patternjavascriptMinor
Pixel manipulation performance
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pixelmanipulationperformance
Problem
I have created a function that adds "shadow" to the content of a canvas. The algorithm for adding shadow follows these rules:
This algorithm runs from top to bottom, left to right. Here's the relevant part, where
As you can see, this function performs per pixel manipulation and is quite slow in performance. What is a good way to improve performance, probably without checking each pixel independently?
- If the either
xoryis 0, ignore.
- If the pixel on the current pixel's top left corner is not completely transparent, set current pixel's alpha value to 0.1.
This algorithm runs from top to bottom, left to right. Here's the relevant part, where
rendered is a canvas that already contain rendered content.function getIndex(x, y){
return (y * width + x) * 4;
}
function isClear(x, y){
var index = getIndex(x, y);
return data[index + 3] === 0;
}
function isInShadow(x, y){
if(x < 1 || y < 1) return false;
if(isClear(x, y) && !isClear(x - 1, y - 1)) return true;
return false;
}
var width = rendered.width, height = rendered.height,
imgData = renderedCtx.getImageData(0, 0, width, height),
data = imgData.data;
for(var x = 0; x < width; x++){
for(var y = 0; y < height; y++){
if(isInShadow(x, y)){
data[getIndex(x, y) + 3] = 256 * 0.1;
}
}
}
renderedCtx.putImageData(imgData, 0, 0);As you can see, this function performs per pixel manipulation and is quite slow in performance. What is a good way to improve performance, probably without checking each pixel independently?
Solution
Use diagonals. The idea is to have only one pixel check, compared to two you have now:
Considering everything should be inlined, the code would be like this:
Another potentially promising idea would be to use asm.js.
- find the first occupied pixel on a diagonal
- skip to an unoccupied one
- fill with 0.1 while unoccupied
- repeat 1-3 until out of bounds
Considering everything should be inlined, the code would be like this:
var x, y, index, index0,
len = width * height * 4,
stride = width * 4;
strideDia = stride + 4; // distance to the sibling pixel towards bottom-right// walk diagonals from the bottom-left corner to the top-left corner
for (x = 0, y = height, index0 = (y * width + x)*4 + 3; --y >= 0; ) {
index = index0 -= stride;
while (index < len && x < width) {
// find an occupied pixel
while (index < len && x < width && !data[index])
index += strideDia, x++;
// skip to an unoccupied
while (index < len && x < width && data[index])
index += strideDia, x++;
// fill with 0.1 all unoccupied
while (index < len && x < width && !data[index]) {
data[index] = 25; // 0.1 * 256
index += strideDia;
x++;
}
}
x = 0;
}// walk diagonals from the top-left corner to the top-right corner
// (0,0) is skipped as it was processed in the previous walk block
for (x = 0, y = 0, index0 = (y * width + x)*4 + 3; ++x < width; ) {
index = index0 += 4;
var x0 = x;
while (index < len && x < width) {
// find an occupied pixel
while (index < len && x < width && !data[index])
index += strideDia, x++;
// skip to an unoccupied
while (index < len && x < width && data[index])
index += strideDia, x++;
// fill with 0.1 all unoccupied
while (index < len && x < width && !data[index]) {
data[index] = 25; // 0.1 * 256
index += strideDia;
x++;
}
}
x = x0;
}Another potentially promising idea would be to use asm.js.
Code Snippets
var x, y, index, index0,
len = width * height * 4,
stride = width * 4;
strideDia = stride + 4; // distance to the sibling pixel towards bottom-right// walk diagonals from the bottom-left corner to the top-left corner
for (x = 0, y = height, index0 = (y * width + x)*4 + 3; --y >= 0; ) {
index = index0 -= stride;
while (index < len && x < width) {
// find an occupied pixel
while (index < len && x < width && !data[index])
index += strideDia, x++;
// skip to an unoccupied
while (index < len && x < width && data[index])
index += strideDia, x++;
// fill with 0.1 all unoccupied
while (index < len && x < width && !data[index]) {
data[index] = 25; // 0.1 * 256
index += strideDia;
x++;
}
}
x = 0;
}// walk diagonals from the top-left corner to the top-right corner
// (0,0) is skipped as it was processed in the previous walk block
for (x = 0, y = 0, index0 = (y * width + x)*4 + 3; ++x < width; ) {
index = index0 += 4;
var x0 = x;
while (index < len && x < width) {
// find an occupied pixel
while (index < len && x < width && !data[index])
index += strideDia, x++;
// skip to an unoccupied
while (index < len && x < width && data[index])
index += strideDia, x++;
// fill with 0.1 all unoccupied
while (index < len && x < width && !data[index]) {
data[index] = 25; // 0.1 * 256
index += strideDia;
x++;
}
}
x = x0;
}Context
StackExchange Code Review Q#137870, answer score: 7
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