Related
I am trying to find all the hex colors in an image and if possible, circle or highlight the X, Y position of where the hex color(s) are. My current code is attempting to find all colors and almost crashes my browser and my attempt to find the X,Y coordinates of each image isn't going good either.
I have two functions doing different things, it's what I have tried to work with to give an example of what has been attempted... Any help would be great!
Any assistance would be amazing!
<canvas id="myCanvas" width="240" height="297" style="border:1px solid #d3d3d3;">
Your browser does not support the HTML5 canvas tag.
</canvas>
<img id="origImage" width="220" height="277" src="loggraph.PNG">
<script>
function getPixel(imgData, index) {
var i = index*4, d = imgData.data;
return [d[i],d[i+1],d[i+2],d[i+3]] // [R,G,B,A]
}
function getPixelXY(imgData, x, y) {
return getPixel(imgData, y*imgData.width+x);
}
function goCheck() {
var cvs = document.createElement('canvas'),
img = document.getElementsByTagName("img")[0];
cvs.width = img.width; cvs.height = img.height;
var ctx = cvs.getContext("2d");
ctx.drawImage(img,0,0,cvs.width,cvs.height);
var idt = ctx.getImageData(0,0,cvs.width,cvs.height);
console.log(getPixel(idt, 852)); // returns array [red, green, blue, alpha]
console.log(getPixelXY(idt,1,1)); // same pixel using x,y
}
function getColors(){
var canvas = document.getElementById("myCanvas");
var devices = canvas.getContext("2d");
var imageData = devices.getImageData(0, 0, canvas.width, canvas.height);
var data = imageData.data;
// iterate over all pixels
for(var i = 0, n = data.length; i < n; i += 4) {
var r = data[i];
var g = data[i + 1];
var b = data[i + 2];
var rgb = "("+r+","+g+","+b+")";
var incoming = i*4, d = imageData.data;
var bah = [d[incoming],d[incoming+1],d[incoming+2],d[incoming+3]];
$('#list').append("<li>"+rgb+"</li>");
colorList.push(rgb);
}
$('#list').append("<li>"+[d[incoming],d[incoming+1],d[incoming+2],d[incoming+3]]+"</li>");
}
}
Must check all pixels
To find a pixel that matches a color will require, in the worst case (pixel of that color not in image), that you step over every pixel in the image.
How not to do it
Converting every pixel to a DOM string is about the worst way to do it, as DOM string use a lot of memory and CPU overhead, especially if instantiated using jQuery (which has its own additional baggage)
Hex color to array
To find the pixel you need only check each pixels color data against the HEX value. You convert the hex value to an array of 3 Bytes.
The following function will convert from CSS Hex formats "#HHH" "#HHHH", "#HHHHHH" and "#HHHHHHHH" ignoring the alpha part if included, to an array of integers 0-255
const hex2RGB = h => {
if(h.length === 4 || h.length === 5) {
return [parseInt(h[1] + h[1], 16), parseInt(h[2] + h[2], 16), parseInt(h[3] + h[3], 16)];
}
return [parseInt(h[1] + h[2], 16), parseInt(h[3] + h[4], 16), parseInt(h[5] + h[6], 16)];
}
Finding the pixel
I do not know how you plan to use such a feature so the example below is a general purpose method that will help and can be modified as needed
It will always find a pixel if you let it even if there is no perfect match. It does this by finding the closest color to the color you are looking for.
The reason that of finds the closest match is that when you draw an image onto a 2D canvas the pixel values are modified slightly if the image has transparent pixels (pre-multiplied alpha)
The function finds the pixel by measuring the spacial distance between the pixel and the hex color (simple geometry Pythagoras). The closest color is the one that is the smallest distance.
It will return the object
{
x, // the x coordinate of the match
y, // the y coordinate of the match
distance, // how closely the color matches the requested color.
// 0 means a perfect match
// to 441 completely different eg black and white
// value is floored to an integer value
}
If the image is tainted (cross origin, local device storage), or you pass something that can not be converted to pixels the function will return undefined
The function keeps a canvas that it uses to get pixel data as it assumes that it will be use many times. If the image is tainted it will catch the error (add a warning to the console), cleanup the tainted canvas and be ready for another image.
