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JavaScript pixel by pixel canvas manipulation

I'm working on a simple web app which simplifies the colours of an uploaded image to a colour palette selected by the user. The script works, but it takes a really long time to loop through the whole image (for large images it's over a few minutes), changing the pixels.
Initially, I was writing to the canvas itself, but I changed the code so that changes are made to an ImageData object and the canvas is only updated at the end of the script. However, this didn't really make much difference.
// User selects colours:
colours = [[255,45,0], [37,36,32], [110,110,105], [18,96,4]];
function colourDiff(colour1, colour2) {
difference = 0
difference += Math.abs(colour1[0] - colour2[0]);
difference += Math.abs(colour1[1] - colour2[1]);
difference += Math.abs(colour1[2] - colour2[2]);
return(difference);
}
function getPixel(imgData, index) {
return(imgData.data.slice(index*4, index*4+4));
}
function setPixel(imgData, index, pixelData) {
imgData.data.set(pixelData, index*4);
}
data = ctx.getImageData(0,0,canvas.width,canvas.height);
for(i=0; i<(canvas.width*canvas.height); i++) {
pixel = getPixel(data, i);
lowestDiff = 1024;
lowestColour = [0,0,0];
for(colour in colours) {
colour = colours[colour];
difference = colourDiff(colour, pixel);
if(lowestDiff < difference) {
continue;
}
lowestDiff = difference;
lowestColour = colour;
}
console.log(i);
setPixel(data, i, lowestColour);
}
ctx.putImageData(data, 0, 0);
During the entire process, the website is completely frozen, so I can't even display a progress bar. Is there any way to optimise this so that it takes less time?

There is no need to slice the array each iteration. (As niklas has already stated).
I would loop over the data array instead of looping over the canvas dimensions and directly edit the array.
for(let i = 0; i < data.length; i+=4) { // i+=4 to step over each r,g,b,a pixel
let pixel = getPixel(data, i);
...
setPixel(data, i, lowestColour);
}
function setPixel(data, i, colour) {
data[i] = colour[0];
data[i+1] = colour[1];
data[i+2] = colour[2];
}
function getPixel(data, i) {
return [data[i], data[i+1], data[i+2]];
}
Also, console.log can bring a browser to it's knees if you've got the console open. If your image is 1920 x 1080 then you will be logging to the console 2,073,600 times.
You can also pass all of the processing off to a Web Worker for ultimate threaded performance. Eg. https://jsfiddle.net/pnmz75xa/

One problem or option for improvement is clearly your slice function, which will create a new array every time it is called, you do not need this. I would change the for loop like so:
for y in canvas.height {
for x in canvas.width {
//directly alter the canvas' pixels
}
}

Finding difference in color
I am adding an answer because you have use a very poor color match algorithm.
Finding how closely a color matches another is best done if you imagine each unique possible colour as a point in 3D space. The red, green, and blue values represent the x,y,z coordinate.
You can then use some basic geometry to locate the distance from one colour to the another.
// the two colours as bytes 0-255
const colorDist = (r1, g1, b1, r2, g2, b2) => Math.hypot(r1 - r2, g1 - g2, b1 - b2);
It is also important to note that the channel value 0-255 is a compressed value, the actual intensity is close to that value squared (channelValue ** 2.2). That means that red = 255 is 65025 times more intense than red = 1
The following function is a close approximation of the colour difference between two colors. Avoiding the Math.hypot function as it is very slow.
const pallet = [[1,2,3],[2,10,30]]; // Array of arrays representing rgb byte
// of the colors you are matching
function findClosest(r,g,b) {
var closest;
var dist = Infinity;
r *= r;
g *= g;
b *= b;
for (const col of pallet) {
const d = ((r - col[0] * col[0]) + (g - col[1] * col[1]) + (b - col[2] * col[2])) ** 0.5;
if (d < dist) {
if (d === 0) { // if same then return result
return col;
}
closest = col;
dist = d;
}
}
return closest;
}
As for performance, your best bet is either via a web worker, or use webGL to do the conversion in realtime.
If you want to keep it simple to prevent the code from blocking the page cut the job into smaller slices using a timer to allow the page breathing room.
The example uses setTimeout and performance.now() to do 10ms slices letting other page events and rendering to do there thing. It returns a promise that resolves when all pixels are processed
function convertBitmap(canvas, maxTime) { // maxTime in ms (1/1000 second)
return new Promise(allDone => {
const ctx = canvas.getContext("2d");
const pixels = ctx.getImageData(0, 0, canvas.width, canvas.height);
const data = pixels.data;
var idx = data.length / 4;
processPixels(); // start processing
function processPixels() {
const time = performance.now();
while (idx-- > 0) {
if (idx % 1024) { // check time every 1024 pixels
if (performance.now() - time > maxTime) {
setTimeout(processPixels, 0);
idx++;
return;
}
}
let i = idx * 4;
const col = findClosest(data[i], data[i + 1], data[i + 2]);
data[i++] = col[0];
data[i++] = col[1];
data[i] = col[2];
}
ctx.putImageData(pixels, 0, 0);
allDone("Pixels processed");
}
});
}
// process pixels in 10ms slices.
convertBitmap(myCanvas, 10).then(mess => console.log(mess));

