation? If Hue is the color component of the image, then this returned image should be without colors right?
This isn't really a matter of one being "correct", it's just a choice of how you want to visualize the data.
Saturation is a [0, 1] value that says how far the color is from gray. You can visualize a [0, 1] value as a grayscale image, it will be dark where the original image was more gray and white where the original image had highly saturated colors. Is this a useful visualization? That's up to you to decide.
Perhaps a way that you would find more intuitive, would be to visualize H by forcing S and V to 1 and converting the image back to RGB to display. That would show you the pure hue of each pixel (at 100% saturation and value). Then, you might visualize H,S together by forcing V to 1 and converting back to RGB. That would show you the hue and saturation (i.e. the chroma) of each pixel. If you compare that to the hue visualization then you may start to get a feel for what saturation means.