I'm having trouble with my Hexadecimal values. I'm trying find a way so as the alpha of my HEX changes over time.
Right now my code looks like:
color c = 0X00202020;
Trying to change the'0x00" to a var as to allow it to be more easily manipulated. I have tried...
int alph = 00;
color = 0xalph202020;
Obviously this does not work.
Is there any other way to use variables inside of a HEX?
I really don’t want to convert the values to RGB. Hoping to find a solution.
You don't mention which language. In most you can "bit shift".
e.g.
int red = 0x20;
int green = 0x20;
int blue = 0x20;
int alpha = 0;
// Shift the alpha value "left" by 24 bits, red by 16 and green by 8.
// Assumes that each value is just 8bits long (0-255)
//
int color = (alpha << 24) + (red << 16) + (green <<8) + blue;
Best way to change only alpha value of your color is using 'color()' methode with hexadecimal value so you do not need to convert value into RGB:
color c = color(0x00202020, alph);
Related
I need to convert 24-bit colors to 4-bit RGBI (1 bit for Red, Green, Blue + Intensity).
Converting to 3-bit RGB is rather simple: set color bit if greater than 127, clear otherwise. However, there's only one intensity bit for all three channels, so what's the correct way to set it (if any)?
First I thought about dividing 8-bit channel to three parts like below:
if 0 <= color <= 85, then clear rgbi-color bit
if 86 <= color <= 170, then set rgbi-color bit
if 171 <= color <= 255, then set rgbi-color bit and intensity
But then I thought that probably the correct way would be to set intensity bit only if two of three channels are greater than 127. But in that case pure R, G, or B will not have intensity ever set (for example, in case of rbg(0,0,200)
Any advice is highly appreciated
A simple way to find the closest 4-bit RGBI approximation of a color is to consider the two possibilities for the intensity bit separately. That is to say, first find the closest RGB0 and RGB1 approximations for the color (which is easy to do, just by dividing each color axis at the appropriate point), and the determine which of these approximations is better.
Here's a simple C-ish pseudocode description of this algorithm:
// find the closest RGBx approximation of a 24-bit RGB color, for x = 0 or 1
function rgbx_approx(red, green, blue, x) {
threshold = (x + 1) * 255 / 3;
r = (red > threshold ? 1 : 0);
g = (green > threshold ? 1 : 0);
b = (blue > threshold ? 1 : 0);
return (r, g, b);
}
// convert a 4-bit RGBI color back to 24-bit RGB
function rgbi_to_rgb24(r, g, b, i) {
red = (2*r + i) * 255 / 3;
green = (2*g + i) * 255 / 3;
blue = (2*b + i) * 255 / 3;
return (red, green, blue);
}
// return the (squared) Euclidean distance between two RGB colors
function color_distance(red_a, green_a, blue_a, red_b, green_b, blue_b) {
d_red = red_a - red_b;
d_green = green_a - green_b;
d_blue = blue_a - blue_b;
return (d_red * d_red) + (d_green * d_green) + (d_blue * d_blue);
}
// find the closest 4-bit RGBI approximation (by Euclidean distance) to a 24-bit RGB color
function rgbi_approx(red, green, blue) {
// find best RGB0 and RGB1 approximations:
(r0, g0, b0) = rgbx_approx(red, green, blue, 0);
(r1, g1, b1) = rgbx_approx(red, green, blue, 1);
// convert them back to 24-bit RGB:
(red0, green0, blue0) = rgbi_to_rgb24(r0, g0, b0, 0);
(red1, green1, blue1) = rgbi_to_rgb24(r1, g1, b1, 1);
// return the color closer to the original:
d0 = color_distance(red, green, blue, red0, green0, blue0);
d1 = color_distance(red, green, blue, red1, green1, blue1);
if (d0 <= d1) return (r0, g0, b0, 0);
else return (r1, g1, b1, 1);
}
Alternatively, you could simply use any generic fixed-palette color quantization algorithm. This may yield better results if your actual color palette is not a pure evenly spaced RGBI palette like the code above assumes, but rather something like e.g. the CGA tweaked RGBI palette.
I want to create a colour scroller effect. I have a function that I give it RGB values (eg. setColor(189,234,45)) and I want to change the colour rapidly but I don't want to get many repeats to create an effect of scrolling through the colours.
I have tried tried the following but it doesn't quite generate the effect that I am looking for.
for (int i = 0; i < 256; i++) {
for (int j = 0; j < 256; j++) {
for (int k = 0; k < 256; k++) {
setColor(i, j, k);
}
}
}
I wanted to know if anyone knows how the colour scroller's colours are arranged next to each other. The arrangement I am looking for looks like the scroll on the right.
The colors you are working with are represented as R,G,B (red green blue) values. However, another
way to think about color is hue, saturation, value. In the scroll image you are trying to emulate,
it is the hue that is changing - the saturation and value (brightness) are unaffected.