Usage
To use the function add it to your code base, it will setup automatically.
Get an image and a hex value and call the function with the image, CSS hex color, and optionally the threshold distance for the color match.
Eg find exact match for #FF0000
const result = findPixel(origImage, "#FF0000", 0); // find exact match for red
if (result) { // only if found
console.log("Found color #FF0000 at pixel " + result.x + ", " + result.y);
} else {
console.log("The color #FF0000 is not in the image");
}
or find color close to
const result = findPixel(origImage, "#FF0000", 20); // find a match for red
// within 20 units.
// A unit is 1 of 256
if (result) { // only if found
console.log("Found closest color within " + result.distance + "units of #FF0000 at pixel " + result.x + ", " + result.y);
}
or find closest
// find the closest, no threshold ensures a result
const result = findPixel(origImage, "#FF0000");
console.log("Found closest color within " + result.distance + "units of #FF0000 at pixel " + result.x + ", " + result.y);
Code
The function is as follows.
const findPixel = (() => {
var can, ctx;
function createCanvas(w, h) {
if (can === undefined){
can = document.createElement("canvas");
ctx = can.getContext("2d");
}
can.width = w;
can.height = h;
}
function getPixels(img) {
const w = img.naturalWidth || img.width, h = img.naturalHeight || img.height;
createCanvas(w, h);
ctx.drawImage(img, 0, 0);
try {
const imgData = ctx.getImageData(0, 0, w, h);
can.width = can.height = 1; // make canvas as small as possible so it wont
// hold memory. Leave in place to avoid instantiation overheads
return imgData;
} catch(e) {
console.warn("Image is un-trusted and pixel access is blocked");
ctx = can = undefined; // canvas and context can no longer be used so dump them
}
return {width: 0, height: 0, data: []}; // return empty pixel data
}
const hex2RGB = h => { // Hex color to array of 3 values
if(h.length === 4 || h.length === 5) {
return [parseInt(h[1] + h[1], 16), parseInt(h[2] + h[2], 16), parseInt(h[3] + h[3], 16)];
}
return [parseInt(h[1] + h[2], 16), parseInt(h[3] + h[4], 16), parseInt(h[5] + h[6], 16)];
}
const idx2Coord = (idx, w) => ({x: idx % w, y: idx / w | 0});
return function (img, hex, minDist = Infinity) {
const [r, g, b] = hex2RGB(hex);
const {width, height, data} = getPixels(img);
var idx = 0, found;
while (idx < data.length) {
const R = data[idx] - r;
const G = data[idx + 1] - g;
const B = data[idx + 2] - b;
const d = R * R + G * G + B * B;
if (d === 0) { // found exact match
return {...idx2Coord(idx / 4, width), distance: 0};
}
if (d < minDist) {
minDist = d;
found = idx;
}
idx += 4;
}
return found ? {...idx2Coord(found / 4, width), distance: minDist ** 0.5 | 0 } : undefined;
}
})();
This function has been tested and works as described above.
Note Going by the code in the your question the alpha value of the image and CSS hex color is ignored.
Note that if you intend to find many colors from the same image this function is not the best suited for you needs. If this is the case let me know in the comment and I can make changes or instruct you how to optimism the code for such uses.
Note It is not well suited for single use only. However if this is the case change the line const findPixel = (() => { to var findPixel = (() => { and after you have used it remove the reference findpixel = undefined; and JS will clean up any resources it holds.
Note If you also want to get the actual color of the closest found color that is trivial to add as well. Ask in the comments.
Note It is reasonably quick (you will be hard pressed to get a quicker result) but be warned that for very large images 4K and above it may take a bit, and on very low end devices it may cause a out of memory error. If this is a problem then another solution is possible but is far slower.
I have an array of "indexed" RGBA color values, and for any given image that I load, I want to be able to run through all the color values of the loaded pixels, and match them to the closest of my indexed color values. So, if the pixel in the image had a color of, say, RGBA(0,0,10,1), and the color RGBA(0,0,0,1) was my closest indexed value, it would adjust the loaded pixel to RGBA(0,0,0,1).
