js高斯模糊算法问题
<!DOCTYPE html>
<html lang="en">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=0" />
<title>高斯模糊</title>
</head>
<body>
<div id="wrap" class="wrap">
<img src="img/12338563.jpg" alt="" id="imageSource" height="500" width="800"/>
<canvas id="myCanvas"></canvas>
</div>
<script type="text/javascript">
//-------------------------------做一个基于canvas的【图片上传,剪切】的插件
//---------------------------------测试需在服务器环境下
window.onload = function() {
var canvas = document.getElementById("myCanvas");
var image = document.getElementById("imageSource");
// re-size the canvas deminsion
console.log(image.width)
canvas.width = image.width;
canvas.height = image.height;
// get 2D render object
var context = canvas.getContext("2d");
context.drawImage(image, 0, 0,canvas.width,canvas.height);
var canvasData = context.getImageData(0, 0, canvas.width, canvas.height); //获取图片信息
//对于一个数组来说,它的每一个元素并不拥有坐标,因此需要写一个方法,也就是告诉【x,y,radius】,就会返还一个数组,当然首先要确定的是【width,height】因为只有这样才能够给数组一个坐标
//对于canvas来说,确定【width,height】就确定了【canvasData.data】的长度为【width*height*4】
//但是坐标范围用width,height来固定
//-------------------------------------------------高斯模糊开始------------------------------------------------------
var radius = 20 ; //定义模糊半径
var quan = getQuan(radius);
var quanSub = ArrSub(quan);
quan = jiaQuan(quan,quanSub); //获取权值
var canvasDataCtrl = canvasData ;
var width = canvas.width ,
height = canvas.height ;
console.log(new Date().getSeconds()) //高斯模糊开始时间
for (var i=0;i<width ;i++ )
{
for (var j=0;j<height ;j++ )
{
var index= (width*4)*j+i*4;
canvasDataCtrl.data[index] = getValue(i,j,0,quan); //r
canvasDataCtrl.data[index+1] = getValue(i,j,1,quan); //g
canvasDataCtrl.data[index+2] = getValue(i,j,2,quan); //b
canvasDataCtrl.data[index+3] = getValue(i,j,3,quan); //a
}
}
context.putImageData(canvasDataCtrl, 0, 0);
console.log(new Date().getSeconds()) //高斯模糊结束时间
//console.log(value)
//在chrome上运行时间大概有30s,运行这么长时间,肯定是算法问题,网上的StackBlur运行时间就很短
//在计算量上,如果图片是像素是500×800,那这张图片所包含的rgba信息就有160,0000 之多
//再加上要对每个像素的rgba值进行高斯转换,计算量就相当之大了,怎么解决呢
console.log('模糊完成')
function gaosi(x,y,a){ //根据高斯二维公式,获取点的高斯值
var e = Math.E ;
var pi = Math.PI ;
return 1/(2*pi*a*a)*Math.pow(e,-((x*x+y*y)/(2*a*a)))
}
//假设radius=x,那么会获得一个(2*x+1)×(2*x+1)的数组矩阵
//数组第一个元素的坐标是【-x,x】
function getQuan(radius){ //获取每个点的高斯值,返回数组
var quan = []
for (var i=-radius;i<=radius ;i++ )
{
for (var j=radius;j>=-radius ;j-- )
{
quan.push(gaosi(i,j,20));
}
}
return quan ;
}
function ArrSub(arr){ //返回高斯数组所有元素的值的和
var sub = 0 ;
for (var i=0,len=quan.length;i<len ;i++ )
{
sub += quan[i] ;
}
return sub ;
}
function jiaQuan(arr,quanSub){ //获取权的数组
for (var i=0,len=arr.length;i<len ;i++ )
{
arr[i] = arr[i]/quanSub;
}
return arr ;
}
function getValue(x,y,index,quan){ //根据坐标以及rgba[index]来算出高斯模糊的最终值