Canvas get only opaque pixels

How to count transperent pixels of the area at image canvas, i wrote this function, but i guess there is something like a another method
ctx = getCanvasImg();
for (var x = selection.x2 * zoom; x > selection.x1 * zoom; x--) {
for (var y = selection.y2 * zoom; y > selection.y1 * zoom; y--) {
var pixel = ctx.getImageData(x, y, 1, 1);
var data = pixel.data;
var rgba = 'rgba(' + data[0] + ',' + data[1] + ',' + data[2] + ',' + (data[3] / 255) + ')';
if (data[0] == 0) {
countWhite++;
}
}
}

Just grab the area in need once:
var idata = ctx.getImageData(startX, startY, widthOfArea, heightOfArea);
To count all opaque pixels, you can iterate using a Uint32Array instead:
var data = new Uint32Array(idata.data.buffer);
// little-endian byte order
for(var i = 0, len = data.length, count = 0; i < len;) {
if (data[i++] >>> 24 === 255) count++; // shifts AABBGGRR to 000000AA, then =255?
}
// count = number of opaque pixels
Update: Just to clarify in regards to comments: this do assume little-endian CPU architecture. Most mainstream/consumer devices nowadays uses this CPU architecture and this is usually not a problem. However, if you suspect your code will come across big-endian systems you have to check for this in advance and reverse the byte-order in the check (for anything using typed arrays views wider than 8-bits as they will use the native systems's architecture).
To check for endianess on a system you can do this:
function isLittleEndian() {
return new Uint16Array(new Uint8Array([255,0]).buffer)[0] === 255
}
And if big-endian, check this way instead using a and mask instead of bit-shift:
// big-endian byte order
for(var i = 0, len = data.length, count = 0; i < len;) {
if (data[i++] & 0xff === 255) count++; // masks RRGGBBAA to 000000AA, then =255?
}
You can always use DataViews and specify endianess, however in this case that is of no benefit due to the slight overhead.

Find regions of similar color in image

I've been working on this problem for some time now with little promising results. I am trying to split up an image into connected regions of similar color. (basically split a list of all the pixels into multiple groups (each group containing the coordinates of the pixels that belong to it and share a similar color).
For example:
http://unsplash.com/photos/SoC1ex6sI4w/
In this image the dark clouds at the top would probably fall into one group. Some of the grey rock on the mountain in another, and some of the orange grass in another. The snow would be another - the red of the backpack - etc.
I'm trying to design an algorithm that will be both accurate and efficient (it needs to run in a matter of ms on midrange laptop grade hardware)
Below is what I have tried:
Using a connected component based algorithm to go through every pixel from top left scanning every line of pixels from left to right (and comparing the current pixel to the top pixel and left pixel). Using the CIEDE2000 color difference formula if the pixel at the top or left was within a certain range then it would be considered "similar" and part of the group.
This sort of worked - but the problem is it relies on color regions having sharp edges - if any color groups are connected by a soft gradient it will travel down that gradient and continue to "join" the pixels as the difference between the individual pixels being compared is small enough to be considered "similar".
To try to fix this I chose to set every visited pixel's color to the color of most "similar" adjacent pixel (either top or left). If there are no similar pixels than it retains it's original color. This somewhat fixes the issue of more blurred boundaries or soft edges because the first color of a new group will be "carried" along as the algorithm progresses and eventually the difference between that color and the current compared color will exceed the "similarity" threashold and no longer be part of that group.
Hopefully this is making sense. The problem is neither of these options are really working. On the image above what is returned are not clean groups but noisy fragmented groups that is not what I am looking for.
I'm not looking for code specifically - but more ideas as to how an algorithm could be structured to successfully combat this problem. Does anyone have ideas about this?
Thanks!