I know PHP has a function imagecolorclosest
int imagecolorclosest( $image, $red, $green, $blue )
Does javascript / p5.js / processing have anything similar? What's the easiest way to compare one color to another. Currently I can read the pixels of the image with this code (using P5.js):
let img;
function preload() {
img = loadImage('assets/00.jpg');
}
function setup() {
image(img, 0, 0, width, height);
let d = pixelDensity();
let fullImage = 4 * (width * d) * (height * d);
loadPixels();
for (let i = 0; i < fullImage; i+=4) {
let curR = pixels[i];
let curG = pixels[i]+1;
let curB = pixels[i+2];
let curA = pixels[i+3];
}
updatePixels();
}
Each color consists 3 color channels. Imagine the color as a point in a 3 dimensional space, where each color channel (red, green, blue) is associated to one dimension. You've to find the closest color (point) by the Euclidean distance. The color with the lowest distance is the "closest" color.
In p5.js you can use p5.Vector for vector arithmetic. The Euclidean distance between to points can be calculated by .dist(). So the distance between points respectively "colors" a and b can be expressed by:
let a = createVector(r1, g1, b1);
let b = createVector(r2, g2, b2);
let distance = a.dist(b);
Use the expression somehow like this:
colorTable = [[r0, g0, b0], [r1, g1, b1] ... ];
int closesetColor(r, g, b) {
let a = createVector(r, g, b);
let minDistance;
let minI;
for (let i=0; i < colorTable; ++i) {
let b = createVector(...colorTable[i]);
let distance = a.dist(b);
if (!minDistance || distance < minDistance) {
minI = i; minDistance = distance;
}
}
return minI;
}
function setup() {
image(img, 0, 0, width, height);
let d = pixelDensity();
let fullImage = 4 * (width * d) * (height * d);
loadPixels();
for (let i = 0; i < fullImage; i+=4) {
let closestI = closesetColor(pixels[i], pixels[i+1], pixels[i+2])
pixels[i] = colorTable[closestI][0];
pixels[i+1] = colorTable[closestI][1];
pixels[i+2] = colorTable[closestI][2];
}
updatePixels();
}
If I understand you correctly you want to keep the colors of an image within a certain limited pallet. If so, you should apply this function to each pixel of your image. It will give you the closest color value to a supplied pixel from a set of limited colors (indexedColors).
// example color pallet (no alpha)
indexedColors = [
[0, 10, 0],
[0, 50, 0]
];
// Takes pixel with no alpha value
function closestIndexedColor(color) {
var closest = {};
var dist;
for (var i = 0; i < indexedColors.length; i++) {
dist = Math.pow(indexedColors[i][0] - color[0], 2);
dist += Math.pow(indexedColors[i][1] - color[1], 2);
dist += Math.pow(indexedColors[i][2] - color[2], 2);
dist = Math.sqrt(dist);
if(!closest.dist || closest.dist > dist){
closest.dist = dist;
closest.color = indexedColors[i];
}
}
// returns closest match as RGB array without alpha
return closest.color;
}
// example usage
closestIndexedColor([0, 20, 0]); // returns [0, 10, 0]
It works the way that the PHP function you mentioned does. If you treat the color values as 3d coordinate points then the closet colors will be the ones with the smallest 3d "distance" between them. This 3d distance is calculated using the distance formula:
I'm trying to make a text effect similar to the effect found at the bottom of this article
My proposed approach is:
Make two canvasses, one is visible, the other is invisible I use this as a buffer.
Draw some text on the buffer canvas
Loop over getImageData pixels
if pixel alpha is not equal to zero (when there is a pixel drawn on the canvas buffer) with a small chance, ie 2%, draw a randomly generated circle with cool effecs at that pixel on the visible canvas.