var imgdata = getInfoArr(x,y,index)
var value = 0 ;
for (var i=0,len=imgdata.length;i<len ;i++ )
{
value += imgdata[i]*quan[i] ;
}
return Math.round(value) ;//返回整数
}
function getInfo(x,y,index){ //返回点的【rgba】值
if (x<0)
{
x=-x
}else if (x>width-1)
{
x=2*(width-1)-x
}
if (y<0)
{
y=-y
}else if (y>height-1)
{
y=2*(height-1)-y
}
var i = (width*4)*y+x*4+index;
return canvasData.data[i]
}
function getInfoArr(x,y,index){ //根据坐标和radius【模糊半径】,返回要和权数组相乘的数组
var arr = [];
for (var i=x-radius;i<=x+radius ;i++ )
{
for (var j=y+radius;j>=y-radius ;j-- )
{
arr.push(getInfo(i,j,index))
}
}
return arr ;
}
};
</script>
</body>
</html>
最近看了阮一峰老师关于实现高斯模糊效果的博客,自己就用js和canvas写了这么一个效果
无奈执行时间太长,看stackblur的源码,又看不太懂
希望大家指教一下
gldmxcs
9 years, 10 months ago
Answers
有人写过高斯模糊的jQuery插件 https://github.com/finom/jQuery-Gaussian-Blur 里面用的是svg
狂拽√龙少
answered 9 years, 10 months ago
高斯模糊有两种方案做:
- 直接用二维公式进行二重循环,复杂度为O(xy(2r)^2)
- 用一维公式分别对x、y循环,复杂度为O(2xy(2r))
测试结果:
- 用二重循环:(500*800,20) 4566ms
- 分别循环:(500*800,20) 237ms
可以发现刚好差20倍左右,也就是
radius
模糊半径的值
结果图:
实现代码如下:
代码较长,建议移步到我的 博客 看代码
html:
html
<!DOCTYPE html> <html> <head lang="en"> <meta charset="UTF-8"> <title>test</title> <script src="GaussianBlur.js"></script> </head> <body> <img src="images/test3.jpg" alt="img source" id="imgSource"> <canvas id="canvas"></canvas> </body> </html>
javascript:
- gaussBlur : 二重循环
- gaussBlur1 : 分别循环
javascript
/** * Created by zhaofengmiao on 15/3/22. */ window.onload = function(){ var img = document.getElementById("imgSource"), canvas = document.getElementById('canvas'), width = img.width, height = img.height; // console.log(width); canvas.width = width; canvas.height = height; var context = canvas.getContext("2d"); context.drawImage(img, 0, 0); var canvasData = context.getImageData(0, 0, canvas.width, canvas.height); //console.log(canvasData); // 开始 var startTime = +new Date(); var tempData = gaussBlur(canvasData, 20); context.putImageData(tempData,0,0); var endTime = +new Date(); console.log(" 一共经历时间:" + (endTime - startTime) + "ms"); } /** * 此函数为二重循环 */ function gaussBlur(imgData, radius, sigma) { var pixes = imgData.data, width = imgData.width, height = imgData.height; radius = radius || 5; sigma = sigma || radius / 3; var gaussEdge = radius * 2 + 1; // 高斯矩阵的边长 var gaussMatrix = [], gaussSum = 0, a = 1 / (2 * sigma * sigma * Math.PI), b = -a * Math.PI; for (var i=-radius; i<=radius; i++) { for (var j=-radius; j<=radius; j++) { var gxy = a * Math.exp((i * i + j * j) * b); gaussMatrix.push(gxy); gaussSum += gxy; // 得到高斯矩阵的和,用来归一化 } } var gaussNum = (radius + 1) * (radius + 1); for (var i=0; i<gaussNum; i++) { gaussMatrix[i] = gaussMatrix[i] / gaussSum; // 除gaussSum是归一化 } //console.log(gaussMatrix); // 循环计算整个图像每个像素高斯处理之后的值 for (var x=0; x<width;x++) { for (var y=0; y<height; y++) { var r = 0, g = 0, b = 0; //console.log(1); // 计算每个点的高斯处理之后的值 for (var i=-radius; i<=radius; i++) { // 处理边缘 var m = handleEdge(i, x, width); for (var j=-radius; j<=radius; j++) { // 处理边缘 var mm = handleEdge(j, y, height); var currentPixId = (mm * width + m) * 4; var jj = j + radius; var ii = i + radius; r += pixes[currentPixId] * gaussMatrix[jj * gaussEdge + ii]; g += pixes[currentPixId + 1] * gaussMatrix[jj * gaussEdge + ii]; b += pixes[currentPixId + 2] * gaussMatrix[jj * gaussEdge + ii]; } } var pixId = (y * width + x) * 4; pixes[pixId] = ~~r; pixes[pixId + 1] = ~~g; pixes[pixId + 2] = ~~b; } } imgData.data = pixes; return imgData; } function handleEdge(i, x, w) { var m = x + i; if (m < 0) { m = -m; } else if (m >= w) { m = w + i - x; } return m; } /** * 此函数为分别循环 */ function gaussBlur1(imgData,radius, sigma) { var pixes = imgData.data; var width = imgData.width; var height = imgData.height; var gaussMatrix = [], gaussSum = 0, x, y, r, g, b, a, i, j, k, len; radius = Math.floor(radius) || 3; sigma = sigma || radius / 3; a = 1 / (Math.sqrt(2 * Math.PI) * sigma); b = -1 / (2 * sigma * sigma); //生成高斯矩阵 for (i = 0, x = -radius; x <= radius; x++, i++){ g = a * Math.exp(b * x * x); gaussMatrix[i] = g; gaussSum += g; } //归一化, 保证高斯矩阵的值在[0,1]之间 for (i = 0, len = gaussMatrix.length; i < len; i++) { gaussMatrix[i] /= gaussSum; } //x 方向一维高斯运算 for (y = 0; y < height; y++) { for (x = 0; x < width; x++) { r = g = b = a = 0; gaussSum = 0; for(j = -radius; j <= radius; j++){ k = x + j; if(k >= 0 && k < width){//确保 k 没超出 x 的范围 //r,g,b,a 四个一组 i = (y * width + k) * 4; r += pixes[i] * gaussMatrix[j + radius]; g += pixes[i + 1] * gaussMatrix[j + radius]; b += pixes[i + 2] * gaussMatrix[j + radius]; // a += pixes[i + 3] * gaussMatrix[j]; gaussSum += gaussMatrix[j + radius]; } } i = (y * width + x) * 4; // 除以 gaussSum 是为了消除处于边缘的像素, 高斯运算不足的问题 // console.log(gaussSum) pixes[i] = r / gaussSum; pixes[i + 1] = g / gaussSum; pixes[i + 2] = b / gaussSum; // pixes[i + 3] = a ; } } //y 方向一维高斯运算 for (x = 0; x < width; x++) { for (y = 0; y < height; y++) { r = g = b = a = 0; gaussSum = 0; for(j = -radius; j <= radius; j++){ k = y + j; if(k >= 0 && k < height){//确保 k 没超出 y 的范围 i = (k * width + x) * 4; r += pixes[i] * gaussMatrix[j + radius]; g += pixes[i + 1] * gaussMatrix[j + radius]; b += pixes[i + 2] * gaussMatrix[j + radius]; // a += pixes[i + 3] * gaussMatrix[j]; gaussSum += gaussMatrix[j + radius]; } } i = (y * width + x) * 4; pixes[i] = r / gaussSum; pixes[i + 1] = g / gaussSum; pixes[i + 2] = b / gaussSum; // pixes[i] = r ; // pixes[i + 1] = g ; // pixes[i + 2] = b ; // pixes[i + 3] = a ; } } //end imgData.data = pixes; return imgData; }
盗光者卡斯托尔
answered 9 years, 10 months ago