You could convert from RGB to HSL to make it easier to calculate the distance between the colors. I'm setting the color difference tolerance in the line:
if (color_distance(original_pixels[i], group_headers[j]) < 0.3) {...}
If you change 0.3, you can get different results.
See it working.
Please, let me know if it helps.
function hsl_to_rgb(h, s, l) {
// from http://stackoverflow.com/questions/2353211/hsl-to-rgb-color-conversion
var r, g, b;
if (s == 0) {
r = g = b = l; // achromatic
} else {
var hue2rgb = function hue2rgb(p, q, t) {
if (t < 0) t += 1;
if (t > 1) t -= 1;
if (t < 1 / 6) return p + (q - p) * 6 * t;
if (t < 1 / 2) return q;
if (t < 2 / 3) return p + (q - p) * (2 / 3 - t) * 6;
return p;
}
var q = l < 0.5 ? l * (1 + s) : l + s - l * s;
var p = 2 * l - q;
r = hue2rgb(p, q, h + 1 / 3);
g = hue2rgb(p, q, h);
b = hue2rgb(p, q, h - 1 / 3);
}
return [Math.round(r * 255), Math.round(g * 255), Math.round(b * 255)];
}
function rgb_to_hsl(r, g, b) {
// from http://stackoverflow.com/questions/2353211/hsl-to-rgb-color-conversion
r /= 255, g /= 255, b /= 255;
var max = Math.max(r, g, b),
min = Math.min(r, g, b);
var h, s, l = (max + min) / 2;
if (max == min) {
h = s = 0; // achromatic
} else {
var d = max - min;
s = l > 0.5 ? d / (2 - max - min) : d / (max + min);
switch (max) {
case r:
h = (g - b) / d + (g < b ? 6 : 0);
break;
case g:
h = (b - r) / d + 2;
break;
case b:
h = (r - g) / d + 4;
break;
}
h /= 6;
}
return [h, s, l];
}
function color_distance(v1, v2) {
// from http://stackoverflow.com/a/13587077/1204332
var i,
d = 0;
for (i = 0; i < v1.length; i++) {
d += (v1[i] - v2[i]) * (v1[i] - v2[i]);
}
return Math.sqrt(d);
};
function round_to_groups(group_nr, x) {
var divisor = 255 / group_nr;
return Math.ceil(x / divisor) * divisor;
};
function pixel_data_to_key(pixel_data) {
return pixel_data[0].toString() + '-' + pixel_data[1].toString() + '-' + pixel_data[2].toString();
}
function posterize(context, image_data, palette) {
for (var i = 0; i < image_data.data.length; i += 4) {
rgb = image_data.data.slice(i, i + 3);
hsl = rgb_to_hsl(rgb[0], rgb[1], rgb[2]);
key = pixel_data_to_key(hsl);
if (key in palette) {
new_hsl = palette[key];
new_rgb = hsl_to_rgb(new_hsl[0], new_hsl[1], new_hsl[2]);
rgb = hsl_to_rgb(hsl);
image_data.data[i] = new_rgb[0];
image_data.data[i + 1] = new_rgb[1];
image_data.data[i + 2] = new_rgb[2];
}
}
context.putImageData(image_data, 0, 0);
}
function draw(img) {
var canvas = document.getElementById('canvas');
var context = canvas.getContext('2d');
context.drawImage(img, 0, 0, canvas.width, canvas.height);
img.style.display = 'none';
var image_data = context.getImageData(0, 0, canvas.width, canvas.height);
var data = image_data.data;
context.drawImage(target_image, 0, 0, canvas.width, canvas.height);
data = context.getImageData(0, 0, canvas.width, canvas.height).data;
original_pixels = [];
for (i = 0; i < data.length; i += 4) {
rgb = data.slice(i, i + 3);
hsl = rgb_to_hsl(rgb[0], rgb[1], rgb[2]);
original_pixels.push(hsl);
}
group_headers = [];
groups = {};
for (i = 0; i < original_pixels.length; i += 1) {
if (group_headers.length == 0) {
group_headers.push(original_pixels[i]);
}
group_found = false;
for (j = 0; j < group_headers.length; j += 1) {
// if a similar color was already observed
if (color_distance(original_pixels[i], group_headers[j]) < 0.3) {
group_found = true;
if (!(pixel_data_to_key(original_pixels[i]) in groups)) {
groups[pixel_data_to_key(original_pixels[i])] = group_headers[j];
}
}
if (group_found) {
break;
}
}
if (!group_found) {
if (group_headers.indexOf(original_pixels[i]) == -1) {
group_headers.push(original_pixels[i]);
}
if (!(pixel_data_to_key(original_pixels[i]) in groups)) {
groups[pixel_data_to_key(original_pixels[i])] = original_pixels[i];
}
}
}
posterize(context, image_data, groups)
}
var target_image = new Image();
target_image.crossOrigin = "";
target_image.onload = function() {
draw(target_image)
};
target_image.src = "http://i.imgur.com/zRzdADA.jpg";
canvas {
width: 300px;
height: 200px;
}
<canvas id="canvas"></canvas>