I'm having trouble at step 4. With the code below, I'm trying to replicate the text on the second canvas, in full red. Instead I get this weird picture.
code
// create the canvas to replicate the buffer text on.
var draw = new Drawing(true);
var bufferText = function (size, textFont) {
// set the font to Georgia if it isn't defined
textFont = textFont || "Georgia";
// create a new canvas buffer, true means that it's visible on the screen
// Note, Drawing is a small library I wrote, it's just a wrapper over the canvas API
// it creates a new canvas and adds some functions to the context
// it doesn't change any of the original functions
var buffer = new Drawing(true);
// context is just a small wrapper library I wrote to make the canvas API a little more bearable.
with (buffer.context) {
font = util.format("{size}px {font}", {size: size, font: textFont});
fillText("Hi there", 0, size);
}
// get the imagedata and store the actual pixels array in data
var imageData = buffer.context.getImageData(0, 0, buffer.canvas.width, buffer.canvas.height);
var data = imageData.data;
var index, alpha, x, y;
// loop over the pixels
for (x = 0; x < imageData.width; x++) {
for (y = 0; y < imageData.height; y++) {
index = x * y * 4;
alpha = data[index + 3];
// if the alpha is not equal to 0, draw a red pixel at (x, y)
if (alpha !== 0) {
with (draw.context) {
dot(x/4, y/4, {fillColor: "red"})
}
}
}
}
};
bufferText(20);
Note that here, my buffer is actually visible to show where the red pixels are supposed to go compared to where they actually go.
I'm really confused by this problem.
If anybody knows an alternative approach, that's very welcome too.
replace this...
index = x * y * 4;
with...
index = (imageData.width * y) + x;
the rest is good :)
So, i'm trying to implement hough transform, this version is 1-dimensional (its for all dims reduced to 1 dim optimization) version based on the minor properties.
Enclosed is my code, with a sample image... input and output.
Obvious question is what am i doing wrong. I've tripled check my logic and code and it looks good also my parameters. But obviously i'm missing on something.
Notice that the red pixels are supposed to be ellipses centers , while the blue pixels are edges to be removed (belong to the ellipse that conform to the mathematical equations).
also, i'm not interested in openCV / matlab / ocatve / etc.. usage (nothing against them).
Thank you very much!
var fs = require("fs"),
Canvas = require("canvas"),
Image = Canvas.Image;
var LEAST_REQUIRED_DISTANCE = 40, // LEAST required distance between 2 points , lets say smallest ellipse minor
LEAST_REQUIRED_ELLIPSES = 6, // number of found ellipse
arr_accum = [],
arr_edges = [],
edges_canvas,
xy,
x1y1,
x2y2,
x0,
y0,
a,
alpha,
d,
b,
max_votes,
cos_tau,
sin_tau_sqr,
f,
new_x0,
new_y0,
any_minor_dist,
max_minor,
i,
found_minor_in_accum,
arr_edges_len,
hough_file = 'sample_me2.jpg',
edges_canvas = drawImgToCanvasSync(hough_file); // make sure everything is black and white!
arr_edges = getEdgesArr(edges_canvas);
arr_edges_len = arr_edges.length;
var hough_canvas_img_data = edges_canvas.getContext('2d').getImageData(0, 0, edges_canvas.width,edges_canvas.height);
for(x1y1 = 0; x1y1 < arr_edges_len ; x1y1++){
if (arr_edges[x1y1].x === -1) { continue; }
for(x2y2 = 0 ; x2y2 < arr_edges_len; x2y2++){
if ((arr_edges[x2y2].x === -1) ||
(arr_edges[x2y2].x === arr_edges[x1y1].x && arr_edges[x2y2].y === arr_edges[x1y1].y)) { continue; }
if (distance(arr_edges[x1y1],arr_edges[x2y2]) > LEAST_REQUIRED_DISTANCE){
x0 = (arr_edges[x1y1].x + arr_edges[x2y2].x) / 2;
y0 = (arr_edges[x1y1].y + arr_edges[x2y2].y) / 2;
