Dynamics AX 2012 R2
I am using Image:rgb(r, g, b); to convert from RGB to integer colors.
When the colors display, they are all shifted--the red and blue values appear to be swapped, but I think it is actually more complicated than that.
255,0,0 displays as blue (should be red)
0,0,255 displays as red (should be blue)
Any colors in-between all come in mixed up as well. I can trade the r & b values, but that doesn't seem like the right answer, considering I'm using a stock AX method for the
conversion that is documented as r, g, b not b, g, r.
It is possible the problem may not be with the Image::rgb method at all. If I manually convert the rgb value and just pass an integer to it, the colors are still reversed.
Any ideas?
Use the WinAPI::RGB2int method to convert from RGB to its internal representation.
Related
I understand that domain or color wheel plotting is typical for complex functions.
Incredibly, I can't find a million + returns on a web search to easily allow me to reproduce some piece of art as this one in Wikipedia:
There is this online resource that reproduces plots with zeros in black - not bad at all... However, I'd like to ask for some simple annotated code in Octave to produce color plots of functions of complex numbers.
Here is an example:
I see here code to plot a complex function. However, it uses a different technique with the height representing the Re part of the image of the function, and the color representing the imaginary part:
Peter Kovesi has some fantastic color maps. He provides a MATLAB function, called colorcet, that we can use here to get the cyclic color map we need to represent the phase. Download this function before running the code below.
Let's start with creating a complex-valued test function f, where the magnitude increases from the center, and the phase is equal to the angle around the center. Much like the example you show:
% A test function
[xx,yy] = meshgrid(-128:128,-128:128);
z = xx + yy*1i;
f = z;
Next, we'll get its phase, convert it into an index into the colorcet C2 color map (which is cyclic), and finally reshape that back into the original function's shape. out here has 3 dimensions, the first two are the original dimensions, and the last one is RGB. imshow shows such a 3D matrix as a color image.
% Create a color image according to phase
cm = colorcet('C2');
phase = floor((angle(f) + pi) * ((size(cm,1)-1e-6) / (2*pi))) + 1;
out = cm(phase,:);
out = reshape(out,[size(f),3]);
The last part is to modulate the intensity of these colors using the magnitude of f. To make the discontinuities at powers of two, we take the base 2 logarithm, apply the modulo operation, and compute the power of two again. A simple multiplication with out decreases the intensity of the color where necessary:
% Compute the intensity, with discontinuities for |f|=2^n
magnitude = 0.5 * 2.^mod(log2(abs(f)),1);
out = out .* magnitude;
That last multiplication works in Octave and in the later versions of MATLAB. For older versions of MATLAB you need to use bsxfun instead:
out = bsxfun(#times,out,magnitude);
Finally, display using imshow:
% Display
imshow(out)
Note that the colors here are more muted than in your example. The colorcet color maps are perceptually uniform. That means that the same change in angle leads to the same perceptual change in color. In the example you posted, for example yellow is a very narrow, bright band. Such a band leads to false highlighting of certain features in the function, which might not be relevant at all. Perceptually uniform color maps are very important for proper interpretation of the data. Note also that this particular color map has easily-named colors (purple, blue, green, yellow) in the four cardinal directions. A purely real value is green (positive) or purple (negative), and a purely imaginary value is blue (positive) or yellow (negative).
There is also a great online tool made by Juan Carlos Ponce Campuzano for color wheel plotting.
In my experience it is much easier to use than the Octave solution. The downside is that you cannot use perceptually uniform coloring.
The CSS3 spec only specifies that:
The format of an HSLA color value in the functional notation is ‘hsla(’ followed by the hue in degrees, saturation and lightness as a percentage, and an , followed by ‘)’.
So am I to understand that these values would be interpreted not as integers but as floats? Example:
hsla(200.2, 90.5%, 10.2%, .2)
That would dramatically expand the otherwise small (relative to RGB) range of colors covered by HSL.
It seems to render fine in Chrome, though I don't know if they simply parse it as an INT value or what.
HSL values are converted to hexadecimal RGB values before they are handed off to the system. It's up to the device to clip any resulting RGB value that is outside the "device gamut" - the range of colors that can be displayed - to a displayable value. RGB values are denoted in Hexadecimal. This is the specified algorithm for browsers to convert HSL values to RGB values. Rounding behavior is not specified by the standard - and there are multiple ways of doing rounding since there doesn't appear to be a built-in rounding function in either C or C++.