You can use "Mean Shift Filtering" algorithm to do the same.
Here's an example.
You will have to determine function parameters heuristically.
And here's the wrapper for the same in node.js
npm Wrapper for meanshift algorithm
Hope this helps!

The process you are trying to complete is called Image Segmentation and it's a well studied area in computer vision, with hundreds of different algorithms and implementations.
The algorithm you mentioned should work for simple images, however for real world images such as the one you linked to, you will probably need a more sophisticated algorithm, maybe even one that is domain specific (are all of your images contains a view?).
I have little experience in Node.js, however from Googling a bit I found the GraphicsMagic library, which as a segment function that might do the job (haven't verified).
In any case, I would try looking for "Image segmentation" libraries, and if possible, not limit myself only to Node.js implementations, as this language is not the common practice for writing vision applications, as opposed to C++ / Java / Python.

I would try a different aproach. Check out this description of how a flood fill algorithm could work:
Create an array to hold information about already colored coordinates.
Create a work list array to hold coordinates that must be looked at. Put the start position in it.
When the work list is empty, we are done.
Remove one pair of coordinates from the work list.
If those coordinates are already in our array of colored pixels, go back to step 3.
Color the pixel at the current coordinates and add the coordinates to the array of colored pixels.
Add the coordinates of each adjacent pixel whose color is the same as the starting pixel’s original color to the work list.
Return to step 3.
The "search approach" is superior because it does not only search from left to right, but in all directions.

You might look at k-means clustering.
http://docs.opencv.org/3.0-beta/modules/core/doc/clustering.html

How to dynamically and efficiently generate texture sizes and dimensions in powers of two

I've fleshed out a loop to dynamically generate texture sizes for each different view I have in a browser-based visualization using shaders. I know the minimum number of pixels I need in order to pass my values to the shaders; however I need to scale them up to a power of 2 size, and then make sure their x and y dimensions are also powers of two with a 1:1, 2:1, or 1:2 ratio. Right now my loop is infinite, and I suppose I will need to continue to increase the overall power of 2 pixel count until I reach a size that satisfies one of my ratios.
My question is: Is there a more efficient or direct way to achieve what I'm trying to do here?
var motifMinBufferSize = 80000;
var bufferSize; // the total number of texels that will be in the computation buffers (must be a power of two)
var dimensions;
function initValues() {
bufferSize = setBufferSize();
dimensions = setPositionsTextureSize();
}
function setBufferSize() {
var buffer = motifMinBufferSize;
// fill out the buffers to a power of two - necessary for the computation textures in the shaders
var powCount = 1;
var powOf2 = 2;
while ( buffer > powOf2 ) {
powOf2 *= 2;
powCount++;
}
while ( buffer < powOf2 ) {
buffer += 1;
}
}
function setPositionsTextureSize() {
var dimensions = {
texWidth : null,
texHeight : null
};
var foundDimensions = false;
var powOf2 = 2;
while (foundDimensions === false) {
var candidateWidth = bufferSize/powOf2;
if ( candidateWidth === powOf2 || candidateWidth/2 === powOf2 || candidateWidth*2 === powOf2 ) {
dimensions.texWidth = candidateWidth;
dimensions.textHeight = powOf2;
foundDimensions = true;
} else {
powOf2 *= 2;
}
}
return dimensions;
}