a = Math.sqrt((arr_edges[x1y1].x - arr_edges[x2y2].x) * (arr_edges[x1y1].x - arr_edges[x2y2].x) + (arr_edges[x1y1].y - arr_edges[x2y2].y) * (arr_edges[x1y1].y - arr_edges[x2y2].y)) / 2;
alpha = Math.atan((arr_edges[x2y2].y - arr_edges[x1y1].y) / (arr_edges[x2y2].x - arr_edges[x1y1].x));
for(xy = 0 ; xy < arr_edges_len; xy++){
if ((arr_edges[xy].x === -1) ||
(arr_edges[xy].x === arr_edges[x2y2].x && arr_edges[xy].y === arr_edges[x2y2].y) ||
(arr_edges[xy].x === arr_edges[x1y1].x && arr_edges[xy].y === arr_edges[x1y1].y)) { continue; }
d = distance({x: x0, y: y0},arr_edges[xy]);
if (d > LEAST_REQUIRED_DISTANCE){
f = distance(arr_edges[xy],arr_edges[x2y2]); // focus
cos_tau = (a * a + d * d - f * f) / (2 * a * d);
sin_tau_sqr = (1 - cos_tau * cos_tau);//Math.sqrt(1 - cos_tau * cos_tau); // getting sin out of cos
b = (a * a * d * d * sin_tau_sqr ) / (a * a - d * d * cos_tau * cos_tau);
b = Math.sqrt(b);
b = parseInt(b.toFixed(0));
d = parseInt(d.toFixed(0));
if (b > 0){
found_minor_in_accum = arr_accum.hasOwnProperty(b);
if (!found_minor_in_accum){
arr_accum[b] = {f: f, cos_tau: cos_tau, sin_tau_sqr: sin_tau_sqr, b: b, d: d, xy: xy, xy_point: JSON.stringify(arr_edges[xy]), x0: x0, y0: y0, accum: 0};
}
else{
arr_accum[b].accum++;
}
}// b
}// if2 - LEAST_REQUIRED_DISTANCE
}// for xy
max_votes = getMaxMinor(arr_accum);
// ONE ellipse has been detected
if (max_votes != null &&
(max_votes.max_votes > LEAST_REQUIRED_ELLIPSES)){
// output ellipse details
new_x0 = parseInt(arr_accum[max_votes.index].x0.toFixed(0)),
new_y0 = parseInt(arr_accum[max_votes.index].y0.toFixed(0));
setPixel(hough_canvas_img_data,new_x0,new_y0,255,0,0,255); // Red centers
// remove the pixels on the detected ellipse from edge pixel array
for (i=0; i < arr_edges.length; i++){
any_minor_dist = distance({x:new_x0, y: new_y0}, arr_edges[i]);
any_minor_dist = parseInt(any_minor_dist.toFixed(0));
max_minor = b;//Math.max(b,arr_accum[max_votes.index].d); // between the max and the min
// coloring in blue the edges we don't need
if (any_minor_dist <= max_minor){
setPixel(hough_canvas_img_data,arr_edges[i].x,arr_edges[i].y,0,0,255,255);
arr_edges[i] = {x: -1, y: -1};
}// if
}// for
}// if - LEAST_REQUIRED_ELLIPSES
// clear accumulated array
arr_accum = [];
}// if1 - LEAST_REQUIRED_DISTANCE
}// for x2y2
}// for xy
edges_canvas.getContext('2d').putImageData(hough_canvas_img_data, 0, 0);
writeCanvasToFile(edges_canvas, __dirname + '/hough.jpg', function() {
});
function getMaxMinor(accum_in){
var max_votes = -1,
max_votes_idx,
i,
accum_len = accum_in.length;
for(i in accum_in){
if (accum_in[i].accum > max_votes){
max_votes = accum_in[i].accum;
max_votes_idx = i;
} // if
}
if (max_votes > 0){
return {max_votes: max_votes, index: max_votes_idx};
}
return null;
}
function distance(point_a,point_b){
return Math.sqrt((point_a.x - point_b.x) * (point_a.x - point_b.x) + (point_a.y - point_b.y) * (point_a.y - point_b.y));
}
function getEdgesArr(canvas_in){
var x,
y,
width = canvas_in.width,
height = canvas_in.height,
pixel,
edges = [],
ctx = canvas_in.getContext('2d'),
img_data = ctx.getImageData(0, 0, width, height);
for(x = 0; x < width; x++){
for(y = 0; y < height; y++){
pixel = getPixel(img_data, x,y);
if (pixel.r !== 0 &&
pixel.g !== 0 &&
pixel.b !== 0 ){
edges.push({x: x, y: y});
}
} // for
}// for
return edges
} // getEdgesArr
function drawImgToCanvasSync(file) {
var data = fs.readFileSync(file)
var canvas = dataToCanvas(data);
return canvas;
}
function dataToCanvas(imagedata) {
img = new Canvas.Image();
img.src = new Buffer(imagedata, 'binary');
var canvas = new Canvas(img.width, img.height);
var ctx = canvas.getContext('2d');