Here is a function that happens to make a hue-cycle gradient like the one in the image you linked to:
int n = 256; // number of steps
float TWO_PI = 3.14159*2;
for (int i = 0; i < n; ++i) {
int red = 128 + sin(i*TWO_PI/n + 0) + 127;
int grn = 128 + sin(i*TWO_PI/n + TWO_PI/3) + 127;
int blu = 128 + sin(i*TWO_PI/n + 2*TWO_PI/3) + 127;
setColor(red, grn, blu);
}
To understand how that function works, I recommend that you read my color tutorial that GreenAsJade linked to.
However, that kind of gradient function isn't quite what you need, because you want to start from a particular color you are passing in, and then go to the next color in the sequence. It's much easier to do this kind of thing if you represent your colors as HSV triplets (or HSB triplets), instead of RGB triplets. Then you can manipulate just the hue component, and get those kind of rainbow effects. In helps to have a set of function that can convert from RGB to HSV and back again.
This site contains a bunch of color conversion source code, including the ones you need for those conversions. Using the two conversion functions supplied on that page, your code might look like:
void cycleMyColor(int *r, int *g, int *b) {
float h,s,v, fr,fg,fb;
RGBtoHSV(*r/255.0,*g/255.0,*b/255.0,&h,&s,&v);
h += 1/256.0; // increment the hue here
h -= (int) h; // and cycle around if necessary
HSVtoRGB(&fr,&fg,&fb,h,s,v);
*r = fr*255; *g = fg*255; *b = fb*255;
setColor(*r,*g,*b);
}
This code is a little more complicated than it needs to be because the color conversions on that site use floating point color components that go from 0-1, instead of integers that go from 0-255, as you were using, so I'm spending a few lines converting between those two representations. You may find it simpler to just keep your color in HSB space, and then convert to RGB when you want to display it.
As you mentioned in your edit, you don't like the sequence of colours, because you start from black an end at white, instead of starting at one end of the rainbow and going to the other.
So you are going to need to work out a sequence of RGB that goes from blue through green and yellow to red. That means you need to start with (0,0,255) and end at (255, 0, 0), and don't pass through (255,255,255) or (0,0,0) - in a nutshell, that's how its done.
There are many ways you could do this and get a pleasing effect - beyond the scope of an answer here. This article explores it in depth:
http://krazydad.com/tutorials/makecolors.php
The project in question: https://github.com/matutter/Pixel2 is a personal project to replace some out of date software at work. What it should do is, the user adds an image and it generates a color palette of the image. The color palette should have no duplicate colors. (thats the only important stuff)
My question is: why do larger or hi-res or complex images not work as well? (loss of color data)
Using dropzone.js I have the user put a picture on the page. The picture is a thumbnail. Next I use jquery to find the src out of a <img src="...">. I pass that src to a function that does this
function generate(imgdata) {
var imageObj = new Image();
imageObj.src = imgdata;
convert(imageObj); //the function that traverses the image data pulling out RGB
}
the "convert" function pulls out the data fairly simply by
for(var i=0, n=data.length; i<n; i+=4, pixel++ ) {
r = data[i];
g = data[i+1];
b = data[i+2];
color = r + g + b; // format is a string of **r, g, b**
}
finally, the last part of the main algorithme filters out duplicate colors, I only want just 1 occurrence of each... here's the last part
color = monoFilter(color); // the call
function monoFilter(s) {
var unique = [];
$.each(s, function(i, el){
if($.inArray(el, unique) === -1) unique.push(el);
});
unique.splice(0,1); //remove undefine
unique.unshift("0, 0, 0"); //make sure i have black
unique.push("255, 255, 255"); //and white
return unique;
}
I'm hoping someone can help me identify why there is such a loss of color data in big files.
If anyone is actually interesting enough to look at the github, the relivent files are js/pixel2.js, js/dropzone.js, and ../index.html
This is probably the cause of the problem:
color = r + g + b; // format is a string of **r, g, b**
This simply adds the numbers together and the more pixels you have the higher risk you run to get the same number. For example, these colors generate the same result:
R G B
color = 90 + 0 + 0 = 90;
color = 0 + 90 + 0 = 90;
color = 0 + 0 + 90 = 90;
even though they are completely different colors.
To avoid this you can do it like this if you want a string:
color = [r,g,b].join();
or you can create an integer value of them (which is faster to compare with than a string):
color = (b << 16) + (g << 8) + r; /// LSB byte-order
Even an Euclidean vector would be better:
color = r*r + g*g + b*b;
but with the latter you risk eventually the same scenario as the initial one (but useful for nearest color scenarios).
Anyways, hope this helps.
"The problem was that I wasn't accounting for alpha. So a palette from an image that uses alpha would have accidental duplicate records."
I figured this out after finding this Convert RGBA color to RGB
I want to fill the intersection of two(or more filled) rectangles with the average color. I have the colors of each rectangle stored as unsigned ints. How can I get the average color?
Thank you for you help!