Your buffer must contain 2^n elements since both the width and height of the buffer are powers of two. The smallest n that satisfies the requirement of holding
at least motifMinBufferSize elements is calculated using logarithms: n = Math.ceil(Math.log2(motifMinBufferSize)).
Let's say height of the buffer is 2^h and the width of the buffer is 2^w. We know that w and h can differ by at most one (due to the restrictions on the ratio of the buffer dimensions). We also know that 2^n = 2^w * 2^h which means n = w + h. Since w and h differ by at most 1, they are both basically half of n. Therefore we can get:
function getBufferDimensions(minBufferSize) {
var n = Math.ceil(Math.log2(minBufferSize));
var w = Math.ceil(n / 2);
var h = n - w;
return {
width: Math.pow(2, w),
height: Math.pow(2, h),
};
}

How to get the average or main color from an image with javascript?

what i want is to the the HEX or the RGB average value from an image to the another div background this color.
So if i upload an image with a ot of red i get something like #FF0000 just as an example.
Let Me know if this is posible :)
Many thanks.

First, draw the image on a canvas:
function draw(img) {
var canvas = document.createElement("canvas");
var c = canvas.getContext('2d');
c.width = canvas.width = img.width;
c.height = canvas.height = img.height;
c.clearRect(0, 0, c.width, c.height);
c.drawImage(img, 0, 0, img.width , img.height);
return c; // returns the context
}
You can now iterate over the image's pixels. A naive approach for color-detection is to simply count the frequency of each color in the image.
// returns a map counting the frequency of each color
// in the image on the canvas
function getColors(c) {
var col, colors = {};
var pixels, r, g, b, a;
r = g = b = a = 0;
pixels = c.getImageData(0, 0, c.width, c.height);
for (var i = 0, data = pixels.data; i < data.length; i += 4) {
r = data[i];
g = data[i + 1];
b = data[i + 2];
a = data[i + 3]; // alpha
// skip pixels >50% transparent
if (a < (255 / 2))
continue;
col = rgbToHex(r, g, b);
if (!colors[col])
colors[col] = 0;
colors[col]++;
}
return colors;
}
function rgbToHex(r, g, b) {
if (r > 255 || g > 255 || b > 255)
throw "Invalid color component";
return ((r << 16) | (g << 8) | b).toString(16);
}
getColors returns a map of color names and counts. Transparent pixels are skipped. It should be trivial to get the most-frequently seen color from this map.
If you literally want an average of each color component, you could easily get that from the results of getColors, too, but the results aren't likely to be very useful. This answer explains a much better approach.
You can use it all like this:
// nicely formats hex values
function pad(hex) {
return ("000000" + hex).slice(-6);
}
// see this example working in the fiddle below
var info = document.getElementById("info");
var img = document.getElementById("squares");
var colors = getColors(draw(img));
for (var hex in colors) {
info.innerHTML += "<li>" + pad(hex) + "->" + colors[hex];
}
See a working example.

Put image on canvas.
Get 2D context.
Loop through pixels, and store each r,g,b value. If you find the same, increment it once.
Loop through stored r,g,b values and take note of largest r,g,b value.
Convert r,g,b to hex.

This is only possible using the canvas tag as described here :
http://dev.opera.com/articles/view/html-5-canvas-the-basics/#pixelbasedmanipulation
Of course this is only available in newer browsers

You might consider using the convolution filters css allows you to apply. This might be able to get the effect you're going for ( assuming you're wanting to present it back into the html). So you could display the image twice , one convolved.
That being said, doesn't really work if you need the information yourself for some purpose.

For finding that average color:
Put Image on Canvas
Resize image to 1px by 1px
Get the color of the resulting pixel(This pixel will be the calculated average)