ctx.patternQuality = "best";
ctx.drawImage(img, 0, 0, img.width, img.height,
0, 0, img.width, img.height);
return canvas;
}
function writeCanvasToFile(canvas, file, callback) {
var out = fs.createWriteStream(file)
var stream = canvas.createPNGStream();
stream.on('data', function(chunk) {
out.write(chunk);
});
stream.on('end', function() {
callback();
});
}
function setPixel(imageData, x, y, r, g, b, a) {
index = (x + y * imageData.width) * 4;
imageData.data[index+0] = r;
imageData.data[index+1] = g;
imageData.data[index+2] = b;
imageData.data[index+3] = a;
}
function getPixel(imageData, x, y) {
index = (x + y * imageData.width) * 4;
return {
r: imageData.data[index+0],
g: imageData.data[index+1],
b: imageData.data[index+2],
a: imageData.data[index+3]
}
}
It seems you try to implement the algorithm of Yonghong Xie; Qiang Ji (2002). A new efficient ellipse detection method 2. p. 957.
Ellipse removal suffers from several bugs
In your code, you perform the removal of found ellipse (step 12 of the original paper's algorithm) by resetting coordinates to {-1, -1}.
You need to add:
`if (arr_edges[x1y1].x === -1) break;`
at the end of the x2y2 block. Otherwise, the loop will consider -1, -1 as a white point.
More importantly, your algorithm consists in erasing every point which distance to the center is smaller than b. b supposedly is the minor axis half-length (per the original algorithm). But in your code, variable b actually is the latest (and not most frequent) half-length, and you erase points with a distance lower than b (instead of greater, since it's the minor axis). In other words, you clear all points inside a circle with a distance lower than latest computed axis.
Your sample image can actually be processed with a clearing of all points inside a circle with a distance lower than selected major axis with:
max_minor = arr_accum[max_votes.index].d;
Indeed, you don't have overlapping ellipses and they are spread enough. Please consider a better algorithm for overlapping or closer ellipses.
The algorithm mixes major and minor axes
Step 6 of the paper reads:
For each third pixel (x, y), if the distance between (x, y) and (x0,
y0) is greater than the required least distance for a pair of pixels
to be considered then carry out the following steps from (7) to (9).
This clearly is an approximation. If you do so, you will end up considering points further than the minor axis half length, and eventually on the major axis (with axes swapped). You should make sure the distance between the considered point and the tested ellipse center is smaller than currently considered major axis half-length (condition should be d <= a). This will help with the ellipse erasing part of the algorithm.
Also, if you also compare with the least distance for a pair of pixels, as per the original paper, 40 is too large for the smaller ellipse in your picture. The comment in your code is wrong, it should be at maximum half the smallest ellipse minor axis half-length.
LEAST_REQUIRED_ELLIPSES is too small
This parameter is also misnamed. It is the minimum number of votes an ellipse should get to be considered valid. Each vote corresponds to a pixel. So a value of 6 means that only 6+2 pixels make an ellipse. Since pixels coordinates are integers and you have more than 1 ellipse in your picture, the algorithm might detect ellipses that are not, and eventually clear edges (especially when combined with the buggy ellipse erasing algorithm). Based on tests, a value of 100 will find four of the five ellipses of your picture, while 80 will find them all. Smaller values will not find the proper centers of the ellipses.