HOW TO RETURN hsl.to.rgb(h, s, l):
SELECT:
l<=0.5: PUT l*(s+1) IN m2
ELSE: PUT l+s-l*s IN m2
PUT l*2-m2 IN m1
PUT hue.to.rgb(m1, m2, h+1/3) IN r
PUT hue.to.rgb(m1, m2, h ) IN g
PUT hue.to.rgb(m1, m2, h-1/3) IN b
RETURN (r, g, b)
From the proposed recommendation
In other words, you should be able to represent the exact same range of colors in HSLA as you can represent in RGB using fractional values for HSLA.
AFAIK, every browser casts them to INTs. Maybe. If I'm wrong you won't be able to tell the difference anyway. If it really matters, why not just go take screenshots an open them in photoshop or use an on-screen color meter. Nobody here is going to have a definitive answer without testing it, and it takes 2 minutes to test... so...
I wouldn't know exactly, but it makes sense to just put in some floating numbers and see if it works? it takes two seconds to try with a decimal, and without..
How to compute similarity between two colors in RGBA color space? (where the background color is unknown of course)
I need to remap an RGBA image to a palette of RGBA colors by finding the best palette entry for each pixel in the image*.
In the RGB color space the most similar color can be assumed to be the one with the smallest euclidean distance. However, this approach doesn't work in RGBA, e.g., Euclidean distance from rgba(0,0,0,0) to rgba(0,0,0,50%) is smaller than to rgba(100%,100%,100%,1%), but the latter looks much better.
I'm using premultiplied RGBA color space:
r = r×a
g = g×a
b = b×a
and I've tried this formula (edit: See the answer below for better formula):
Δr² + Δg² + Δb² + 3 × Δa²
but it doesn't look optimal — in images with semitransparent gradients it finds wrong colors that cause discontinuities/sharp edges. Linear proportions between opaque colors and alpha seem fishy.
What's the optimal formula?
*) for simplicity of this question I'm ignoring error diffusion, gamma and psychovisual color spaces.
Slightly related: if you want to find nearest color in this non-Euclidean RGBA space, vp-trees are the best.
Finally, I've found it! After thorough testing and experimentation my conclusions are:
The correct way is to calculate maximum possible difference between the two colors.
Formulas with any kind of estimated average/typical difference had room for discontinuities.
I was unable to find a working formula that calculates the distance without blending RGBA colors with some backgrounds.
There is no need to take every possible background color into account. It can be simplified down to blending maximum and minimum separately for each of R/G/B channels:
blend the channel in both colors with channel=0 as the background, measure squared difference
blend the channel in both colors with channel=max as the background, measure squared difference
take higher of the two.
Fortunately blending with "white" and "black" is trivial when you use premultiplied alpha.
The complete formula for premultiplied alpha color space is:
rgb *= a // colors must be premultiplied
max((r₁-r₂)², (r₁-r₂ - a₁+a₂)²) +
max((g₁-g₂)², (g₁-g₂ - a₁+a₂)²) +
max((b₁-b₂)², (b₁-b₂ - a₁+a₂)²)
C Source including SSE2 implementation.
Several principles:
When two colors have same alpha, rgbaDistance = rgbDistance * ( alpha / 255). Compatible with RGB color distance algorithm when both alpha are 255.
All Colors with very low alpha are similar.
The rgbaDistance between two colors with same RGB is linearly dependent on delta Alpha.
double DistanceSquared(Color a, Color b)
{
int deltaR = a.R - b.R;
int deltaG = a.G - b.G;
int deltaB = a.B - b.B;
int deltaAlpha = a.A - b.A;
double rgbDistanceSquared = (deltaR * deltaR + deltaG * deltaG + deltaB * deltaB) / 3.0;
return deltaAlpha * deltaAlpha / 2.0 + rgbDistanceSquared * a.A * b.A / (255 * 255);
}
My idea is integrating once over all possible background colors and averaging the square error.
i.e. for each component calculate(using red channel as example here)
Integral from 0 to 1 ((r1*a1+rB*(1-a1))-(r2*a2+rB*(1-a2)))^2*drB
which if I calculated correctly evaluates to:
dA=a1-a2
dRA=r1*a1-r2*a2
errorR=dRA^2+dA*dRA+dA^2/3
And then sum these over R, G and B.