Technically, you might be running on a color-map device, which means you need to go through X11 color management for all of this. You need to query the XColor for your two input colors, compute the average, then look up the closest representable color:
// Query XColor for both input colors
XColor xcol1, xcol2, outcol;
xcol1.pixel = color1;
xcol2.pixel = color2;
XQueryColor(display, colormap, &xcol1);
XQueryColor(display, colormap, &xcol2);
// Average red/green/blue and look up nearest representable color
outcol.red = (xcol1.red + xcol2.red) / 2;
outcol.green = (xcol1.green + xcol2.green) / 2;
outcol.blue = (xcol1.blue + xcol2.blue) / 2;
XAllocColor(display, colormap, &outcol);
// outcol.pixel is now the color to use
On a paletted device, you also need to free the color afterwards etc. - it's a mess, basically.
But in all likelihood you're on a 32-bit truecolor device, which means the integer is just a bitfield of r, g, b and a (not necessarily in that order). You can compute their average like this:
UInt out_color = 0;
for (int i=0; i < 4; i++) {
// Extract channel i from both input colors
UInt in1 = (color1 >> (i*8)) & 0xff;
UInt in2 = (color2 >> (i*8)) & 0xff;
// Compute the average and or it into the output color
out_color |= ((in1 + in2) / 2) << (i*8);
}
Color color1 = Color.FromArgb(UInt1);
Color color2 = Color.FromArgb(UInt2);
Color averageColor = Color.FromArgb(255,(color1.R + color2.R)/2,(color1.G + color2.G)/2,(color1.B + color2.B)/2);
This is assuming that you need a fully opaque average color.
Given two colors and n steps, how can one calculate n colors including the two given colors that create a fade effect?
If possible pseudo-code is preferred but this will probably be implemented in Java.
Thanks!
Divide each colour into its RGB components and then calculate the individual steps required.
oldRed = 120;
newRed = 200;
steps = 10;
redStepAmount = (newRed - oldRed) / steps;
currentRed = oldRed;
for (i = 0; i < steps; i++) {
currentRed += redStepAmount;
}
Obviously extend that for green and blue.
There are two good related questions you should also review:
Generating gradients programatically?
Conditional formatting — percentage to color conversion
Please note that you're often better off doing this in the HSV color space rather than RGB - it generates more pleasing colors to the human eye (lower chance of clashing or negative optical properties).
Good luck!
-Adam
If you want a blend that looks anything like most color picker GUI widgets, you really want to translate to HSL or HSV. From there, you're probably fine with linear interpolation in each dimension.
Trying to do any interpolations directly in RGB colorspace is a bad idea. It's way too nonlinear (and no, gamma correction won't help in this case).
For those looking for something they can copy and paste. Made a quick function for RGB colors. Returns a single color that is the amount of ratio closer to rgbColor2.
function fadeToColor(rgbColor1, rgbColor2, ratio) {
var color1 = rgbColor1.substring(4, rgbColor1.length - 1).split(','),
color2 = rgbColor2.substring(4, rgbColor2.length - 1).split(','),
difference,
newColor = [];
for (var i = 0; i < color1.length; i++) {
difference = color2[i] - color1[i];
newColor.push(Math.floor(parseInt(color1[i], 10) + difference * ratio));
}
return 'rgb(' + newColor + ')';
}
The quesiton is what transformation do you want to occur? If you transpose into the HSV colourspace and given
FF0000 and 00FF00
It will transition from red through yellow to green.
However, if you define "black" or some other shade as being the mid-point of the blend, you have to shade to that colour first ff0000->000000->00ff00 or via white : ff0000 -> ffffff -> 00ff00.
Transforming via HSV however can be fun because you have to use a bit of trig to map the circular map into the vector components.
The easiest thing to do is linear interpolation between the color components (see nickf's response). Just be aware that the eye is highly nonlinear, so it won't necessarily look you're making even steps. Some color spaces attempt to address this (CIE maybe?), so you might want to transform into another color space first, interpolate, then transform back to RGB or whatever you're using.
How about this answer
- (UIColor *)colorFromColor:(UIColor *)fromColor toColor:(UIColor *)toColor percent:(float)percent
{
float dec = percent / 100.f;
CGFloat fRed, fBlue, fGreen, fAlpha;
CGFloat tRed, tBlue, tGreen, tAlpha;
CGFloat red, green, blue, alpha;
if(CGColorGetNumberOfComponents(fromColor.CGColor) == 2) {
[fromColor getWhite:&fRed alpha:&fAlpha];
fGreen = fRed;
fBlue = fRed;
}
else {
[fromColor getRed:&fRed green:&fGreen blue:&fBlue alpha:&fAlpha];
}
if(CGColorGetNumberOfComponents(toColor.CGColor) == 2) {
[toColor getWhite:&tRed alpha:&tAlpha];
tGreen = tRed;
tBlue = tRed;
}
else {
[toColor getRed:&tRed green:&tGreen blue:&tBlue alpha:&tAlpha];
}
red = (dec * (tRed - fRed)) + fRed;
green = (dec * (tGreen - fGreen)) + fGreen;
blue = (dec * (tBlue - fBlue)) + fBlue;
alpha = (dec * (tAlpha - fAlpha)) + fAlpha;
return [UIColor colorWithRed:red green:green blue:blue alpha:alpha];
}