function random_rgba() { var o = Math.round, r = Math.random, s = 255; return ‘rgba(‘ + o(r()*s) + ‘,’ + o(r()*s) + ‘,’ + o(r()*s) + ‘,’ + r().toFixed(1) + ‘)’; } var color = random_rgba(); g.fillStyle = color; g.strokeStyle = color; FIDDLE
Yes, invertible transformations exist. equasys GmbH posted invertible transformations from RGB to YUV, YCbCr, and YPbPr, along with explanations of which situation each is appropriate for, what this clamping is really about, and links to references. (Like a good SO answer.) For my own application (jpg images, not analog voltages) YCbCr was appropriate, so I … Read more
Lets say we have a 3 Channel (RGB) image given by some matrix A A = [[[198 218 227] [196 216 225] [196 214 224] … … [185 201 217] [176 192 208] [162 178 194]] and a blur kernal as K = [[0.1111, 0.1111, 0.1111], [0.1111, 0.1111, 0.1111], [0.1111, 0.1111, 0.1111]] #which is actually … Read more
Actually, I’d like turn the problem around, and say I’m surprised that so many different platforms and tools actually produce the same values. 🙂 JPEG is lossy First of all, JPEG is a lossy image compression method. This means that reproducing the exact data of the original is not possible. Or, if you like, several … Read more
def rgb(minimum, maximum, value): minimum, maximum = float(minimum), float(maximum) ratio = 2 * (value-minimum) / (maximum – minimum) b = int(max(0, 255*(1 – ratio))) r = int(max(0, 255*(ratio – 1))) g = 255 – b – r return r, g, b
What you are looking for is called Delta-E. http://www.colorwiki.com/wiki/Delta_E:_The_Color_Difference It is the distance between two colors in LAB color space. It is said that the human eye cannot distinguish colors below 1 DeltaE (I find that my eyes can find differences in colors below 1 DeltaE, each person is different.) There are 4 formulas for … Read more
Well, if I understood correctly, you want to iterate through the pixels in the image, perform some kind of test, and if it passes you want to store that pixel in an array. Here´s how you could do that: using System.Drawing; Bitmap img = new Bitmap(“*imagePath*”); for (int i = 0; i < img.Width; i++) … Read more