First of all, a very interesting problem :)
I don't have a full solution (at least not yet), but there are 2 obvious extreme cases we should consider:
When Δa==0 the problem is similiar to RGB space
When Δa==1 the problem is only on the alpha 1-dim space
So the formula (which is very similar to the one you stated) that would satisfy that is:
(Δr² + Δg² + Δb²) × (1-(1-Δa)²) + Δa² or (Δr² + Δg² + Δb²) × (1-Δa²) + Δa²
In any case, it would probably be something like (Δr² + Δg² + Δb²) × f(Δa) + Δa²
If I were you, I would try to simulate it with various RGBA pairs and various background colors to find the best f(Δa) function. Not very mathematic, but will give you a close enough answer
I've never done it, but theory and practice say that converting the RGB values in the image and the palette to luminance–chrominance will help you find the best matches. I'd leave the alpha channel alone, as transparency should have little to nothing to do with the 'looking better' part.
This xmass I made some photomosaics for presents using open-source software that matches fragments of the original image to a collection of images. That seems like a harder problem than the one you're trying to solve. One of them programs was metapixel.
Lastly, the best option should be to use an existing library to convert the image to a format, like PNG, in which you can control the palette.
I want to have buttons that are switching from one color to another.
Eg. 1000 buttons where the first one is yellow and the last one is green and all the between will slowly move from yellow to green.
How can I generate all the hex codes for colors (eg. #8a3a3a) between these two colors?
Split the two input colors into red, green, blue components and convert them to float. Subtract source components from destination components, divide each by 1000 and call them f.ex. deltaRed, deltaGreen, deltaBlue. Start with source components, convert those into a "#rrggbb" string 1000 times, each loop adding the deltas. If you want to actually reach the destination color, you have to loop from 0 to 1000, ie. 1001 times.
Yes, it is. You can compute it as follows:
Imagine the colors are points in a three-dimensional space, with each component (red, green, blue) representing one dimension. Depending on how many color tones you want between the two colors, you can try to evenly divide the differences between the two colors, for each component separately. For instance, if rA is the red component of color A, and rB the red component of color B, and if you want 10 steps in between, then the red component of the second step is r2 = (rB - rA) * 2 / 10.
Convert the components to decimal first (e.g. 8a => 138), and you should probably write a small program for the computation. I don't think you need so many tones though, because each component has only a range from 0 to 255 (rounding necessary), and the human eye cannot differentiate between so many colors anyway.
This may be what you need: http://sandbox.leigeber.com/fader/fader.html
Code: http://www.leigeber.com/wp-content/uploads/2008/05/fader.zip
I want to write a function in ruby that given a number between 1 and 500 will output a 6 digit hex color code that gets linearly darker for higher numbers. This doesn't seem that hard but I'm not sure where to begin. How can I implement this?
edit
Hue seems like a more reliable way to go. I'd like to give a reference color, say a shade of green, and then darken or lighten it based on the input number.
input: 10
output: color code (in rgb or HSV) that is a light shade of the reference color
input: 400
output: color code (in rgb or HSV) that is a fairly dark shade of the reference color
edit 2
The only reason I need to use between 1 and 500 is because that's the input I have to work with. It's alright if some numbers that are close together map to the same color.
The 6 digit hex color code is in RGB.
What you want is to work in HSV: pick a Hue and Saturation, and gradually decrease the Value.
Convert from HSV to RGB to output the color.
See here for an example.
Basic linear interpolation?
// Pseudocode
function fade_colour(source, factor)
const max = 500
const min = 1
foreach component in source
output[component] = round(source[component] * (max - value) / (max - min))
endforeach
return output
endfunction
Why not just return a gray level then, #ffffff to #000000? 500 levels of darkness aren't really distinguishable anyway, and grays give you 256 levels.
If you only want to darken your reference color, it's easy. Given an R,G,B color that is the brightest you want to go, multiply each of the 3 values by (500-input) and divide by 499. Convert each of the values to 2 hex digits and append them with a # at the front.