Sample image is not black & white
Despite the comment, sample image is not exactly black and white. You should convert it or apply some threshold (e.g. RGB values greater than 10 instead of simply different form 0).
Diff of minimum changes to make it work is available here:
https://gist.github.com/pguyot/26149fec29ffa47f0cfb/revisions
Finally, please note that parseInt(x.toFixed(0)) could be rewritten Math.floor(x), and you probably want to not truncate all floats like this, but rather round them, and proceed where needed: the algorithm to erase the ellipse from the picture would benefit from non truncated values for the center coordinates. This code definitely could be improved further, for example it currently computes the distance between points x1y1 and x2y2 twice.
I'm trying to implement ColorPicker using Canvas just for fun. But i seem lost. as my browser is freezing for a while when it loads due to all these for loops.
I'm adding the screenshot of the result of this script:
window.onload = function(){
colorPicker();
}
function colorPicker(){
var canvas = document.getElementById("colDisp"),
frame = canvas.getContext("2d");
var r=0,
g=0,
b= 0;
function drawColor(){
for(r=0;r<255;r++){
for(g=0;g<255;g++){
for(b=0;b<255;b++){
frame.fillStyle="rgb("+r+","+g+","+b+")";
frame.fillRect(r,g,1,1);
}
}
}
}
drawColor();
Currently , i only want a solution about the freezing problem with better algorithm and it's not displaying the BLACK and GREY colors.
Please someone help me.
Instead of calling fillRect for every single pixel, it might be a lot more efficient to work with a raw RGBA buffer. You can obtain one using context.getImageData, fill it with the color values, and then put it back in one go using context.putImageData.
Note that your current code overwrites each single pixel 255 times, once for each possible blue-value. The final pass on each pixel is 255 blue, so you see no grey and black in the output.
Finding a good way to map all possible RGB values to a two-dimensional image isn't trivial, because RGB is a three-dimensional color-space. There are a lot of strategies for doing so, but none is really optimal for any possible use-case. You can find some creative solutions for this problem on AllRGB.com. A few of them might be suitable for a color-picker for some use-cases.
If you want to fetch the rgba of the pixel under the mouse, you must use context.getImageData.
getImageData returns an array of pixels.
var pixeldata=context.getImageData(0,0,canvas.width,canvas.height);
Each pixel is defined by 4 sequential array elements.
So if you have gotten a pixel array with getImageData:
// first pixel defined by the first 4 pixel array elements
pixeldata[0] = red component of pixel#1
pixeldata[1] = green component of pixel#1
pixeldata[2] = blue component of pixel#1
pixeldata[4] = alpha (opacity) component of pixel#1
// second pixel defined by the next 4 pixel array elements
pixeldata[5] = red component of pixel#2
pixeldata[6] = green component of pixel#2
pixeldata[7] = blue component of pixel#2
pixeldata[8] = alpha (opacity) component of pixel#2
So if you have a mouseX and mouseY then you can get the r,g,b,a values under the mouse like this:
// get the offset in the array where mouseX,mouseY begin
var offset=(imageWidth*mouseY+mouseX)*4;
// read the red,blue,green and alpha values of that pixel
var red = pixeldata[offset];
var green = pixeldata[offset+1];
var blue = pixeldata[offset+2];
var alpha = pixeldata[offset+3];
Here's a demo that draws a colorwheel on the canvas and displays the RGBA under the mouse:
http://jsfiddle.net/m1erickson/94BAQ/
A way to go, using .createImageData():
window.onload = function() {
var canvas = document.getElementById("colDisp");
var frame = canvas.getContext("2d");
var width = canvas.width;
var height = canvas.height;
var imagedata = frame.createImageData(width, height);
var index, x, y;
for (x = 0; x < width; x++) {
for (y = 0; y < height; y++) {
index = (x * width + y) * 4;
imagedata.data[index + 0] = x;
imagedata.data[index + 1] = y;
imagedata.data[index + 2] = x + y - 255;
imagedata.data[index + 3] = 255;
}
}
frame.putImageData(imagedata, 0, 0);
};
http://codepen.io/anon/pen/vGcaF